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• Before 1/7   ⊞ HW  
  • Quiz 1 (on 1/7) will cover the following problem. Consider the linear mapping L(x,y) = (y,x). Give the 2×2 matrix of this mapping, and describe the motion of the plane taking (x,y) to L(x,y).
• 1/7 Linear mappings, diagonalization   Notes   ⊞ HW  
  1. For each linear motion L of the plane, give its matrix [L] and the formula for L(x,y).
     1. Reflection across the y-axis
     2. Reflection across the line y = −x
     3. Rotation by 90° clockwise
  2. For each matrix, describe the corresponding motion of the plane, and give its formula L(x,y).
     1. [

        −2	0
        0	1

        ]
     2. [

        √2⁄2	−√2⁄2
        √2⁄2	√2⁄2

        ]
  3. Diagonalize the following matrices: first find eigenvalues by solving p(λ) = det(L-λI) = 0; then for a given λ, find the eigenvector v = (x,y) by solving the linear system (L−λI)v = 0. Describe the corresponding motion of the plane.
     1. [

        3	4
        4	−3

        ]
     2. [

        5	2
        1	4

        ]
  4. Extra Credit: A plane rotation matrix R (other than R = ±I) clearly has no real eigenvectors. Nevertheless, diagonalize it by finding its complex eigenvalues λ ∈ C and eigenvectors v₁, v₂ ∈ C².
  ⊞ Soln  

  1a. The columns of [L] are L(1,0) = (−1,0) and L(0,1) = (1,0), so [L]  =  [

  −1	0
  0	1

  ]   and L(x,y) = (−x,y).

  1b. Sketching the line of reflection y = −x shows that L(1,0) = (0,−1) and L(0,1) = (−1,0), so [L]  =  [

  0	−1
  −1	0

  ]   and L(x,y) = (−y,−x).

  1c. Rotating 90° clockwise shows that L(1,0) = (0,−1) and L(0,1) = (1,0), so [L]  =  [

  0	1
  −1	0

  ]   and L(x,y) = (y,−x).

  2a. The columns of the matrix are L(1,0) = (−2,0), so L flips over the the x-axis and stretches it by a factor of 2; and L(0,1) = (0,1), fixing the y-axis. L(x,y) = (−2x, y).

  2b. The values of L(1,0) and L(0,1) are 45° counterclockwise rotations of the x- and y-axes, so L rotates all vectors by 45° counterclockwise. L(x,y) = ^√2⁄₂(x−y, x+y).

  3a. Follow the example in the Notes. WolframAlpha says: L fixes the vector (2,1) and flips over the perpendicular vector (−1,2); thus, the motion is an orthogonal reflection across the line y = ½x.

  3b. W|A: The motion stretches the vector (−1,1) by a factor of 3, and the vector (2,1) by a factor of 6.

• 1/9 Rotations & reflections, plane symmetry   Notes   ⊞ HW  
  1. Let R be a counterclockwise rotation of the plane around the origin by ¹⁄₃ of a full turn (120°).
     1. Explain why R is a linear mapping, and find its 2×2 matrix. Hint: The columns are R(1,0) and R(0,1).
     2. Check that the matrix R = [v₁|v₂] is orthogonal: first check that its columns form an orthonormal basis, with
        v₁·v₁ = v₂·v₂ = 1   and   v₁·v₂ = 0;
        then, equivalently, check that R^T·R = I, where R^T is the transposed matrix with columns exchanged for rows.
     3. Consider the equilateral triangle centered at the origin with one vertex at v₁ = (2,2). Explicitly compute the other two vertices as v₂ = R(v₁), v₃ = R(v₂) = R²(v₁).
  2. Let the reflection R be defined by the matrix [

     √2⁄2	√2⁄2
     √2⁄2	−√2⁄2.

     ]
     1. Check that the matrix is orthogonal.
     2. Find the eigenvalues and eigenvectors of R. Use these to describe R geometrically: what reflection is it?
     3. In the Notes, we gave the general formula for a reflection across the fixed line R(cos α, sin α): it is [

        cos 2α	sin 2α
        sin 2α	−cos 2α

        ]. Show that R above is of this form for the correct value of α. Match this value of alpha with the line of reflection corresponding to the eigenvector of λ = 1 above.
  3. Consider each mapping R below. As in the notes, draw an asymmetric object X_o, and construct X = X_o ∪ R(X_o), an object having R as a twofold symmetry.
     1. R = reflection across the y-axis
     2. R = 180° rotation
  4. Install Mathematica. Unzip and run this demo to draw a polygon with the reflection symmetry from the previous problem, as the union of two symmetric polygons. Do the same for the rotation symmetry. Experiment with different points to get interesting polygons.
  5. Extra Credit: Show that a composition of two plane reflections is equal to a rotation by twice the angle between the two lines of reflection. Is the rotation clockwise or counterclockwise?
  ⊞ Soln  

  1a. R is a linear mapping because R(v+w) = R(v) + R(w), since the rotation of a parallelogram is a parallelogram; and R(cv) = c R(v), since the rotation of a stretched vector is a stretched vector. R rotates the standard basis vectors e₁ = (1,0), e₂ = (0,1) by ^2π⁄₃ radians to v₁ = (cos ^2π⁄₃, sin ^2π⁄₃), v₂ = (−sin ^2π⁄₃, cos ^2π⁄₃), giving the matrix:

  R  =  Rot_2π/3  =  [

  cos 2π⁄3	−sin 2π⁄3
  sin 2π⁄3	cos 2π⁄3

  ]  =  [

  −1⁄2	−√3⁄2
  √3⁄2	−1⁄2

  ].

  1b. The matrix product R^T·R gives the matrix of dot products of the column vectors of R = [v₁ | v₂].

  R^T·R  =  [

  v1

  v2

  ]·[v₁ | v₂]  =  [

  v1·v1	v1·v2
  v2·v1	v2·v2

  ]
  
   =  [

  1⁄21⁄2+√3⁄2√3⁄2	1⁄2(−√3⁄2)+√3⁄21⁄2
  −√3⁄21⁄2+1⁄2√3⁄2	−√3⁄2(−√3⁄2)+1⁄21⁄2

  ]  =  [

  1	0
  0	1

  ].

  Thus by definition, R is orthogonal.

  1c. Wolfram gives v₂ = R(2,2) = (−1−√3, −1+√3) and v₃ = R(v₂) = (−1+√3, −1−√3). Note the symmetry across y = x.

  2a. It is easy to check R^T·R = I, so R is orthogonal.

  2b. Wolfram gives eigenvalue λ₁ = 1 with eigenvector v₁ = (1+√2, 1) and λ₂ = −1 with v₂ = (1−√2, 1). Thus, R is the orthogonal reflection across the fixed line in direction v₁, having slope  ¹⁄_(1+√2)  =  −1 + √2.

  2c. If  (cos 2α, sin 2α) = (^√2⁄₂, ^√2⁄₂),  then  2α = ^π⁄₄ = 45°  and  α = ^π⁄₈ = 22.5°. The line in this direction has slope  tan ^π⁄₈  =  ¹⁄_(1+√2) using the tangent half-angle formula. This agrees with part (b).

  3a,b. See Notes p.5.

• 1/11 Spatial symmetry   Notes   ⊞ HW  
  1. 3D rotation matrices
     1. Write the matrix R_x for a 90° rotation around the x-axis, and similarly R_y for the y-axis. Hint: The column vectors are the images of the standard basis.
     2. Mutliply these matrices to get a new linear mapping L = R_x·R_y. Find the eigenvalues of L to describe its motion geometrically.
     3. Find the eigenvector of λ = 1 in part (b) to determine the axis of rotation.
  2. 3D reflection matrices

     R1  =

     1   0   0  0 1 0 0 0 −1

     R2  =

     1⁄3

     1  −2  −2  −2   1  −2  −2  −2   1

     1. Describe R₁ as a motion of space.
     2. Show that R₂ is an orthogonal twofold symmetry by checking R^T·R = I and R² = I.
     3. Find the eigenvalues of R₂ to describe its motion geometrically.
     4. Multiply L = R₁·R₂. Find the eigenvalues of L to describe it geometrically. Show it is a rotation, and determine its axis. What about its angle?
     5. Extra Credit: Explain the dot product formula for the reflection R_p which flips over the vector p and fixes the plane orthogonal to p:
        R_p(v)  =  v − 2(^v·p⁄_p·p) p.
        Here v = (x,y,z) is the input vector. Derive the matrices R₁ and R₂ from this.
  3. Hand In Mon 1/14: Choose ONE of the following, NOT both.
     1. Out of wire or cardboard, construct a 3D curve or shape with spatial point symmetry. You might make the two halves with different colors to emphasize the construction as X = X₀ ∪ R(X₀). Be creative: make it at least a little different from twisted glasses. Tag it with your name!
     2. Adapt the Mathematica demo from the previous HW to make a 3D polyhedron with spatial point symmetry, instead of just a 2D polygon. (Your shape should have no other symmetry.) Hand in a printout of your input and output, including a picture of the polyhedron and a list of its defining vertices.
  ⊞ Soln  

  1a. According to the Righthand Rule, if you point your thumb along the x-axis and turn ¼ in the direction of your fingers, this takes e₁ to e₁; e₂ to e₃; e₃ to −e₂. Similarly for the y-axis. Thus, the rotation matrices are:

           

  Rx  =

  1   0   0 0 0 −1 0 1  0

  Ry  =

  0   0   1  0 1 0 −1  0 0

  1b. The product matrix is:

           

  L  =  Rx·Ry  =

  0   0   1  1 0 0 0 1  0

  Its characteristic polynomial is p(λ) = det(L−λI) = −λ³ + 1 = −(λ−1)(λ²−λ+1). (Recall the formulas for computing a 3×3 determinant.) The λ₁ = 1 eigenvector v₁ is a fixed axis for L. The orthogonal motion of the perpendicular plane has no fixed directions, since (λ−1) only divides p(λ) once, so it must rotate that plane, and L must be a rotation around v₁.

  Since L cyclically permutes the three coordinate axes e₁→e₂→e₃→e₁, we have L³ = I, so it must be a ¹⁄₃ rotation. Another way to see this is by looking at the complex roots of p(λ), which are the complex cube roots of 1, namely λ₂, λ₃ = cos(^2π⁄₃) ± i sin(^2π⁄₃). These are the complex eigenvalues of a ¹⁄₃ rotation.

  Note that taking the product in the other order would give the opposite rotation:

  R_y·R_x  =  R_y⁻¹·R_x⁻¹  =  (R_x·R_y)⁻¹  =  L⁻¹.

  1c. Use Gaussian elimination (row reduction) to solve (L−I)v = 0 for v = (x,y,z) to find v = (c,c,c) for any c ∈ R, so we may take v₁ = (1,1,1). In fact, L is ¹⁄₃ rotation around v₁ in the righthanded sense.

  2a. R₁ is reflection of the z-axis across the xy-plane.

  2b. Note that R = R₂ satisfies R = R^T (that is, R₂ is a symmetric matrix), so the equation R^T·R = I is equivalent to R·R = I. This will happen for any orthogonal twofold symmetry, though not for a general orthogonal matrix.

  2c. Wolfram gives R₂  =  PDP⁻¹ for:

  D  =  J  =

  −1   0   0  0 1 0 0 0 1

  ,                     P  =  S  =

  1   1   1  1 −1 0 1 0 −1

  .

  The eigenvalues in D show that R₂ flips over one direction, and leaves the orthogonal plane fixed, so R₂ is a reflection. The columns of P give the eigenvector v₁ = (1,1,1) for λ₁ = −1, and a two-dimensional eigenspace for λ₂ = λ₃ = 1, spanned by (1,−1,0) and (1,0,−1). The last two eigenvectors are not unique: any linear combination of them is also an eigenvector with the same eigenvalue.

  2d. We have:

  L  =  R1·R2  =   1⁄3

  1  −2  −2  −2  1 −2  2 2 −1

  ,

  whose only real eigenvalue is λ = 1, which must be an axis of rotation. That is, L rotates space around some axis.

  The axis must be the intersection of the fixed planes of the two reflections, namely the direction orthogonal to both flipped directions p₁ = (0,0,1) and p₂ = (1,1,1). This axis can be obtained by solving v·p₁ = 0 and v·p₂ = 0, i.e. the system z = 0, x + y + z = 0; or equivalently take the 3D vector cross product p₁ × p₂  =  (−1,1,0). Thus the axis is the line R(−1,1,0).

  We can find the angle just as for plane reflections: it is twice the angle between the fixed planes, or between the flipped lines p₁, p₂. This latter angle is θ  =  arccos ^(p₁·p₂)⁄_|p1||p2|  =  arccos ¹⁄_√3  ≈  54.74°. Thus the angle of rotation is 2 arccos ¹⁄_√3  ≈  109.5°.

• 1/14 Ch 7.1A Intro to groups, permutations   Notes   ⊞ HW  
  Reading: Hungerford Ch 7.1A, pp. 183-196.    

  1. Work out the symmetry group G = Sym(X) of the rigid square X below, including rotations and reflections. (We call this the dihedral group D₄.) There are 8 symmetries, including the identity.
     
     1. Sketch how each symmetry L(x,y) moves the square, and write its matrix, which consists of the column vectors [ L(1,0) | L(0,1) ]. For example, the 90° counter-clockwise rotation R is [

        0	−1
        1	0

        ].
     2. For each symmetry, write the corresponding permutation of the four vertices (numbered as shown). For example, the above rotation R is (

        1	2	3	4
        2	3	4	1

        ), where the columns mean vertex 1 is moved to vertex 2, . . . , vertex 4 is moved to vertex 1.
     3. Show that the square cannot have more than 8 symmetries. Hint: Choose where vertex 1 goes, then vertex 2, etc., carefully noting geometric restrictions (all sides of the square have to go to sides).
     4. Give each symmetry a letter: the identity I, the rotations R, R², R³, and reflections A, B, C, D. Write out the 8×8 multiplication table of G under the composition operation, computing entries either as products of matrices or compositions of permutations.
  2. For the equilateral triangle T below, find the coordinates of its vertices v₁, v₂, v₃, and work out the symmetry group G = Sym(T) as in #1. There are 6 symmetries, including the identity.
     

  ⊞ Soln  

  1a,b. The symmetries of the square are:
  
  

  I	Identity mapping, no motion	[  1   0  0 1 ]	(  1   2   3   4  1 2 3 4 )
  R	90° rotation counterclockwise	[  0  −1  1 0 ]	(  1   2   3   4  2 3 4 1 )
  R2	180° rotation	[ −1   0  0 −1  ]	(  1   2   3   4  3 4 1 2 )
  R3	270° = −90° rotation	[  0   1  −1  0 ]	(  1   2   3   4  4 1 2 3 )
  A	reflection across y-axis	[ −1   0   0   1  ]	(  1   2   3   4  2 1 4 3 )
  B	reflection across x-axis	[  1   0   0  −1  ]	(  1   2   3   4  4 3 2 1 )
  C	reflection across y = x	[  0   1   1   0  ]	(  1   2   3   4  1 4 3 2 )
  D	reflection across y = −x	[  0  −1  −1   0  ]	(  1   2   3   4  3 2 1 4 )

  
  1c. Claim: The above 8 are the only symmetries of the square.
  Proof: Consider how an arbitrary symmetry S moves the vertices 1,2,3,4.

  • Choose S(1) = 1,2,3,4.
  • Given S(1), choose S(2) = S(1) ± 1, one of the two neighbors of X(1).
  • S(3) must be the unique unused neighbor of S(2)
  • S(4) must be the unique unused neighbor of S(3).

  
  Thus, there are at most 4×2 = 8 choices of symmetry, and we have found all of them above.

  1d. The group table of G:
  
  

  ⚬	I	R	R2	R3	A	B	C	D
  I	I	R	R2	R3	A	B	C	D
  R	R	R2	R3	I	D	C	A	B
  R2	R2	R3	I	R	B	A	D	C
  R3	R3	I	R	R2	C	D	B	A
  A	A	C	B	D	I	R2	R	R3
  B	B	D	A	C	R2	I	R3	R
  C	C	B	D	A	R3	R	I	R2
  D	D	A	C	B	R	R3	R2	I

  
  To compute compositions, we can multiply the corresponding matrices, or compose permutations. For example, doing A after R:

  A ⚬ R  =  (

  1	2	3	4
  2	1	4	3

  ) ⚬ (

  1	2	3	4
  2	3	4	1

  )   =   (

  1	2	3	4
  1	4	3	2

  )   =   C,

  since:

  A(R(1)) = A(2) = 1  ,  A(R(2)) = A(3) = 4  ,  A(R(3)) = A(4) = 3  ,  A(R(4)) = A(1) = 2.

  2. The coordinates of the vertices are given by repeatedly applying the rotation symmetry R = Rot_2π/3 to v₁:

  v₁  =  (1,0)
  v₂  =  (cos(θ), sin(θ))  =  (⁻¹⁄₂ , ^√3⁄₂)  for  θ = 120°
  v₃  =  (cos(θ), sin(θ))  =  (⁻¹⁄₂ , ^−√3⁄₂)  for  θ = −120°.

  The symmetries of the triangle are:
  
  

  I	Identity mapping, no motion	[  1   0  0 1 ]	(  1   2   3  1 2 3 )
  R	120° rotation Rot2π/3	[ −1/2  −√3/2 √3/2 −1/2 ]	(  1   2   3  2 3 1 )
  R2	240° rotation Rot4π/3	[ −1/2   √3/2 −√3/2 −1/2 ]	(  1   2   3  3 1 2 )
  A	reflection across x-axis	[  1   0   0  −1  ]	(  1   2   3  1 3 2 )
  B	reflection across y = −√3 x	[ −1/2   −√3/2 −√3/2 1/2 ]	(  1   2   3  3 2 1 )
  C	reflection across y = √3 x	[ −1/2   √3/2 √3/2 1/2 ]	(  1   2   3  2 1 3 )

  
  Note that A(v₁) = v₁, B(v₂) = v₂, C(v₃) = v₃.

  Could there be other symmetries of T? No: G can be defined by its permutations of the three vertices, so G ⊂ S₃; but |S₃| = 3! = 6, so there cannot be more elements of G than the 6 above.

• 1/16 Ch 7.1 Abstract groups, C_n, D_n, S_n, Cubic rotations S₄   ⊞ Notes  
  • Reading: Hungerford Ch 7.1A, pp. 183-196.    
  • Revised Notes
  • Standard symmetry groups: cyclic C_n ⊂ dihedral D_n ⊂ symmetric S_n
    |C_n| = n,   |D_n| = 2n,   |S_n| = n!
  • Abstract group axioms: closed operation, associative, identity element, inverses (not necessarily commutative)
  • Group theory has two branches:
    • Abstract groups: study groups G only in terms of their multiplication tables
    • Representation theory: given an abstract group G, find all objects X having G = Sym(X)
  ⊞ HW  
  1. Rotations of a cube: Let X be a cube in space and G = Sym(X) be the group of its rotation symmetries only. Mirror reflections are not allowed, only the symmetries obtained by moving the cube in your hands. The easiest way to deal with this group is to consider it as permutations of the 6 faces of the cube (rather than the 8 vertices), so that G ⊂ S₆. It is helpful to picture X using dice, so that faces 1,2,3 share a corner, and each pair of opposite faces total 7 (1 is opposite 6, etc.).
     1. Show that |G| ≤ 24: there are at most 24 rotations of the cube (counting the identity). Hint: Think where face 1 can be moved, then faces 2,3.
     2. In fact, there are exactly 24 symmetries in G. To find them, consider that G is generated by the one-quarter rotations around the x, y, and z axes, which we may write as permutations in cycle notation: A = (

        1	2	3	4	5	6
        1	3	5	2	4	6

        ), B = (

        1	2	3	4	5	6
        3	2	6	1	5	4

        ), C = (

        1	2	3	4	5	6
        2	6	3	4	1	5

        ). That is, A takes faces 1 and 6 to themselves because the x-axis runs through these faces. These elements and their powers account for 10 symmetries in G. For example, A² is a one-half rotation around the x-axis.
        Problem: Find the other 14 symmetries.
        Hint: There are two more types of rotation symmetry, each of which can be centered around several axes. You can find these by multiplying known symmetries, and describing the result geometrically.
     3. Extra Credit: Imagine relabeling the faces with each of the stickers 1,2,...,6, making a new configuration of dice. How many ways are there to do this? How many different sticker configurations if we count all rotations of a configuration as the same? That is, how many physically different types of dice are possible?
     4. Extra Credit: The fact that |G| = 24 = 4! should make you suspect it is the "same" as S₄, the group of all permutations of 4 objects. (Technically, we say the groups are isomorphic.) In fact, there is a way to picture each cube rotation as switching around 4 objects constructed inside the cube, with all 4! permutations appearing. How?
  2. Consider the ring (Z₅ , +, ·), modular arithmetic modulo 5, which you learned in Algebra I. That is, integer addition and multiplication as usual, except that in Z₅ we have 5 = 0 = −5, 6 = 1 = −4, etc.
     1. Forgetting addition, consider G = (Z₅)^× = {1,2,3,4} as a group under the multiplication operation. Check the four group axioms, as well as the commutative axiom, using the definition of modular multiplication. Write the multiplication table of G.
     2. Forgetting multiplication, consider H = Z₅ as a group under the addition operation. Check the four group axioms, remembering that the multiplicative notation a·b in the axioms is written a+b for a,b ∈ H, and the inverse a⁻¹ in the axioms is written −a ∈ H. Write the operation table for H (that is, the addition table). Is this group commutatitive (abelian)?
     3. In the multiplication table of H = {0,1,2,3,4}, replace the elements by those of C₅ = {I,R,R²,R³,R⁴}. This produces the multiplication table of C₅ ! Abstractly, the group H is actually C₅ in disguise. We say these groups are isomorphic.
     4. Extra Credit: The group G is isomorphic to the cyclic group C₄. Find a correspondence between the elements of G = {1,2,3,4} and C₄ = {I,R,R²,R³} which turns the multiplication table of G into that of C₄ (with the elements in a different order).
  3. Linear groups
     1. Check the four group axioms for the general linear group GL_n(R) = {all invertible n×n matrices}, under matrix multiplication. Is this group commutatitive (abelian)? (Use your knowledge of matrix multiplication from previous courses.)
     2. Do the same for the orthogonal group O(n) = {all orthogonal n×n matrices}. Particularly check that the multiplication is closed.
  ⊞ Soln  

  1a. A cube symmetry T can rotate face 1 to any of the 6 faces, so there are 6 choices for T(1). Then T(2) can be any of the 4 faces adjacent to T(1). Now T(3) must be one of the two faces adjacent to T(1) and T(2), but in fact it must be the one that makes the path T(1) → T(2) → T(3) → T(1) a clockwise circle around their common corner (since the path 1 → 2 → 3 → 1 circles clockwise around their common corner). Finally, T(4) is the face opposite T(3), etc. This makes a total of 6 × 4 = 24 choices, so that |G| ≤ 24. However, this does not prove that all 24 choices lead to symmetries.

  1b. The cube has 1 identity symmetry I = (1)(2)...(6). Also 3 axes through opposite pairs of faces, each with ¹⁄₄, ¹⁄₂, and ³⁄₄ rotations, such as A and A². Also 4 axes through opposite pairs of corners, each with ¹⁄₃ and ²⁄₃ rotations, such as: (

  1	2	3	4	5	6
  2	3	1	6	4	5

  ). Also 6 axes through midpoints of opposite pairs of edges, each with a ¹⁄₂ rotation, such as: (

  1	2	3	4	5	6
  2	1	4	3	6	5

  ). Total: 1 + (3)(3) + (4)(2) + (6)(1) = 24, giving all the possible symmetries of G according to part (a).

  2a. G = (Z₅)^× = {1,2,3,4} is a closed under multiplication, since a product of non-zero elements mod 5 (i.e. not divisible by 5) will never be zero mod 5 (i.e. divisible by 5). (This is because 5 is a prime number.) G is associative because integer multiplication is associative: (ab)c = a(bc). G has the identity element 1 (mod 5). We can check that each element of G has a multiplicative inverse (reciprocal): 1⁻¹ = 1, 2⁻¹ = 3, 3⁻¹ = 2, 4⁻¹ = 4, since 2·3 = 6 = 1 (mod 5), etc. (In ring terms, this means Z₅ is a field.)

  2b. H = {0,1,2,3,4} is clearly closed and associative under the addition operation +. That is, we write a+b for a combination rather than a·b. Also, H has the identity element 0, since operating by 0 has no effect: 0 + a = a + 0 = a. (This is the additive notation for the axiom 1·a = a·1 = a.) Finally, each element a has an inverse which cancels it to give the identiy element: a + (−a) = (−a) + a = 0. (This is the additive notation for the axiom a·a⁻¹ = a^{−1·a = 1.)}

  2c. The operation tables of H and C₅ are:

  +	0	1	2	3	4
  0	0	1	2	3	4
  1	1	2	3	4	0
  2	2	3	4	0	1
  3	3	4	0	1	2
  4	4	0	1	2	3

               

  ·	I	R	R2	R3	R4
  I	I	R	R2	R3	R4
  R	R	R2	R3	R4	I
  R2	R2	R3	R4	I	R
  R3	R3	R4	I	R	R2
  R4	R4	I	R	R2	R3

  Clearly, relabeling k ∈ H by R^k ∈ C₅, with R⁰ = I, turns one table into the other.

  3a. We show GL_n(R) is an abstract group. The matrix A lies in GL_n(R) whenever det(A) ≠ 0, i.e. when A represents an invertible linear mapping. Closed: We know from linear algebra that if det(A), det(B) ≠ 0, then det(A·B) = det(A) det(B) ≠ 0; or alternatively, that a composition of invertible mappings is invertible. Associative: One can check algebraically that (A·B)·C = A·(B·C), but this follows immediately from considering · as composition of mappings, since (A·B)(C(x)) = A(B(C(x)) = A((B·C)(x)). Identity element: the n×n identity matrix I. Inverses: By definition, A ∈ GL_n(R) is invertible, and its inverse is in GL_n(R).

  3b. We show O(n) is an abstract group. The matrix A lies in O(n) whenever A^TA = I, i.e. when A represents an orthogonal linear mapping. Closed: If A^TA = B^TB = I, then:

  (A·B)^T(A·B) = B^TA^TAB = B^TIB = B^TB = I.

  Thus A·B is orthogonal. Alternatively, a composition of orthogonal mappings still preserves distance and dot product, so it is still orthogonal. Associativity: follows from that for all matrices. Identity: The matrix I is in O(n). Inverses: An orthogonal matrix has A⁻¹ = A^T, which is also orthogonal.
• 1/18 Ch 7.2 Properties of groups, lemmas   ⊞ Notes  
  • Reading: Ch 7.2, pp. 196-198.    
  • Abstract group (G, · )
    • Defined solely by its multiplication table, without considering its representations as symmetries.
    • The only essential properties of abstract groups are the four axioms below: any operation with these properties turns out to be a symmetry group.
    • The main aim of abstract group theory is to classify all abstract groups, and determine what kind of symmetries are possible.
  • Group Axioms. For any a,b,c ∈ G:
    • Closed operation:  a·b ∈ G
    • Associative:  a·(b·c) = (a·b)·c
    • Identity e:  a·e = e·a = a
    • Inverses:  each a has a⁻¹ with a·a⁻¹ = a⁻¹·a = e.
    G is a commutative or abelian group if it obeys:
    • Commutative:  ab = ba.
  • Group properties: In the propositions below, we prove some immediate consequences of the four axioms, not assuming any other properties, so the results are valid for all possible groups. Here a,b,c ∈ G are arbitrary elements of a group G.
    • Uniqueness of identity: There can be only one identity element in G.
      Proof: Suppose e, e' ∈ G both obey the Identity axiom. Then ee' = e because e' is an identity, and ee' = e' because e is an identity. Thus e = e' (by transitivity of equality), and there is only one identity.
    • Cancellation law: If ab = ac, then b = c.
      Proof: If ab = ac, then a⁻¹(ab) = a⁻¹(ac) (by substitution of equal quantities), so (a⁻¹a)b = (a⁻¹a)c (by Associativity), and eb = ec (by Inverses), and b = c (by Identity).
    • Inverse of a product: (ab)⁻¹ = b⁻¹a⁻¹.
      Proof: We have (ab)(b⁻¹a⁻¹) = a(bb⁻¹)a⁻¹ = aa⁻¹ = e. Thus b⁻¹a⁻¹ is inverse to ab. (Here I assumed you can see how each step is justified by an axiom.)
  • Isomorphism  G ≅ H
    • Groups G and H are isomorphic when relabeling the multiplication table of G produces the table of H, so the tables are structurally the same.
    • Formally, there exists an isomorphism, an invertible mapping φ : G → H such that  φ(ab) = φ(a) φ(b)  for all  a,b ∈ G. (Invertible means bijective: one-to-one and onto.) We often use Greek letters for isomorphisms, such as φ (phi) and ψ (psi).
    • A key isomorphism is between the group of integers modulo n under the addition operation, (Z_n , + ), and the cyclic group C_n = {e, r, r², . . . , rⁿ⁻¹} with rⁿ = e. The isomorphism is φ(i) = rⁱ. This is well defined since any other integer representative i + kn = i mod n gives the same output:
      φ(i+kn)  =  r^i+kn  =  rⁱ(rⁿ)^k  =  rⁱe^k  =  rⁱ.
      We also have the structure equation relating the + operation in Z_n to the multplication in C_n:
      φ(i+j)  =  r^i+j  =  rⁱr^j  =  φ(i) φ(j).

    • Taking n = 3, we can see that the isomorphism φ(i) = rⁱ relabels the (addition) operation table of Z₃ = {0, 1, 2} to give the (multiplication) operation table of C₃ = {e, r, r²}.

      +	0	1	2
      0	0	1	2
      1	1	2	0
      2	2	0	1

                   

      ·	e	r	r2
      e	e	r	r2
      r	r	r2	e
      r2	r2	e	r

  • Isomorphism of a finite linear group
    • Let G = GL₂(F) be the group of invertible 2×2 matrices with entries in the two-element modular arithmetic ring F = Z₂ = {0,1} with 2 = 0. There are 6 such matrices, all except the matrices with all 0's or all 1's.
    • G is naturally the symmetry group of the vector space F², the "plane" whose vectors are (0,0), v₁ = (1,0), v₂ = (0,1), v₃ = (1,1), with (2,0) = (0,0), (3,0) = (1,0), etc.
    • A matrix L ∈ G represents a linear mapping which takes the zero vector to itself, and permutes the three non-zero vectors. The 6 matrices in G correspond to all 3! = 6 permutations in S₃, the group of all permutations of {1,2,3}.
    • Formally, we show G ≅ S₃ by defining an isomorphism φ : G → S₃. If L(v₁) = v_i , L(v₂) = v_j , L(v₃) = v_k , we let φ(L)  =  (

      1	2	3
      i	j	k

      ).
  ⊞ HW  
  1. Find the number of elements in G = GL₂(F), invertible matrices with entries in the field F = Z_p for a prime integer p. How many elements in G?
     Hint: The columns of an invertible matrix make a basis of the vector space F². Count the possible choices for each basis vector.
  2. Prove the Uniqueness of inverses: Any group element a ∈ G has only one inverse a⁻¹. That is, if there are elements b,b',c,c' ∈ G with ab = ab' = e and ca = c'a = e, then these partial inverses are all equal: b = b' = c = c' = a⁻¹.
     Note: It is this proposition which justifies the notation a⁻¹: otherwise, we might have several inverse elements to choose from. In particular, the proposition implies a matrix A ∈ G = GL_n(R) has only one inverse matrix.
  3. Cyclic supgroups. Consider an arbitrary group element a ∈ G, and let:
     ⟨a⟩  =  {aⁱ  for  i ∈ Z}  =  { . . . , a⁻² , a⁻¹, e, a, a², . . . }.
     Here  a⁰ = e , a² = aa, a⁻² = a⁻¹a⁻¹, etc.
     1. Show that ⟨a⟩ is itself a group, i.e. a subgroup of G.
     2. If G is finite, show that there must be some k > 0 with a^k = e.
     3. Let n be the smallest such k, and define the order |a| = n. Show that:
        ⟨a⟩  =  {e, a, a², . . . , aⁿ⁻¹}  ≅  C_n.
        That is, ⟨a⟩ is an isomorphic copy of the cyclic group C_n inside G, and |⟨a⟩| = |a|.
  ⊞ Soln  

  1. Choosing an invertible matrix with entries in the field F = Z_p is the same thing as choosing a basis {v₁, v₂} of the vector space F². The first vector v₁ can be any non-zero element of F², meaning p² − 1 choices. The second vector can be any vector linearly independent from v₁, i.e. anything outside the line Fv₁ having p elements; this means p² − p choices. The total number of possible bases is |GL₂(Z_p)| = (p² − 1)(p² − p).

  2. Proof: By hypothesis, suppose we have two right-inverses ab = ab' = e and two left-inverses ca = c'a = e. We proved above that we can cancel a in the equation ab = ab', so b = b', and similarly c = c'. Now:

  b  =  eb  =  cab  =  ce = c.

  Here the first and last equalities are by Identity, the second and third by hypothesis.

  3a. The set  H = ⟨a⟩  satisfies the four group axioms. It is closed because for any  aⁱ, a^j ∈ H, we have  aⁱa^j = a^i+j ∈ H. It is associative because the group G itself is associative. It contains the identity e = a⁰. Finally, the inverse of aⁱ is a⁻ⁱ ∈ H.

  3b. Since G is finite, the elements { . . . , a⁻² , a⁻¹, e, a, a², . . . } cannot all be different: we must have some coincidence  aⁱ = a^j  for some i < j, so that  aⁱ = aⁱa^j−i. Cancelling aⁱ from both sides gives  e = a^j−i, so we can take k = j−i > 0 with a^k = e.

  3c. Suppose n = min{k > 0  with  a^k = e}. Then the elements {e, a, a², . . . , aⁿ⁻¹} must all be distict. Otherwise we would have  aⁱ = a^j  for some 0 ≤ i < j ≤ n−1, so that  a^j−i = e  for  0 < j−i < n, which contradicts the choice of n as the smallest with aⁿ = e.
  Now define an isomorphism φ : ⟨a⟩ → C_n by φ(aⁱ) = rⁱ. This is one-to-one and onto by the above, and clearly obeys the structural equation φ(aⁱa^j) = rⁱr^j = φ(aⁱ) φ(a^j). Thus φ is an isomorphism giving ⟨a⟩ ≅ C_n .

• 1/23 Ch 7.3 Subgroups, cyclic subgroups   ⊞ Notes  
  • Reading: Ch 7.3, pp. 203-207.    
  • Product of Groups: For groups G and H, define a new group
    G×H  =  {(g,h)  for  g ∈ G, h ∈ H}
    with componentwise multiplication (g₁ , h₁) (g₂ , h₂) = (g₁g₂ , h₁h₂).
  • Subgroup: For a group G, a subset H ⊂ G is a subgroup if H is itself a group satisfying the four group axioms.
    In fact, it is enough to check the first axiom (if a,b ∈ H, then ab ∈ H) and the fourth axiom (if a ∈ H, then a⁻¹ ∈ H), and second and third axioms follow automatically.
  • Cyclic subgroup: Consider an arbitrary group element a ∈ G, and define the cyclic subgroup generated by a:
    ⟨a⟩  =  {aⁱ  for  i ∈ Z}  =  { . . . , a⁻² , a⁻¹, e, a, a², . . . }.
    Here  a⁰ = e , a^k = a···a with k factors, and a^−k = a⁻¹···a⁻¹ with k factors.
  • Theorem: If ⟨a⟩ is finite, then there is some n > 0, called the order of a and denoted n = |a|, with
    ⟨a⟩  =  {e, a, a², . . . , aⁿ⁻¹}  ≅  C_n , where  aⁿ = e.
    That is, ⟨a⟩ is an isomorphic copy of the cyclic group C_n inside G, and |⟨a⟩| = |a|.
    Proof: Assuming ⟨a⟩ is finite, the sequence a¹, a², . . . cannot have infinitely many distinct elements, so we must have aⁱ = a^j = aⁱa^j−i for some 0 < i < j. Therefore a^j−i = e, and the set {k > 0  with  a^k = e} is non-empty and has a smallest element n = |a|. Then all the elements {e,a,a², . . . , aⁿ⁻¹} are distinct, since otherwise aⁱ = a^j with 0 < i < j < n, so a^j−i = e with 0 < j−i < n, and this contradicts the choice of n. This makes it clear that ⟨a⟩ ≅ C_n.
  • If ⟨a⟩ is infinite, we say a has infinite order, |a| = ∞, and ⟨a⟩ ≅ C_∞ ≅ (Z, +), where C_∞ = {aⁱ for i ∈ Z} with distinct elements aⁱ ≠ a^j for i ≠ j.
  ⊞ HW  
  1. Show that in a product group G×H, the subgroup G×e = {(g,e) for g ∈ G} is isomorphic to G, and similarly e×H ≅ H. (There are obvious isomorphism mappings.)
  2. Recall the mulitplicative group Z_n^× = {i ∈ Z_n  with  gcd(i,n) = 1}, with the usual multiplication operation mod n.
     1. Show that Z₉^× ≅ C₆. Find a generator c of order 6 in Z₉^×, so that ⟨c⟩ fills the whole group, and define an isomorphism mapping φ : C₆ → Z₉^× with φ(r^k) = c^k.
     2. Show that Z₁₂^× ≅ C₂ × C₂, by defining an explicit isomorphism.
     3. Show that C₂ × C₃ ≅ C₆. Hint: Find elements of order 2 and 3 in C₆, and use these to define an isomorphism φ : C₂ × C₃ → C₆. Alternatively, find an element of order 6 in C₂ × C₃ and use it to define an isomorphism ψ : C₂ × C₃ → C₆ . (Here the Greek letters phi φ and psi ψ denote inverse isomorphisms, which undo each other.)
     4. Show that C₂ × C₂ is not isomorphic to C₄. That is, there no possible isomorphism between them. Hint: Show that the multiplication table of C₂ × C₂ has a basically different property from C₄, so there cannot be a relabeling which turns one into the other.
  3. Consider the dihedral group D₃ from previous HW, having rotation subgroup {e, r, r²} and reflections {a, b, c}, with r³ = a² = b² = c² = e.
     1. Show that the reflections can be written as {a, ar, ar²}, and that ra = ar⁻¹, which determines the multiplication rule rⁱ·ar^j = ar^j−i. Write the multiplication table in this notation.
     2. Write all cyclic subgroups of D₃, and determine the order of each element.
     3. Is there a cyclic subgroup ⟨c⟩ which fills all of D₃, or is it not generated by any single element? That is, do we have an isomorphism S₃ ≅ C₆?
     4. D₃ has cyclic subgroups of order 2 and 3. Can you use these to define an isomorphism ψ : C₂ × C₃ → D₃ , similarly to #2(c) above?
  ⊞ Soln  

  1. The subset G×e ⊂ G×H is a subgroup, since it is clearly closed under multiplication and inverses: for example, (g,e)⁻¹ = (g⁻¹, e) ∈ G×e. Define an isomorphism φ : G → G×e by φ(g) = (g,e). This is clearly invertible (with inverse ψ(g,e) = g), and it obeys the stuctural equation:

  φ(g₁g₂)  =  (g₁g₂ , e)  =  (g₁,e) (g₂,e)  =  φ(g₁) φ(g₂).

  Thus, φ is an isomorphism, and its input and output groups are isomorphic.
  Similarly H ≅ e×H ⊂ G×H.

  2a. In the multiplicative group Z₉^× = {1,2,4,5,7,8}, we find the cyclic subgroup:

  ⟨2⟩ = {1 , 2 , 2² = 4 , 2³ = 8 , 2⁴ = 7 , 2⁵ = 5},

  with 2⁶ = 1 mod 9; thus 2 has order 6, denoted |2| = 6. But this fills up the whole group, so we get a natural isomorphism φ : C₆ → ⟨2⟩ = Z₉^× defined by φ(r^k) = 2^k, where we denote C₆ = {e, r, r², . . . , r⁵}.
  Instead of 2, we could also have used 2⁻¹ = 5, another element of order 6.

  2b. In the multiplicative group Z₁₂^× = {1,5,7,11}, we find that every element has a² = e, so that a has order 2, i.e. |a| = 2. Choosing the two elements 5 and 7, we can define an isomorphism

  φ : C₂ × C₂ → Z₁₂^×   by   φ(r,e) = 5 , φ(e,r) = 7 , φ(r,r) = 5·7 = 11.

  We can easily check the structural equation for φ, using the fact that (r,e)(e,r) = (e,r)(r,e) = (r,r) and (e,r)² = (r,e)² = (r,r)² = (e,e), which correspond to 5·7 = 7·5 = 11 and 5² = 7² = 11² = 1.

  2c. The cyclic group C₆ = {e, r, r², . . . , r⁵} has r³ as an element of order 2, and r² as an element of order 3. Thus, writing C₂ = {e,s} and C₃ = {e, t, t²} we can define an isomorphism

  φ : C₂ × C₃ → C₆   by   φ(sⁱ, t^j)  =  (r³)ⁱ (r²)^j  =  r^3i+2j.

  This is easily seen to obey the structure equation, and we check directly that it is one-to-one and onto:

  C2×C3	(e,e)	(e,t)	(e,t2)	(s,e)	(s,t)	(s,t2)
  C6	e	r2	r4	r3	r5	r

  Alternatively, we can see from the above table that (s,t²) is an element of order 6, so we can define an isomorphism:

  ψ : C₆ → C₂ × C₃   by   φ(rⁱ)  =  (s,t²)ⁱ = (sⁱ,t²ⁱ).

  This is in fact the inverse isomorphism to φ, so that ψ(φ(sⁱ, t^j)) = (sⁱ, t^j) and φ(ψ(rⁱ)) = rⁱ.

  2d. The groups C₂ × C₂ has all elements of order 2, i.e. g² = e for all g, but C₄ has elements of order 2 and 4. But any isomorphism preserves the order of each element, so there can be no such isomorphism.
  More formally, if there were an isomorphism φ : C₄ → C₂ × C₂, then the structural equation would force φ(r²) = φ(r)² = e = φ(e), even though r² ≠ e ∈ C₄; thus φ could not be one-to-one, a contradiction.

  3a. As in the previous HW, the dihedral group permutes the three vertices of the triangle, making D₃ ≅ S₃, with the isomorphism:

  φ(r)  =  (

  1	2	3
  2	3	1

  )     φ(a)  =  (

  1	2	3
  1	3	2

  )     φ(b)  =  (

  1	2	3
  3	2	1

  )     φ(c)  =  (

  1	2	3
  2	1	3

  ) .

  We can then multiply permutations to check the desired relations b = ar , c = ar² , ra = ar⁻¹, as well as r³ = a² = b² = c² = e. These relations give a new way to compute products, such as:

  ar² · ar  =  ar(ra)r  =  ar(ar⁻¹)r  =  a(ra)r⁻¹r  =  a(ar⁻¹)r⁻¹r  =  a²r⁻¹  =  r².

  We get the multiplication table:

  ·	e	r	r2	a	ar	ar2
  e	e	r	r2	a	ar	ar2
  r	r	r2	e	ar2	a	ar
  r2	r2	e	r	ar	ar2	a
  a	a	ar	ar2	e	r	r2
  ar	ar	ar2	a	r2	e	r
  ar2	ar2	a	ar	r	r2	e

  Here the entry in row x, column y is the product x·y.

  3b. The cyclic subgroups are:

  ⟨r⟩ = {e,r,r²}       ⟨a⟩ = {e,a}       ⟨ar⟩ = {e,ar}       ⟨ar²⟩ = {e,ar²} .

  3c. Evidently, there are no elements of order 6, so D₃ is not isomorphic to C₆. In fact, D₃ has two elements of order 3 and three elements of order 2, whereas C₆ has two elements of order 3, one of order 2, and two of order 6. Another difference which prevents an isomorphism is that D₃ is not abelian (commutative), whereas C₆ is abelian.

  3d. If we had C₂ × C₃ ≅ D₃, and by #2c we know C₂ × C₃ ≅ C₆, this would mean D₃ ≅ C₆, contradicting #3c; thus we cannot have C₂ × C₃ ≅ D₃.
  Still what if we take a,r ∈ D₃ of order 2 and 3, and try defining φ : C₂ × C₃ → D₃ by φ(sⁱ, t^j) = aⁱr^j . This is an invertible mapping, and the structural equation holds in many cases, for example φ((s,e)(e,t)) = φ(s,t) = ar = φ(s,e) φ(e,t). But it must fail somewhere! Namely, φ((e,t)(s,e)) = φ(s,t) = ar, but φ(e,t) φ(s,e) = ra : this reflects the non-commutativity of D₃.

• 1/25 Ch 7.4 Isomorphisms, classify |G| = 4   ⊞ Notes  
  • Reading: Ch 7.4, pp. 214-221.    
  • Classification of 4-element groups: Any group G with 4 elements is isomorphic to C₄ or C₂ × C₂.
    Proof: Let N = max{|a| for a ∈ G}, the size of the largest cyclic subgroup of G.
    • If N = 4, then there is an element of order 4, and G = ⟨a⟩ = {e, a, a², a³} with a⁴ = 1. Thus G ≅ C₄ .
    • If N = 3, then we can write the four distinct elements as G = {e, a, a², b} with a³ = e. Now, what is ab? If ab = e, then ab = a³, and we can cancel a (multiply both sides by a⁻¹) to get b = a². If ab = a, then b = e. If ab = a², then b = a. If ab = b, then a = e. All of these conclusions force two elements to be the same, which contradicts |G| = 4. Thus, it is impossible to complete the multiplication table in a way that satisfies the group axioms, and this case is impossible.
    • If N = 2, then all non-identity elements have order 2, and we may write G = {e, a, b, c} with a² = b² = c² = e. Then as above, we cannot have ab = e, a, or b, so the only remaining possibility is ab = c, and similarly ba = c. This fixes the multiplication table as:

      ·	e	a	b	ab
      e	e	a	b	ab
      a	a	e	ab	b
      b	b	ab	e	a
      ab	ab	b	a	e

      In fact, there is an isomorphism φ : G → C₂ × C₂, where C₂ = {e, r}, defined by φ(e) = (e,e), φ(a) = (r,e), φ(b) = (e,r), φ(ab) = (r,r).
    • If N = 1, this would mean a¹ = e for every a ∈ G, which is impossible.
  ⊞ HW  
  1. Center subgroup. For a group G, we say an element c ∈ G is central if it commutes with all elements:  cg = gc for all g ∈ G. The center is the set of all central elements:
     Z(G) = {c ∈ G  with  cg = gc for all g ∈ G}.
     1. Show that the center Z(G) is always a subgroup of G. Hint: You must show that if c,d ∈ Z(G), then cd ∈ Z(G) and c⁻¹ ∈ Z(G).
     2. Find the center of the dihedral group D₄, the rotation and reflection symmetries of the square. Hint: A scalar matrix λI commutes with all other matrices. Which elements of D₄ are represented by scalar matrices?
  2. Show that any group with 5 elements must be isomorphic to the cyclic group C₅.
  3. Go back through our previous assignments to note all the groups with 6 elements. How many different types are there, counting isomorphic groups as the same?
  4. Inverses
     1. Show that if ab = e, then ba = e.
     2. Extra Credit: Show that the order of ab is the same as the order of ba; that is, |ab| = |ba|.
  5. Prove: If every non-identity element of a group G has order 2, then G must be abelian.
  6. Prove: The only subgroups of the infinite cyclic group (Z, +) are ⟨n⟩ = nZ = {nk for k ∈ Z}.
  7. Extra Credit: Show that if |G| is even, then G must contain an element of order 2. Hint: Split G into pairs {a, a⁻¹}. What happens if a² = e?
  ⊞ Soln  

  1a. Proposition: For any group G, the center Z(G) is a subgroup.
  Proof: We know that a subset of G is a subgroup if it is closed under multiplication and inverses (the other two group axioms follow automatically). Thus, let c,d ∈ Z(G), meaning cg = gc and dg = gd for all g ∈ G. Then:

  (cd)g  =  c(dg)  =  c(gd)  =  (cg)d  =  (gc)d  =  g(cd);

  thus cd commutes with all g, and cd ∈ Z(G).
  Also, cg = gc implies c⁻¹(cg)c⁻¹ = c⁻¹(gc)c⁻¹, which simplifies to gc⁻¹ = c⁻¹g, so c⁻¹ ∈ Z(G). Since Z(G) is closed under multiplication and inverses, it is a subgroup.

  1b. Considering D₄ as symmetries of a square in the plane, we can check that the only matrix which commutes with all the others is the 180° rotation R² = −I. Thus, Z(D₄) = {I, R²}.

  2. Proposition: If G is a group with |G| = 5, then G ≅ C₅.
  Proof: Again let N = max{|a| for a ∈ G}, the size of the largest cyclic subgroup of G.

  • If N = 5, then there is an element a of order 5, so that G = {e, a, a², a³, a⁴} ≅ C₅ with a⁵ = e. We must show the other cases are impossible.
  • If N = 4, we can write G = {e, a, a², a³, b} with a⁴ = e and b ∉ ⟨a⟩. Now what is ab? If ab = aⁱ, for i = 0,1,2,3, then multiplying by a⁻¹ gives b = aⁱ⁻¹ ∈ ⟨a⟩, contradicting our choice of b. If ab = b, then a = e, contradicting our choice of a.
  • If N = 3, then G = {e, a, a², b, c} with a³ = e. As in the previous case, we can only have ab = c, and similarly ac = b. But then b = ac = a(ab), and e = a², impossible.
  • If N = 2, then every non-identity element of G has order 2, and G = {e, a, b, c, d} with a² = b² = c² = d² = e. As previously, we must have ab = c or d; but c,d have the same assumed properites, so the same arguments will hold in the two cases, and we may assume ab = c. Then ac = a(ab) = b. Now what is ad? We cannot have ad = e, a, d; and ad = b = ac leads to d = c; and ad = c = ab leads to d = b. Thus, there is no possible value for ad, and this case is again impossible.
  • N = 1 would again mean a = e for all a ∈ G, impossible.

  3. We have seen the following groups with 6 elements in the previous HW:

  C₆  ≅  C₂ × C₃  ,  D₃  ≅  S₃  ≅  GL₂(Z₂).

  We will show later that these are the only isomorphism types of 6-element groups.

  4a. If ab = e, then  b(ab)b⁻¹ = bb⁻¹ = e, which simplifies to ba = e.
  This also follows from the uniqueness of inverses proved previously.

  5. Prop: If every element of a group G has order 2, then G must be abelian.
  Proof: For a,b ∈ G, suppose by hypothesis that a² = b² = (ab)² = e, i.e. a = a⁻¹, b = b⁻¹, and abab = e. We manipulate abab = e so as to leave only ba on the left side:

  a⁻¹(abab)b⁻¹  =  a⁻¹b⁻¹,
  (a⁻¹a)ba(bb⁻¹)  =  ab,
  ba  =  ab.

  This holds for all a,b, so G is abelian.

  6. Let H be any subgroup of (Z, +). If we do not have H = {0} = ⟨0⟩, then there is some i ∈ H with i > 0. (If i < 0, we have the inverse −i ∈ H.) Let n = min{i ∈ H with i > 0}, and take any i ∈ H. Then it is clear that ⟨n⟩ ⊂ H.
  Applying the division algorithm to i divided by n, we have i = nk + r for 0 ≤ r < n. But then r = nk − i = ±(n + ··· + n) − i ∈ H. By the choice of n, we must have r ≥ n, which is false, or r ≤ 0, which leaves only the possibility r = 0. Thus i = nk, and i ∈ ⟨n⟩. Therefore H ⊂ ⟨n⟩.

• 1/28 & 2/1 Ch 8.1 Congruence, cosets, Lagrange Thm, classify |G| = 6   ⊞ Notes  
  • Reading: Ch 8.1, pp. 237-245.    
  • Congruence classes: Recall that in clock arithmetic modulo n, each integer a ∈ Z represents a congruence class:
    [a]  =  a + nZ  =  { . . . , a−n, a, a+n, a+2n, . . . }.  =  {b ∈ Z  with  b ≡ a mod n}.
    Thus a ≡ b mod n whenever [a] = [b], and the modular ring is the set of these classes: Z_n  =  {[0], [1], . . . , [n−1]}. The integers are the disjoint union of all the modular equivalence classes:
    Z  =  [0] ∪ [1] ∪ ··· ∪ [n−1]  =  nZ ∪ (1 + nZ) ∪ ··· ∪ (n−1 + nZ).
    We will generalize this construction from the subgroup nZ ⊂ (Z, +) to a general subgroup H ⊂ G.
  • Cosets: For a subgroup H ⊂ G, define the left coset of H at the element a ∈ G to mean:
    aH  =  {ah  for  h ∈ H}.
    The coset aH is the shift of H under multiplication by a.
  • Example: G = D₃ = {e, r, r², a, ar, ar²}, with r³ = a² = e and ra = ar⁻¹.
    • For the subgroup H = ⟨a⟩ = {e,a}, we have [G : H] = 3, and the 3 cosets are:
      D₃  =  H ∪ rH ∪ r²H  =  {e, a} ∪ {r, ra = ar²} ∪ {r², r²a = ar}.

    • For the subgroup K = ⟨r⟩ = {e, r, r²}, we have [G : K] = 2, and the 2 cosets are:
      D₃  =  H ∪ aH  =  {e, r, r²} ∪ {a, ar, ar²}.

  • Coset Lemmas:
    1. Two cosets are disjoint or equal: either aH ∩ bH = {} or aH = bH.
    2. Any cosets has the same number of elements as H: that is, |aH| = |H|
    3. The group is a disjoint union of cosets: G = H ∪ a₁H ∪ a₂H ∪ ··· , corresponding to representatives e, a₁, a₂, . . . ∈ G.
    Proof:
    1. Suppose aH and bH are not disjoint, so there is some c ∈ aH ∩ bH. That is, c = ah = bh' for h, h' ∈ H. Now consider any ah" ∈ aH: we have
       ah"  =  ahh⁻¹h"  =  bh'h⁻¹h"  ∈  bH,
       so aH ⊂ bH, and similarly bH ⊂ aH, so that aH = bH.
    The proofs of (ii), (iii) are exercises below.
  • Index [G:H]: The number n of distinct cosets in G = H ∪ a₁H ∪ ··· ∪ a_n−1H is called the index of H in G, denoted n = [G:H].
  • Lagrange's Theorem: For a subgroup H of a finite group G, the number of elements of H is a divisor of the number of elements of G: that is, |H| divides |G|. In fact, |G| = |H| [G : H].
    Proof: Letting H, a₁H, . . . , a_n−1H be the distinct cosets, where n = [G : H], we have the disjoint union:
    G  =  H ∪ a₁H ··· ∪ a_n−1H.
    There are the same number of elements on both sides:
    |G|  =  |H| + |a₁H| + ··· + |a_n−1H|  =  |H| n  =  |H| [G : H].

  • Classification 6-element groups: A group of 6 elements is isomorphic to C₆ or D₃.
    Proof: Let N = max{|a|  for  a ∈ G}, the size of the largest cyclic subgroup. By Lagrange's Theorem, all elements have order 1, 2, 3, or 6, the divisors of |G| = 6.
    1. If N = 6, then G has an element c of order 6, and G = {e, c, c², c³, c⁴, c⁵} ≅ C₆.
    2. If N = 3, then G has an element r of order 3, and a subgroup H = {e, r, r²} with r³ = e. Taking an element a ∉ H, we have the cosets G = H ∪ aH. Now, a² ∉ aH, since otherwise a² = arⁱ and a = rⁱ, contrary to our choice of a. Thus a² ∈ H, i.e. a² = e, r, or r². If a² = r, then e, a, a² = r, a³ = ar, a⁴ = r², a⁵ = ar² are all distinct, and |a| = 6, contrary to N = 3. Similarly for a² = r². Thus we must have a² = e.
       Finally ra ≠ e, r, r², a so ra = ar or ar². If ra = ar, then we would clearly have |ar| = 6, contrary to N = 3. Hence ra = ar², and G ≅ D₃.
    3. If N = 2, then g² = e for all g ∈ G. This means G is abelian: ba = (aa)ba(bb) = a(abab)b = ab for any a,b. In this case, any two non-identity elements a,b ∈ G generate a subgroup ⟨a,b⟩ = {e, a, b, ab = ba}, but this is impossible since 4 is not a divisor of 6.
  ⊞ HW  
  1. Prove Coset Lemmas (ii) and (iii) above: Why do the cosets all have the same size as H, and why does every element of G lie in some coset of H?
     Hints: For (ii), consider the mapping f : H → aH given by f(h) = ah. For (iii), explicitly find the coset aH which contains a particular element g ∈ G.
  2. For each group G below, find all subgroups H ⊂ G, and decomposeinto cosets: G = H ∪ a₁H ∪ ··· ∪ a_n−1H.
     1. Cyclic group G = C₁₅ = {e, r, r², . . . , r¹⁴} with r¹⁵ = e.
     2. Additive group G = (Z₁₅ , +) = {0, 1, 2, . . . , 14}.
     3. Dihedral group G = D₄, the symmetries of the square, written as D₄ = {e, r, r², r³, a, ar, ar², ar³} obeying r⁴ = a² = e and ra = ar⁻¹.
  3. Use Lagrange's Theorem to very easily classify groups of order |G| = 5, 7, and 11. Generalize these cases!
  ⊞ Soln  

  1. Coset Lemma (ii): For any a ∈ G, we have |H| = |aH|.
  Proof: The mapping f : H → aH, h ↦ ah has inverse function ah ↦ h = a⁻¹ ah. Since it is invertible, it must be one-to-one and onto, so its domain and codomain have the same size: |H| = |aH|.

  Coset Lemma (iii): G is a disjoint union of cosets, G = H ∪ a₁ ∪ a₂ ∪ ··· .
  Proof: By Lemma (i), each g ∈ G lies in at most one coset; but any element is in its own coset, g = ge ∈ gH, so every element lies in exactly one coset; and the cosets disjointly cover G.

  2a. As for all groups, G = C₁₅ has the trivial subgroup H = {e}, whose cosets are the individual elements of G; and H = G, which has only one coset containing all elements. We have the following non-trivial decompositions of C₁₅ into cosets:

  C₁₅  =  {e, r⁵, r¹⁰} ∪ {r,r⁶,r¹¹} ∪ {r²,r⁷¹²} ∪ {r³,r⁸¹³} ∪ {r⁴,r⁹¹⁴}

  C₁₅  =  {e, r³, r⁶, r⁹, r¹²} ∪ {r, r⁴, r⁷, r¹⁰, r¹³} ∪ {r², r⁵, r⁸, r¹¹, r¹⁴}.

  2b. Since (Z₁₅ , +) ≅ C₁₅, this is the same problem as 2a, just in a different notation.

  Z₁₅  =  {0,5,10} ∪ {1,6,11} ∪ {2,7,12} ∪ {3,8,13} ∪ {4,9,14}

  Z₁₅  =  {0,3,6,9,12} ∪ {1,4,7,10,13} ∪ {2,5,8,11,14}.

  2c. G = D₄ (symmetries of a square) has, besides its trivial subgroups {e} and G, the rotation subgroup {e, r, r², r³} and the reflection subgroups {e, a}, {e ar}, {e, ar²}, {e, ar³}, which give the following decompositions into cosets:

  D₄  =  {e, r, r², r³} ∪ {a, ar, ar², ar³}

  D₄  =  {e, a} ∪ {r, ra = ar³} ∪ {r², r²a = ar²} ∪ {r³, r³a = ar}

  D₄  =  {e, ar} ∪ {r, rar = a} ∪ {r², r²ar = ar³} ∪ {r³, r³ar = ar²}

  D₄  =  {e, ar²} ∪ {r, rar² = ar} ∪ {r², r²ar² = a} ∪ {r³, r³ar² = ar³}

  D₄  =  {e, ar³} ∪ {r, rar³ = ar²} ∪ {r², r²ar³ = ar} ∪ {r³, r³ar³ = a} .

  We also have the partial rotation subgroup ⟨r²⟩ = {e, r²}, which combines with the reflection subgroups to give {e, a, r², ar²} and {e, ar, r², ar³}.

  3. If |G| = 5, by Lagrange's Theorem any subgroup H must have 5 = |H| [G : H], so we can only have |H| = 1 or 5 since these are the only factors of the prime number 5. Now, taking H = ⟨a⟩, we find the order |a| = |H| = 1 or 5; hence there must be an element of order 5, and G = ⟨a⟩ ≅ C₅. That is, the only group of order 5 is a cyclic group.

  This argument goes through the same way for |G| = p, any prime number. We must have G ≅ C_p.

• 2/4 & 2/6 Ch 8.2 Normal subgroup, quotient group   ⊞ Notes  
  • Reading: Ch 8.2, pp. 248-252.
  • Modular addition group. Recall that in clock arithmetic, we decompose the integers (Z, + ) into cosets of nZ:
    Z   =   [0] ∪ [1] ∪ ··· ∪ [n−1]   =   nZ ∪ (1 + nZ) ∪ ··· ∪ (n−1 + nZ).
    The cosets represent the clock-numbers under arithmetic modulo n: Z/nZ = {[0], [1], . . . , [n−1]}. For example, taking n = 3, we compute [1] + [2] = [3] = [0].
    Each element is equal to a set of representatives, any of which can be used to define the element:
    a ≡ a' mod n   ⇔   [a] = [a']   ⇔   a + nZ = a' + nZ   ⇔   a − a' ∈ nZ.   ⇔   a'  ∈  a + nZ.
    The last condition means we can use any element a' in the coset a + nZ as a representative for the coset.
  • Addition is well-defined: Different representatives of the same classes in Z/nZ will always give the same sum: that is, if [a] = [a'] and [b] = [b'], then [a+b] = [a'+b'].
    Example: In Z/3Z, we have [1] = [−5] and [2] = [11], and we compute:

    [1] + [2]	=	[3]  =  [0]
    [−5] + [11]	=	[6]  =  [0].

    Proof: The standard definition of modular addition is: [a] + [b] = [a+b]; but we get the same definition by adding all elements in the corresponding cosets. For example, define:
    [0] + [0]  =  nZ + nZ  =  {ni + nj  for  i,j ∈ Z}  =  nZ = [0].
    (In fact, ni + nj = n(i+j) ∈ nZ, so that nZ + nZ ⊂ nZ, and the reverse inclusion is obvious.)
    Now define:
    [a] + [b]  =  a + nZ + b + nZ  =  {a+ni + b+nj  for  i,j ∈ Z}.
    Then we clearly have a + nZ + b + nZ  =  a + b + nZ + nZ  =  a + b + nZ  =  [a+b], as before.
    But now it is clear by definition that if the cosets are the same, [a] = [a'] and [b] = [b'], then their sum is the same, [a] + [b] = [a'] + [b'].
  • Quotient group (attempt). Now we try to adapt this idea to an arbitrary subgroup H ⊂ G. We split G into cosets:
    G   =   H ∪ g₁H ∪ ··· ∪ g_n−1H,
    and we want to define the quotient group G/H to be the set whose elements are all cosets [g] = gH:
    G/H   =   {[e], [g₁], . . . , [g_n−1]}.
    As before, we have equivalence of elements of G modulo H:
    [g] = [g']   ⇔   gH = g'H   ⇔   g⁻¹g' ∈ H   ⇔   g' ∈ gH.
    We will try to define [a][b] = [ab].
  • Couterexample. Unfortunately, this operation is usually not well-defined when G is a non-abelian group. Taking the dihedral group G = D₃ = {e, r, r², a, ar, ar²} with r³ = a² = e and ra = ar⁻¹, and the subgroup H = {e, a}, we have:
    D₃   =   H ∪ rH ∪ r²H   =   {e, a} ∪ {r, ra = ar²} ∪ {r², r²a = ar},
    so that G/H = {[e], [r], [r²]}. However, if we attempt to multiply [r][r²] using different representatives in rH and r²H, namely [r] = [ar²] and [r²] = [ar], we get two different products:

    [r][r2]  =  [r3]  =  [e]

    [ar2][ar]  =  [ar2ar]  =  [r2].

    The source of the trouble here is that the left cosets gH are not equal to the right cosets Hg:
    D₃   =   H ∪ Hr ∪ Hr²   =   {e, a} ∪ {r, ar} ∪ {r², ar²}.

  • If we try this for the subgroup N = {e, r, r²} ⊂ D₃, we have:
    D₃   =   N ∪ aN   =   N ∪ Na   =   {e, r, r²} ∪ {a, ar, ar²}.
    In this case, multiplication in G/N = {[e], [a]} is well-defined, and obeys the group axioms. In fact, [a]² = [e], and G/N ≅ C₂ .
  • Normal subgroup: We say a subgroup N ⊂ G is normal if gN = Ng for all g ∈ G; or equivalently, gNg⁻¹ = N.
  • Quotient group: For a normal subgroup N ⊂ G, we define G/N = {[g] = gN  for  g ∈ G}, the set of all cosets; and we multiply [a][b] = [ab]. Then this gives a well-defined group operation on the elements of G/N.
    Proof. To see this is well-defined, we use the equivalent definition aN bN = {anbn'  for  n,n' ∈ N}, which agrees with the definition using representatives:
    [a][b]  =  aN bN  =  a(Nb)N  =  a(bN)N  =  abN  =  [ab],
    since NN = N. Thus, if [a] = [a'] and [b] = [b'], i.e. aN = a'N and bN = b'N, then [a][b] = aN bN = a'N b'N = [a'][b'].
    The proof that G/N obeys the group axioms is an exercise below.
  ⊞ HW  
  1. Prove that, for any subgroup H ⊂ G, and for HH = {hh'  for  h,h' ∈ H}, we have HH = H.
     Hint: Separately prove HH ⊂ H and H ⊂ HH.
  2. Show that for a normal subgroup N ⊂ G, the quotient G/N, as defined in the notes, satisfies the four group axioms.
  3. Index of a subgroup: Recall that n = [G : H] means the number of left cosets, so that G  =  H ∪ g₁H ∪ ··· ∪ g_n−1H.
     1. Show that for any subgroup H ⊂ G, the number of left cosets is equal to the number of right cosets. Here we assume [G : H] is finite, but H and G might be infinite, as for nZ ⊂ Z.
        Hint: Find a bijective mapping φ : G → G which takes left cosets to right cosets.
     2. Show that if H ⊂ G is a subgroup with [G : H] = 2, then H must be a normal subgroup.
  4. For G = D₄, determine whether each of the 10 subgroups in the previous HW is a normal subgroup, and if so write the multiplication table for the quotient group G/H.
  5. Extra Credit: Show that if a subgroup H ⊂ G is not normal, then multiplication of cosets is not well-defined: there exist some [a] = [a'] and [b] = [b'] with [ab] ≠ [a'b'].
  ⊞ Soln  

  1. Proposition: For a subgroup H ⊂ G, we have HH = H.
  Proof: First, if hh' ∈ HH, then hh' ∈ H by the closure axiom for the subgroup H; thus HH ⊂ H. Second, if h ∈ H, then h = he ∈ HH by the identity axiom for H; thus H ⊂ HH. Therefore, we have HH = H.

  2. The quotient G/N, whose elements are cosets [g] = gN, satisfies the four group axioms because the group G satisfies these axioms.
  Closure: The multiplication [a][b] = [ab] is well-defined (proved in notes).
  Associative: ([a][b])[c] = [(ab)c] = [a(bc)] = [a]([b][c]).
  Identity: [a][e] = [ae] = [a].
  Inverses: [a][a⁻¹] = [aa⁻¹] = [e].

  3a. Proposition: For any subgroup H ⊂ G, the number of left cosets is equal to the number of right cosets.
  Proof: Define the inverse mapping φ : G → G by φ(g) = g⁻¹, and denote φ(H) = {φ(h) for h ∈ H}. For h ∈ H, we have φ(h) = h⁻¹ ∈ H, so φ(H) ⊂ H. Also, h = (h⁻¹)⁻¹ = φ(h⁻¹), so H ⊂ φ(H); and therefore φ(H) = H.
  Applying φ to any left coset gH = {gh for h ∈ H} gives:

  φ(gH)  =  {(gh)⁻¹ = h⁻¹g⁻¹ for h ∈ H}  =  φ(H)g⁻¹  =  Hg⁻¹.

  Similarly φ(Hg) = g⁻¹H, so φ gives an invertible (bijective) correspondence between left and right cosets. Thus, there must be the same number of each.
  Note: This gives the following correspondence between left and right cosets:

  G   =   H ∪ g₁H ∪ ··· ∪ g_n−1H   =   H ∪ H(g₁)⁻¹ ∪ ··· ∪ H(g_n−1)⁻¹.

  However, e, g₁, . . . , g_n−1 are not necessarily representatives for the right cosets, as we can see in the case G = D₃ , H = {e, a}, rH = {r, ra = ar²}, so that D₃ = H ∪ ar²H ∪ r²H, but Har² = Hr².

  3b. Proposition: For a subgroup H ⊂ G, if [G : H] = 2, then H is a normal subgroup.
  Proof: Take any g ∈ G. If g ∈ H, then gH = H = Hg. On the other hand, if g ∉ H, the hypothesis [G : H] = 2 means there are just two left cosets, and by part (a) there are also just two right cosets:

  G  =  H ∪ gH  =  H ∪ Hg.

  Then gH and Hg are both equal to the complement G−H = {g ∈ G with g ∉ H}, so that again gH = Hg. Thus, H satisfies the definition of normality.

  4. The normal subgroups of G = D₄ = {e, r, r², r³, a, ar, ar², ar³} with a² = r⁴ = e and ra = ar⁻¹, are the trivial subgroups {e} and G, and:

  N = ⟨r²⟩ = {e, r²},   ⟨r⟩ = {e, r, r², r³},   ⟨a, r²⟩ = {e, a, r², ar²},   ⟨ar, r²⟩ = {e, ar, r², ar³}.

  We can see this for the first subgroup because r² ∈ Z(G), the center, meaning r² commutes with all g ∈ G (geometrically, r² is a 180° rotation with matrix −I). The others all have index [G : H] = 2, so they must be normal by #3b above. The other 2-element subgroups, each generated by a reflection, are not normal: H = ⟨a⟩, ⟨ar⟩, ⟨ar²⟩, ⟨ar³⟩. For each, it is easy to find a coset with gH ≠ Hg.
• 2/8 Ch 8.4 Homomorphisms   ⊞ Notes  
  • Reading: Ch 8.3, pp. 263-266.
  • Homomorphism: A mapping between groups φ : G → G' which satisfies the structure equation:
    φ(a·b)  =  φ(a)·φ(b)   for all  a,b ∈ G.
    Thus, each multiplication a·b in G corresponds to a mulitplication φ(a)·φ(b) in G'.
    A homomorphism need not be one-to-one or onto, but an isomorphism is a homomorphism which is one-to-one and onto (bijective, invertible).
  • Kernel: The elements of G which are sent to the identity e' ∈ G':   Ker(φ) = {k ∈ G with φ(k) = e'}.
    Image: The elements g' ∈ G' which are outputs of the homomorphism: Im(φ) = {g' = φ(g) for g ∈ G}.
  • Projection homomorphism: The most common type of homomorphism (other than isomorphisms) is the natural mapping from a group G to a quotient group G/N:
    π : G → G/N   with   π(g) = [g] = gN ∈ G/N.
    
    The kernel contains k with π(k) = [k] = [e], which means k ∈ N; that is, Ker(π) = N.
    The mapping π is clearly onto: each class is the class [g] of some element g ∈ G; so Im(π) = G/N.
  • Example: Exponential mapping. Define φ : (Z , + ) → (R^×, · ) by φ(i) = 2ⁱ. Here R^× = R−{0} is the multiplicative group of non-zero real numbers; and the structural equation is the addition-of-exponents rule:
    φ(i + j)  =  2^i+j  =  2ⁱ 2^j  =  φ(i) φ(j).

    Here the operation on inputs G = Z is + , while that on the outputs R^× is multiplication.
    Kernel: trivial, since φ(i) = 2ⁱ = 1 only for i = 0; Ker(φ) = {0}.
    Image: the set of integer powers of 2, which is the cyclic subgroup generated by 2

    Im(φ) = { . . . , ¹⁄₄, ¹⁄₂, 1, 2, 4, . . . } = ⟨2⟩.

  • Example: Power mapping. Define φ : (R^×, · ) → (R^×, · ) by φ(x) = x². The structural equation is the rule that powers distribute over products:
    φ(xy)  =  (xy)²  =  x²y²  =  φ(x) φ(y).

    Note that x,y are real numbers with their usual commutative multiplication.
    Kernel: the solutions of φ(x) = x² = 1;  Ker(φ) = {1, −1} = ⟨−1⟩.
    Image: the real numbers of the form φ(x) = x², i.e. the positive numbers;  Im(φ) = R_>0 .

  • Example: Determinant. Define det : GL₂(R) → (R^×, · ) by the usual formula for det(A). The structural equation is the rule that the determinant of a product is the product of determinants:
    det(AB)  =  det(A) det(B).
    Kernel: the subgroup of matrices having determinant 1, which is called the special linear group SL₂:  Ker(det)  =  {A ∈ GL₂(R) with det(A) = 1} = SL₂(R). That is, (R).
    Image: all non-zero reals, Im(det) = R^×
  • Homomorphism Theorem: Let φ : G → G' be any homomorphism of groups. Then:
    1. Ker(φ) is a normal subgroup of G.
    2. Im(φ) is a subgroup of G'.
    3. The image of φ is isomorphic to the quotient of G by the kernel of φ:
       Im(φ)  ≅  G/Ker(φ).

    Part (iii) is known as the First Isomorphism Theorem.
  • In the diagram, the kernel K = Ker(φ) has cosets gK which are the elements of G/K. The whole coset gK maps to an element g' = φ(g) ∈ G', and the quotient group G/K is isomorphic to the image subgroup Im(φ) ⊂ G'.
  • Examples
    • For a projection homomorphism π : G → G/N, the theorem just says that G/N = G/Ker(φ), since N = Ker(φ).
    • Exponential mapping φ : (Z , + ) → (R^×, · ), φ(i) = 2ⁱ. The output group Im(φ) = ⟨2⟩ is isomorphic to Z/⟨0⟩ = Z; indeed, both of these are infinite cyclic groups, isomorphic to C_∞.
    • Power mapping φ : R^× → (R^×, φ(x) = x². The output group of positive reals, Im(φ) = R_>0, is isomorphic to R^×/{±1}, i.e. non-zero reals ignoring signs (counting ±x as the same number).
  • Proof of (iii): Given the homomorphism φ : G → G', define the isomorphism Φ : G/Ker(φ) → Im(φ) by Φ([g]) = φ(g).
    • Φ is well-defined, since if we take different representatives with [g₁] = [g₁], this means g₂ = g₁k for k ∈ Ker(φ), so:
      Φ([g₂])  =  φ(g₂)  =  φ(g₁k)  =  φ(g₁) φ(k)  =  φ(g₁) e'  =  φ(g₁)  =  Φ([g₁]).

    • Φ is one-to-one since Φ([g₁]) = Φ([g₂]) means φ(g₁) = φ(g₂) and φ(g₁⁻¹g₂) = φ(g₁)⁻¹ φ(g₂) = e', so that g₁⁻¹g₂ = k ∈ Ker(φ) and g₂ = g₁k and [g₂] = [g₁] ∈ G/Ker(φ).
    • Φ is clearly onto Im(φ): the outputs of Φ are the same as those of φ, namely φ(g) for g ∈ G.
    • The structure equation holds:  Φ([a][b])  =  φ(ab)  =  φ(a) φ(b)  =  Φ([a]) Φ([b]).
  ⊞ HW  
  1. Recall the complex numbers C of the form a+bi for real numbers a,b ∈ R and i² = −1. We let (C^×, · ) be the multiplicative group of non-zero complex numbers. Define the square-length mapping φ : C^× → R^× by φ(a+bi) = a²+b².
     1. Show that φ is a homomorphism.
     2. Determine the kernel and image of φ.
     3. Verify the First Isomorphism Theorem in this case.
  2. Let φ : G → G' be any homomorphism, satisfying φ(ab) = φ(a) φ(b). Prove that φ(e) = e' (the identity in G goes to the identity in G'), and φ(g⁻¹) = φ(g)⁻¹.
  3. Prove that Ker(φ) = {e} if and only if φ is injective (one-to-one). Hint: Prove (⇒) and (⇐) directions separately.
  4. Prove the Homomorphism Theorem parts (i), (ii). Hint: To show normality, it is enough to show g Ker(φ) g⁻¹ ⊂ Ker(φ) for any g ∈ G; that is, φ(gkg⁻¹) = e'.
  5. Extra Credit. Prove the Second Isomorphism Theorem: Consider subgroups H,N ⊂ G with N normal in G. Then:
     1. N ∩ H is a normal subgroup of H
     2. HN = {hn for h ∈ H, n ∈ N} is a group, with N as a normal subgroup.
     3. We have the isomorphism  H/(H∩N)  ≅  HN/N.
     You may consult a reference, but write up the proof in your own words.
  ⊞ Soln  

  1a. You only need to check the structure equation:

  φ((a+bi)(c+di))  =  φ((ac−bd) + (ad+bc)i)  =  (ac−bd)² + (ad+bc)²
   =  a²c² + a²d² + b²c² + b²d²  =  (a²+b²)(c²+d²)  =  φ(a+bi) φ(c+di).

  1b. The kernel is:

  Ker(φ)  =  {a+bi with φ(a+bi) = a²+b² = 1}  =  {cos(θ) + i sin(θ) for 0 ≤ θ ≤ 2π}  =  Cir,

  the unit circle in the plane of complex numbers, which forms a subgroup under multiplication.
  The image is the positive real numbers: Im(φ) = R_>0

  1c. The theorem says R_>0 ≅ C^×/Cir. In fact, any complex number can be written in polar form as  a+bi  =  r (cos(θ) + i sin(θ)), with r = √(a²+b²). Taking this mod Cir, the second factor becomes irrelevant, and [a+bi] = (a+bi)Cir = r Cir, and clearly this is isomorphic to multiplication of positive reals.

  2. Proposition: Any homomorphism has φ(e) = e' and φ(g⁻¹) = φ(g)⁻¹.
  Proof: First, for any g ∈ G, we have g = ge, so φ(g) = φ(ge) = φ(g)φ(e). Multiplying both sides by φ(g)⁻¹ gives e' = φ(e). Second, φ(g)φ(g⁻¹) = φ(gg⁻¹) = φ(e) = e', so φ(g) and φ(g⁻¹) multiply to e', and are inverses.

  3. Proposition: Ker(φ) = {e} if and only if φ is injective.
  Proof: (⇒) Suppose Ker(φ) = {e}, i.e. φ(k) = e' only for k = e. Suppose φ(g₁) = φ(g₂); then e' = φ(g₁)⁻¹φ(g₂) = φ(g₁⁻¹g₂), and g₁⁻¹g' ∈ Ker(φ), so g₁⁻¹g₂ = e, so g₁ = g₂. Thus two different elements of G cannot have the same output under φ.
  (⇐) Suppose φ is one-to-one. Since φ(e) = e', there can be no other k ∈ G with φ(k) = e', so Ker(φ) = {e}.

  4. Given a group homomorphism φ : G → G'.

  Proposition: Ker(φ) is a normal subgroup of G.
  Proof: First we check that Ker(φ) is a subgroup. If k₁,k₂ ∈ Ker(φ), then φ(k₁k₂) = φ(k₁) φ(k₂) = e'e' = e', so k₁k₂ ∈ Ker(φ). Hence Ker(φ) is closed under multiplication. Second, if k ∈ Ker(φ), then φ(k⁻¹) = φ(k)⁻¹ = (e')⁻¹ = e', so k⁻¹ ∈ Ker(φ). Hence Ker(φ) has inverses. These two axioms imply Ker(φ) is a subgroup.
  Next, we show Ker(φ) is normal by checking g Ker(φ) g⁻¹ = Ker(φ). Let k ∈ Ker(φ) and g ∈ G. Then:

  φ(gkg⁻¹)  =  φ(g) φ(k) φ(g)⁻¹  =  φ(g) e' φ(g)⁻¹  =  e',

  so gkg⁻¹ ∈ Ker(φ), and g Ker(φ) g⁻¹ ⊂ Ker(φ). Since this is true for any g, we may restate it with g replaced by g⁻¹, giving g⁻¹ Ker(φ) g ⊂ Ker(φ); and multiplying by g on the left and g⁻¹ on the right gives Ker(φ) ⊂ g Ker(φ) g⁻¹. Hence g Ker(φ) g⁻¹ = Ker(φ), and Ker(φ) is normal.

  Proposition: Im(φ) is a subgroup of G'.
  Proof: Take any two elements of the image, φ(g₁), φ(g₂) ∈ Im(φ). Then φ(g₁) φ(g₁) = φ(g₁g₂) ∈ Im(φ), and Im(φ) is closed under multiplication. Also, if φ(g) ∈ Im(φ), then φ(g)⁻¹ = φ(g⁻¹) ∈ Im(φ), so Im(φ) has inverses. Hence Im(φ) is a subgroup.

• 2/11 Subgroups as symmetries, decorated shapes, normal decorations  ⊞ Notes  
  • We have seen how a abstract group G can be analyzed purely in terms of its multiplication table, with structures such as subgroups, normal subgroups, and quotient groups. But groups arise as symmetries, and their abstract structures can be interpreted in terms of symmetries.
  • Let G = Sym(X), the symmetries of a geometric object X, including polygons and polyhedra, perhaps with markings and colors, and more general collections of points.
  • Let H ⊂ G be a subgroup. Then we can construct an object X⁺, a "decoration" of X, with H = Sym(X⁺); that is, H contains those symmetries of X which also preserve the decorations.
    To construct X⁺, first add some patterns such as arrows or colors to X so as to break all symmetry, obtaining a figure X^# with Sym(X^#) = {e}, the trivial subgroup; that is, gX^# ≠ X^# for all non-trivial g ∈ G. Now we let:
    X⁺  =  ∪_h∈H hX^#,
    the union of the translates of X^# by elements of H. Then this has H = Sym(X⁺).
  • Normal subgroups may also be characterized in this framework. The symmetry group of the translation gX⁺ is the conjugate subgroup gHg⁻¹. Thus H is normal in G whenver:
    gHg⁻¹  =  Sym(gX⁺)  =  Sym(X⁺)  =  H   for all   g ∈ G.
    That is, H is normal whenever the decoration of X⁺ may be drawn in any orientation, with respect to any side or feature of X, and still yield the same symmetry group gHg⁻¹ = H.
  • Example: G = D₃ = Sym(X), where X is an equilateral triangle; this has elements {e, r, r², a, b, c}, where r, r² are ⅓ rotations, and a, b = ar, c = ar² are reflections. Then the asymmetric X^# can be chosen to be X decorated with a single arrow, and each subgroup H is shown with the corresponding X⁺ with H = Sym(X⁺): The subgroup H = {e, r, r²} is normal, since the symmetries of the oriented triangle X⁺ are the same for both clockwise and counterclockwise orientations:  Sym(X⁺) = Sym(gX⁺)  for all g ∈ G.
    On the other hand, H = {e, a} is not normal, since its X⁺ (double arrow on bottom edge) has translates such as bX⁺ with a different symmetry group:  Sym(bX⁺) = {beb⁻¹, bab⁻¹} = {e, c}.
• 2/13 & 2/15 Cayley graphs   ⊞ Notes  
  • For an abstract group G, there is a general way to produce an object X which has G as its symmetry group:
    G  ≅  Sym(X).
    This X is a combinatorial graph, a set of vertices connected by arrows (directed curves), also known as nodes in a network. A graph is defined solely by which vertices are connected to which others; it does not contain any geometric data about length, angle, etc.
  • Cayley graph: Suppose the group G is generated by multiplying elements a,b, . . . (and their inverses). Define the graph X to have a vertex for every g ∈ G, and an a-colored arrow from each g to ga, b-colored arrow from g to gb, etc: g

    a

    →

    ga   ,   g

    b

    →

    gb   ,    . . . . This graph encodes the multiplication table of G. To multiply g·h, locate g, h, and the identity e as vertices of X. Follow a sequence of arrows from e to h; the sequence of arrow labels shows how to write h in terms of the generators a,b, . . . . Finally, start from g and follow the same sequence of labeled arrows to end at the vertex gh.
  • A graph symmetry of X is a permutation of the vertices, a bijection f : G → G, such that each a-colored edge g →^a ga is taken to vertices which have a similar edge, f(g) →^a f(ga). Since f permutes the elements of G = {e, g₁, . . . , g_n−1}, where n = |G|, we may consider f ∈ S_n, and Sym(X) ⊂ S_n.
  • Proposition: The symmetries of the Cayley graph X are isomorphic to the group:  Sym(X) ≅ G.
    Proof:
    • Each g ∈ G corresponds to a permutation f_g : G → G with f_g(h) = gh;  this is a graph symmetry, since it takes h →^a ha to gh →^a gha.
    • I claim these are the only graph symmetries of X. Let f : G → G be any graph symmetry, and g = f(e); then I claim f = f_g. First, f(e) = g = f_g(e). Now, suppose we know f(h) = f_g(h) for some h. Then the arrow h →^a ha is taken to an a-arrow by both f and f_g, namely f(h) →^a f(ha) and f_g(h) →^a f_g(ha). But since these arrows begin at the same vertex f(h) = f_g(h), and there is only one a-arrow from this vertex, the arrows must also end at the same vertex: f(ha) = f_g(ha). Thus, the vertices h where f and f_g agree spread out inductively from h = e to include the whole graph, so f = f_g.
    • Now consider the mapping φ : G → S_n given by φ(g) = f_g . Then φ is a homomorphism, since:
      f_gg'(h)  =  gg'h  =  g(g'h)  =  f_g(f_g'(h))  =  (f_g∘ f_g')(h),
      which means φ(gg') = φ(g) ∘ φ(g'). (Recall that composition ∘ is the group operation on permutations in S_n, or in any symmetry group.)
    • Furthermore, φ is injective: φ(g) = φ(g') means f_g = f_g', so that gh = g'h for all h; but then g = g'. This means Ker(φ) = {e}.
    • Now by the above combined with the Homomorphism Theorem, we have:
      Sym(X)  =  {f_g for g ∈ G}  =  Im(φ)  ≅  G/Ker(φ)  =  G/{e}  ≅  G.

  • Example:  G = D₃ = ⟨r,a⟩ = {e, r, r², a, ar, ar²}  with  r³ = a² = e  and  ra = ar⁻¹}. The Cayley graph X is: We can use the graph to multiply elements such as:  g · h  =  r²a · r⁻¹.
    • Starting at e, follow the sequence of arrows colored by the generator factors gh = r·r·a·r⁻¹, where r⁻¹ means to go backward along an r-arrow:
      e

      r

      →

      r

      r

      →

      r²

      a

      →

      ar

      r

      ←

      a.

    • The endpoint is the product:  r²a·r⁻¹ = a.
  • D₃ is the symmetry group of the colored graph:  G = Sym(X). Although the graph is not geometric, we can choose a nice drawing of X as a triangular prism, as above, and see G as the geometric symmetries of this solid, obtaining a new geometric model of G.
    • r is ¹⁄₃ rotation around a horizontal axis
    • a is ¹⁄₂ rotation around a vertical axis
    Both these mappings preserve the blue and red arrows; but note that reflection across a vertical plane would reverse the blue arrows, so it is not a symmetry of the decorated prism. We can also say G is the group of only rotational symmetries of the undecorated prism.
  ⊞ HW  
  1. Draw another Cayley graph of G = D₃, this time with respect to the reflection generators a and b = ar. That is, the vertices are the same, but the two types of arrows are  g →^a ga  and  g →^b gb .
  2. Draw two Cayley graphs of C₆ = {e, c, . . ., c⁵}.
     1. With respect to the generator c.
     2. With respect to the two generators r = c² and a = r³. Compare this to the first Cayley graph of D₃.
     3. The graph in (b) can be seen as the Cartesian product of an oriented triangle with a double-arrow interval. What does this mean about the structure of the group G = C₆?
        Extra Credit: Give a general definition of a Cartesian product graph, and show that a product of Cayley graphs is the Cayley graph of the product group.
     4. Draw the graph in (b) as a decorated triangular prism. How do the generators r,a act geometrically?
  ⊞ Soln  

  1. Recall ra = ar², r³ = a³ = e.

  2a. The Cayley graph of C₆ = ⟨c⟩ is just an oriented 6-cycle, with only c-arrows going counterclockwise (or clockwise).

  2b. The Cayley graph of C₆ = ⟨r=c², a=c³⟩ is the same as that of D₃ = ⟨r,a⟩, except the three-cycles of r-arrows have the same orientation (rather than opposite).

  2c. The Cayley graph is a product because the group is isomorphic to a product: C₆ ≅ C₃ × C₂, where r = c² and a = c³ map to the generators of the two factors.

  2d. In the geometric model of C₆ as the symmetries of a triangular prism, r again acts as a ¹⁄₃ rotation around a horizontal axis, but a acts as a reflection across a vertical plane. These two mappings move orthogonal subspaces because of the Cartesian product structure described in (c).

• 2/18 Ch 7.5 Symmetric group S_n, permutahedron & S₄   ⊞ HW  
  Reading: Ch 7.5, pp. 227-231. A useful video.

  1. Consider the following Cayley graph of G = S₄ with respect to the generators written in cycle notation as a = (12), b = (23), c = (34). That is: a(1) = 2, a(2) = 1, a(3) = 3, a(4) = 4; and similarly for b,c. The vertices have the permutations written in one-line notation as f(1) f(2) f(3) f(4), so f = 4213 means f(1) = 4, f(2) = 2, etc.) Label each vertex with its cycle notation, and verify that the arrows of each type are correct: g →^a ga, etc.
  2. The Cayley graph is drawn to suggest a polyhedron in space, called the permutohedron. Determine how many vertices in this polyhedron, how many edges of each color, and how many faces of each type. What do these geometric features correspond to in terms of group theory?
  3. What is the simplest geometric solid whose symmetry group is S₄? What is the relation of this simple solid to the permutohedron?
  ⊞ Soln  

  1. Since a² = b² = c² = e, all edges are bi-directional ↔. In the picture, the bottom blue edge is e = 1234 ↔ 2143 = (12) = a, so a-edges are blue ↔^a, and similarly green b-edges ↔^b and red c-edges ↔^c.
  To check that the edges are correct, note that that the edge f ↔^a fa is written in one-line notation as:

  f(1) f(2) f(3) f(4)  

  a

  ↔

    f(2) f(1) f(3) f(4),

  so that multiplying on the right by a = (12) switches the first two entries; and similarly for b,c.

  2. The vertices correspond to group elements, so |S₄| = 4! = 24. The edges of each color pair off the 24 vertices, so there are 12 of each color, representing multiplication by a generator. There are three types of faces: 6 squares corresponding to the relation ac = ca; 4 hexagons corresponding to aba = bab; and 4 hexagons corresponding to bcb = cbc.

  3. Let X be a regular tetrahedron, a pyramid with three sides and triangular bottom, all of them equilateral triangles. This has 4 vertices, and the geometric symmetries permute them in all possible ways, so Sym(X) = S₄.
  We can obtain a permutohedron by slicing off the tip of each corner of the tetrahedron, exposing 4 new triangles and chopping the original 4 triangle faces into hexagons; then shaving off a sliver from each of the original 6 edges, exposing 6 rectangle faces and chopping the 4 new triangle faces into hexagons. It is clearer in this picture of the permutahedron, with unequal edge-lengths.

  

• 2/20 & 2/22 Review: Geometric symmetries, cubic rotations S₄   ⊞ HW  
  Reading: First week notes.

  1. Consider G = D₃ = S₃ as the symmetries of the equilateral triangle with center at (0,0) and one vertex at (1,0).
     1. Find the two other vertices using trigonometry.
     2. Write matrix of the rotation R = Rot_2π/3. Recall that the column vectors of the matrix [R] are the outputs of the standard basis vectors R(1,0) and R(0,1).
     3. Verify that R² = R⁻¹.
     4. Write the matrix of A, the vertical reflection across the x-axis, and verify that A = A⁻¹, and that RA = AR²
     5. Find the other two reflections B = AR and C = AR².
     6. Write each symmetry as a permutation of the three vertices in cycle notation. Compute with permutations to verify again that A² = R³ = I and RA = AR⁻¹.
     7. Starting with the asymmetric point P = (6,2), plot the 6 points gP for each g ∈ G. Connect these with lines to form a hexagon whose symmetries are the 6 linear mappings of G.
     8. Label each hexagon vertex gP with the corresponding g ∈ G, and each edge with the appropriate generator, turning the outline of the hexagon into a Cayley graph of G (with respect to the appropriate generators).
  2. Let G be the group of rotational symmetries of the cube. (see HW 1/16 #1 and HW 1/11 #1). Label the faces of the cube 1, 2, . . . , 6 as shown:
     1. Let R₁ and R₂ be ¹⁄₄ rotation around the x and y axes. Write each as a permutation of the 6 faces, in cycle notation; and also as a 3×3 matrix. Hint: Each should be a 4-cycle.
     2. Compute the permutations of the symmetries P = R₁R₂ and Q  =  R₁² R₂.
     3. Since P and Q are orthogonal matrices with determinant 1, they must be rotations. (Explain.) Look at the picture and identify the axis and angle for each.
     4. Write the matrices of P and Q above, by multiplying the matrices of R₁ and R₂ appropriately.
     5. Find the eigenvalues and eigenvectors of P and Q. Do they agree with your visual results from (c)?
  ⊞ Soln  

  1a. The vertices are P₁ = (1,0) and P₂, P₃ = (cos(^2π⁄₃), ±sin(^2π⁄₃)). See W|A.

  1b. See first week HW.

  R  =  Rot_2π/3  =  [

  cos 2π⁄3	−sin 2π⁄3
  sin 2π⁄3	cos 2π⁄3

  ]  =  [

  −1⁄2	−√3⁄2
  √3⁄2	−1⁄2

  ].

  1c. Compute R² by matrix multiplication, and recall the formula for inverse of a 2×2 matrix: [

  a	c
  b	d

  ]⁻¹   =   ¹⁄_(ad−bc) [

  d	−c
  −b	a

  ]

  1d. A   =   [

  1	0
  0	−1

  ]

  1e. W|A computes: B  =  AR  =  [

  −1⁄2	−√3⁄2
  −√3⁄2	1⁄2

  ],         C  =  AR²  =  [

  −1⁄2	√3⁄2
  √3⁄2	1⁄2

  ].

  1f. Letting P₁ = (1,0), P₂ = R(1,0), P₃ = R²(1,0), the symmetries give the permutations:

  I = id     R = (123)     R² = (321) = (132)     A = (23)     B = AR = (13)     C = AR² = (12).

  1g. W|A can compute the points and draw the hexagon:

  

  1h. The short edges correspond to the generator A, the long edges to C = AR².

  2a.Taking each ¹⁄₄ rotation according to the right-hand rule, with thumb along the axis, we get:

  P  =  R1  =  (  1   2   3   4   5   6  3 2 6 1 5 4 )   =   (1364)   =

  1   0   0  0 0 −1 0 1 0

  Q  =  R2  =  (  1   2   3   4   5   6  2 6 3 4 1 5 )   =   (1265)   =

  0   0   1  0 1 0 −1 0 0

  .

  2b. By inputing i = 1, 2, . . . on the right side of the composed permutation R₁ ∘ R₂, we compute:

  R₁R₂   =   (1364)(1265)   =   (124)(365)
  
  R₁² R₂   =   (16)(34) (1265)   =   (12)(34)(56).

  2c. An orthogonal transformation R is a linear mapping which preserves distances, angles, and the dot product. Its matrix [R] has column vectors which form an orthonormal basis, or equivalently [R]^T R = I. Any orthogonal matrix has determinant ±1; and in 3 dimensions, det[R] = 1 means R is a rotation, and det[R] = −1 means R is a roto-reflection (reflection possibly composed with a rotation whose axis is the reflection direction).

  In our case, we are composing rotations, i.e. multiplying orthogonal matrices of determinant 1, so the result is also orthogonal of det 1, and is a rotation.

  Looking at how the faces move on the cube, we see that P = (124)(365) is a ¹⁄₃ rotation around the diagonal axis (1,1,1), which is the common point of faces 1,2,4. Similarly, Q = (12)(34)(56) is a ¹⁄₂ rotation around the partial diagonal axis (1,0,1), which is the midpoint of the common edge of faces 1,2. These are somewhat surprising symmetries of the cube: try them out on dice or some other physical model.

  2d. By multiplying matrices, we get:

  P  =  R1R2  =

  0   0   1  1 0 0 0 1 0

  Q  =  R12 R2  =

  0   0   1  0 −1 0 1 0 0 .

  We could get these matrices from the geometric descriptions in (c), noting that P permutes the x,y,z axes cyclically; while Q switches the x,z axes and flips over the y axis.

  2e. From the geometric description of P, we see that its rotation axis is an eigenvector of eigenvalue 1, and there are no other real eigenvectors. Applying the general method, we find the eigenvalues as roots of the characteristic polynomial p(λ) = det(R−λI) = 1 − λ³. The only real number solution is λ = 1, and we find its eigenvector v = (x,y,z) by solving the linear system given by (P−I)(x,y,z) = (0,0,0). This gives the axis (x,y,z) = (1,1,1), as in (c).

  P also has complex eigenvalues and eigenvectors. The non-real roots of p(λ) = 1 − λ³ are the cube roots of unity λ = cos(^2π⁄₃) ± i sin(^2π⁄₃). These indicate that P is a rotation by angle ^2π⁄₃ .

  Similarly for Q, the rotation axis (1,0,1) is an eigenvector of eigenvalue 1, and the vectors perpendicular to this are are all eigenvectors of eigenvalue −1 = cos(π) ± i sin(π). Since sin(π) = 0, all the roots of the characteristic polynomial are real:  p(λ) = −λ³ − λ² + λ + 1 = −(λ−1)(λ+1)².

• 2/25 Review: Abstract groups, classify |G| = 8   ⊞ HW  
  Classify all abstract groups of order 8 (i.e. |G| = 8 elements).

  1. For g ∈ G, what do we mean by a cyclic subgroup ⟨g⟩, and its order |g|?
  2. Let N = max{|g|  for  g ∈ G}, the largest order of a cyclic subgroup inside G. Explain why N > 1.
  3. What major theorem guarantees that |g| = 1, 2, 4, or 8 for every g ∈ G, so that N = 2, 4, or 8?
  4. Show that if N = 8, then G ≅ C₈ .
  5. Show that if N = 2, then G must be abelian, and in fact G ≅ C₂ × C₂ × C₂ .
     Hints: Examine the equation abab = (ab)² = e. Then find independent elements a,b,c ∈ G, and show that G = {e, a, b, c, ab, ac, bc, abc}.
  6. Consider the final case N = 4. Take an element r with |r| = 4, and an element a ∉ ⟨r⟩.
     1. Show that we have two cosets:
        G  =  ⟨r⟩ ∪ a⟨r⟩  =  {e, r, r², r³, a, ar, ar², ar³};
        and that ⟨r⟩ is a normal subgroup.
     2. Show that a² ∈ ⟨r⟩, and in fact a² = e or r².
     3. Show that ra ∈ a⟨r⟩ and in fact ra = ar or ar³. Hint: If ra = ar², what would be r²a?
     4. Suppose ra = ar. Show that G ≅ C₄ × C₂ .
     5. Suppose ra = ar³ and a² = e. Show that G ≅ D₄.
     6. Suppose ra = ar³ and a² = r². Show that G ≅ Q = {±1, ±i, ±j, ±k}, the quaternion group with i² = j² = k² = 1 and ij = k = −ji, jk = i = −kj, ki = j = −ik.
  ⊞ Soln  

  1. The cyclic subgroup of g contains all powers gⁱ including negative powers and g⁰ = e, so that ⟨g⟩ = {e, g, g², . . . , gⁿ⁻¹}, where the order n is the smallest positive number with gⁿ = e:

  n  =  |g|  =  #⟨g⟩  =  min{k > 0  with  g^k = e}.

  For infinite G, we could get an infinite cyclic group ⟨g⟩ with |g| = ∞, meaning g^k ≠ e for any k > 0.

  2. We have |g| = 1 exactly when g = g¹ = e. Any g ≠ e has |g| > 1, and G has 7 such elements, so the maximum |g| is N > 1.

  3. Lagrange's Theorem says that for any subgroup H ⊂ G, the number of elements |H| is a divisor of |G|. This is because G is split into distinct cosets gH, each with |H| elements; the number of cosets is denoted  [G : H]  =  ^|G|⁄_|H| .

  Now, a cyclic subgroup H = ⟨g⟩ has |H| = |g| elements, so the maximum order N = |g| is a divisor of |G| = 8, namely 1,2,4,8, but we know N > 1.

  4. If |g| = 8, then G contains ⟨g⟩ = {e, g, g², . . . , g⁷} ≅ C₈, but since G only has 8 elements, we have G = ⟨g⟩ ≅ C₈.

  5. We saw in HW 1/25 #5 that if g² = e for all g ∈ G, or equivalently g = g⁻¹, then G is abelian. This is because, for any a,b ∈ G, we have ab = (ab)⁻¹ = b⁻¹a⁻¹ = ba.

  Suppose this is the case for G, with N = 2. Take any e ≠ a ∈ G, and any b ≠ a,e. Since a² = b² = e and ba = ab, we have a subgroup ⟨a,b⟩ = {e, a, b, ab} with two cosets:  G = ⟨a,b⟩ ∪ c⟨a,b⟩ for any c ∉ ⟨a,b⟩. Then G = {e, a, b, ab, c, ac, bc, abc}, which is clearly isomorphic to:

  C₂ × C₂ × C₂  =  {(e,e,e), (r,e,e), (e,r,e), (r,r,e), (e,e,r), (r,e,r), (e,r,r), (r,r,r)},

  where r² = e ∈ C₂.

  6a. For r ∈ G with |r| = 4, the number of cosets of H = ⟨r⟩ = {e, r, r², r³} is [G : ⟨r⟩]  =  ^|G|⁄_|⟨r⟩|  =  ⁸⁄₄  =  2.

  We saw in HW 2/4 #3b that if H is any subgroup with [G : H] = 2, then H is normal. This is because the complement set G − H is equal to the coset gH for any g ∉ H, and also equal to Hg, so gH = Hg. Thus H = ⟨r⟩ is normal.

  Choosing any a ∉ ⟨r⟩, we have  G = {e, r, r², r³, a, ar, ar², ar³}.

  6b. We have the normal subgroup H = ⟨r⟩ with the quotient group G/H = {H, aH} ≅ C₂, so a²H = (aH)² = H. That is, that a²H = H, which means a² ∈ H = {e, r, r², r³}.

  If we had a² = r, this would mean a⁸ = r⁴ = e and |a| = 8, contradicting N = 4. (We can easily check aⁱ ≠ e for i = 1,2,3,4,5,6,7.) Thus a² = r is impossible. Similarly, a² = r³ = r⁻¹, with |r⁻¹| = 4, is also impossible. The remaining possiblilities are a² = e and a² = r².

  6c. Since ⟨r⟩ is normal by part (a), we have ⟨r⟩a = a⟨r⟩, so that ra ∈ {a, ar, ar², ar³}. Clearly ra ≠ a.

  If we had ra = ar², then rra = rar² = ar⁴ = a, but then canceling a would give r² = e, which is untrue since |r| = 4. Thus ra = ar² is impossible. The remaining possiblities are ra = ar or ar³

  6d. Define C₄ × C₂ = ⟨r,s⟩ with r⁴ = s² = e and rs = sr. In G, assume ar = ra, so that G is abelian. If we have a² = e, then we get an obvious isomorphism φ : C₄ × C₂ → G defined by φ(r) = r and φ(s) = a.

  The other possibility is a² = r², so that |a| = |r| = 4. Then we can define an isomorphism φ : C₄ × C₂ → G by φ(r) = r and φ(s) = ar. We check the crucial case of the structure equation:

  φ(s²)  =  φ(e) = e,         φ(s)²  =  (ar)²  =  a²r²  =  r²r²  =  e.

  In either case, G ≅ C₄ × C₂ .

  6e. If G = {e, r, r², r³, a, ar, ar², ar³} with r⁴ = a² = e and ra = ar³, then this is just our usual presentation of D₄, which means they have the same multiplication table:  G ≅ D₄

  6f. We defined the quaternion group as Q = {±1, ±i, ±j, ±k} with i² = j² = k² = 1 and ij = k = −ji, jk = i = −kj, ki = j = −ik. In G, assume that ra = ar³ and a² = r², along with r⁴ = e. Then we can define an isomorphism φ : Q → G by φ(i) = r, φ(j) = a, φ(k) = ar³ = ar⁻¹, φ(−1) = r². This obeys the structure equation because φ(i), φ(j), φ(k), φ(−1) in G obey all the defining relations among i, j, k, −1 in Q. For example, note that ra = ar⁻¹ implies r⁻¹a = ar, so that:

  φ(k)²  =  (ar⁻¹)²  =  ar⁻¹ar⁻¹  =  a²rr⁻¹  =  a²  =  r²  =  φ(−1)  =  φ(k²).

  Also note that −1 commutes with all elements of Q, and similarly φ(−1) = r² commutes with all elements of G, since r²a = rar³ = ar⁶ = ar². In a general group, we say an element c ∈ G is central if it commutes with all other elements, that is cg = gc for all g ∈ G. The set of all central elements is always a normal subgroup, denoted Z(G). Under an isomorphism, a central element always goes to a central element; here Z(Q) = {1, −1}, and the unknown 8-element group G has Z(G) = {e, r²}.

• 3/11 Ch 9.1 Product groups, conjugacy, disjoint normal subgroups lemma   ⊞ Notes  
  • Reading: Ch 9.1, pp. 281-285.
  • Recall that for a subgroup H ⊂ G and an element g ∈ G, we have the left coset gH = {gh for h ∈ H} and the right coset Hg = {hg for h ∈ H}. We have gH = g'H whenever g' ∈ gH, and we get the coset decomposition:
    G  =  H ∪ g₁H ∪ ··· ∪ g_n−1H,
    where the number of distinct cosets is called the index, denoted n = [G : H].
  • Normal subgroups. For normal subgroups N ⊂ G only, we can form the quotient group G/N whose elements are cosets gN ∈ G/N.
    The subgroup is normal whenever the left cosets are equal to the right cosets: gN = Ng for all g ∈ G.
  • The conjugate of a subgroup H is the set gHg⁻¹ = {ghg⁻¹ for h ∈ H}. This is also a subgroup.
    A group N is normal if and only if it is equal to all its conjugates: gNg⁻¹ = N for all g ∈ G.
  ⊞ HW  
  1. Prove the following facts about the conjugates gHg⁻¹ of a subgroup H ⊂ G.
     1. gHg⁻¹ is always a subgroup.
     2. H is normal if and only if it gHg⁻¹ = H for all g ∈ G.
  2. Conjugate subgroups. List all the subgroups of the following groups, and identify which ones are conjugate to each other. That is, collect them into classes of conjugate subgroups.
     1. D₃ = {e, r, r², a, ar, ar²}
     2. S₃ = {(1), (12), (23), (13), (123), (132)} in cycle notation.
        Extra Credit: For general permutations g, h ∈ S_n, explain how the cycle notation for g relates to the cycle notation of the conjugate element ghg⁻¹.
  3. Prove: For subgroups H, K ⊂ G, define the set HK = {hk for h ∈ H and k ∈ K}. Then HK is a subgroup if and only if HK = KH. Hint: Prove separately the (⇒) and (⇐) parts of the "if and only if".
  4. Proposition: Suppose we have subgroups H,K ⊂ G satisfying:
     • H and K are normal
     • H ∩ K = {e}
     • |G| = |H| |K|
     Then the group is isomorphic to the direct product of its subgroups:  G ≅ H × K.
     The following exercises lead to a proof of the Proposition.
     1. First, a counterexample: G = D₃, H = {e, r, r²}, K = {e, a}, but G is not isomorphic to H × K. (Why is that obvious?) Nevertheless, the Proposition is true, because this case violates one of the hypotheses. Which one?
     2. Given the hypotheses of the Proposition, prove that hk = kh for all h ∈ H, k ∈ K. Hint: Apply the hypotheses to show hkh⁻¹k⁻¹ = e.
     3. Define the mapping φ : H × K → G by φ(h,k) = hk. Show that φ is a homomorphism.
     4. Show that φ is one-to-one and onto, and thus an isomorphism.
  ⊞ Soln  

  1a. The subset gHg⁻¹ is a subgroup, since it is closed under multiplication and inverses:

  ghg⁻¹·gh'g⁻¹  =  gh(g⁻¹g)h'g⁻¹  =  ghh'g⁻¹  ∈  gHg⁻¹,
  
  (ghg⁻¹)⁻¹  = (g⁻¹)⁻¹h⁻¹g⁻¹  =  gh⁻¹g⁻¹  ∈  gHg⁻¹.

  1b. H is normal  ⇔  gH = Hg  ∀g∈G  ⇔  (gH)g⁻¹ = (Hg)g⁻¹  ∀g∈G  ⇔  gHg⁻¹ = H  ∀g∈G.

  2a. In G = D₃, the trivial subgroups {e} and G are normal, equal to all their conjugates. We also have the normal subgroup N = {e,r,r²}, since any subgroup of index [G : N] = 2 is normal. Also, we have the conjugate subgroups:

  H = {e,a},   rHr⁻¹ = {e, ar},   r²Hr⁻² = {e, ar²}.

  2b. Because S₃ ≅ D₃, this is just a different notation for the previous problem. The normal subgroup is N = {(1), (123), (132)}, and the conjugate subgroups are:

  H = {(1), (12)},   (123)H(123)⁻¹ = {(1), (23)},   (132)H(132)⁻¹ = {(1), (13)}.

  3. Prop: For subgroups H,K ⊂ G, we have HK a subgroup if and only if HK = KH.

  Pf: (⇒)  Suppose HK is a subgroup. Then k = ek and h = he are both elements of the subgroup HK, which is closed under multiplication, so k·h ∈ HK. Thus KH ⊂ HK, and similarly HK ⊂ KH, so HK = KH.

  (⇐)  Conversely, suppose HK = KH. Then any kh ∈ KH is also in HK, and can be written as kh = h'k', where h,h' ∈ H and k,k' ∈ K. Thus:

  h₁k₁ h₂k₂  =  h₁(k₁h₂)k₂  =  h₁(h'k')k₂  =  h₁h' k'k₂  ∈  HK.

  Also:

  (hk)⁻¹  =  k⁻¹h⁻¹  ∈  KH  =  HK.

  Thus HK is closed under multiplication and inverses, and is a subgroup.

  4a. Since D₃ is not commutative, it is clearly not isomorphic to H × K = C₃ × C₂. We have H ∩ K = {e} and |G| = |H| |K| = 6. Also H is a normal subgroup, but K is not, as we can see from its conjugate subgroups in #2a.

  4b. Assume that H,K are normal subgroups with H ∩ K = {e}. Since K is closed, we have KK = K, and since it is normal, we have gKg⁻¹ = K. Now consider the element:

  hkh⁻¹k⁻¹  =  (hkh⁻¹) k⁻¹  ∈  hKh⁻¹ K  =  KK  =  K

  But similarly this element hkh⁻¹k⁻¹ = h (kh⁻¹k⁻¹) ∈ H, and hence this element lies in H ∩ K = {e}. Thus hkh⁻¹k⁻¹ = e, which means hk = kh.

  4c. The mapping φ : H × K → G defined by φ(h,k) = hk satisfies:

  φ((h,k)·(h',k'))  =  φ((hh',kk'))  =  hh'kk'   and   φ(h,k)·φ(h',k')  =  hkh'k'  =  hh'kk'

  since kh' = h'k by the previous problem. Thus φ is a homomorphism.

  4d. To show φ is one-to-one, suppose φ(h,k) = φ(h',k'). Then hk = h'k' and (h')⁻¹h = k'k⁻¹. The left side of the last equation lies in H, while the right side lies in K, so both sides must lie in H ∩ K = {e}. That is, (h')⁻¹h = k'k⁻¹ = e, so h = h' and k = k'. This means φ is one-to-one.

  Now, assuming |H| |K| = |G|, this means φ is a one-to-one (injective) mapping between two sets with the same number of elements: thus it must also be onto (surjective). Hence φ is a bijective homomorphism, i.e. an isomorphism.

• 3/13 Ch 9.3 Sylow Theorems   ⊞ Notes  
  • Reading: Ch 9.3, pp. 298-302.
  • A p-subgroup is a subgroup H ⊂ G whose order is a prime power: |H| = p^k for p a prime number and k ≥ 1.
  • A Sylow p-subgroup is a p-subgroup whose order is the largest power of p dividing |G| = n: that is, |H| = p^m with p^m | n but p^m+1 ∤ n.
  • Sylow Theorems
    1. If a prime power p^k divides |G|, then G has a p-subgroup of order p^k.
    2. All Sylow p-subgroups are conjugate subgroups: that is, if H, H' ⊂ G are Sylow p-subgroups, then H' = gHg⁻¹ for some g ∈ G.
    3. If s is the number of Sylow p-subgroups, then divides the order |G| = n, and s is equal to 1 modulo p: that is, s | n and s = 1 + kp for k ≥ 0.
  • Classification of pq-groups: Let G be a group of order n = pq, for primes p > q with q ∤ p−1. Then G is cyclic:  G ≅ C_pq ≅ C_p × C_q .
    Proof: For concreteness, we prove the case n = 15, p = 5, q = 3, which satisfy 3 ∤ 5−1.
    • Assuming the Sylow Theorems, let s be the number of Sylow 5-subgroups. By Sylow III, we must have s | |G| = 15, so that s = 1,3,5, or 15. Also, s ≡ 1 mod 5, so s = 1,6,11, 16, . . . , and the only intersection of these possibilities is s = 1. That is, there is a unique Sylow 5-subgroup H, whose conjugates gHg⁻¹, also Sylow 5-subgroups, must all be equal to H. That is, H is a 5-element normal subgroup.
    • Similarly, let t be the number of Sylow 3-subgroups. Then t = 1, 3, 5, 15 and t = 3k+1 = 1, 4, 7, 10, 13, . . . , so the only possibility is t = 1. (This is where we use q ∤ p−1, so that t = qk+1 ≠ p.) As before, we get a normal subgroup K of order 3.
    • The intersection H ∩ K is a subgroup of H and of K, so its order divides |H| = 5 and |K| = 3, so it must be 1; thus H ∩ K = {e}.
    • Now we have all the conditions to apply the Proposition from #4 in the previous HW: that is, H,K are normal subgroups whose intersection is {e}, and |G| = |H| |K|. We conclude that G ≅ H × K. Since H,K have prime order, they are both cyclic, and we have G ≅ C₅ × C_3.
  ⊞ HW  
  1. Name some p-groups we have seen among our standard group examples, including abelian and non-abelian ones: that is, G with |G| = p^k.
  2. Sylow examples
     1. Identify all the Sylow p-subgroups of D₃, C₁₂, and D₆ for each possible p.
     2. Verify the three Sylow Theorems by direct inspection for each of the above cases.
  3. Find the Sylow 2-subgroups for S₄.
  4. Using all our classification theorems, for which values of |G| ≤ 20 can we say that G must be a product of cyclic groups? For which do we know non-abelian examples? For which are we unsure?
  5. Proposition: If |G| = 2p for a prime p > 2, then either:
     • G ≅ C₂ × C_p ≅ C_2p, or
     • G ≅ D_2p, the dihedral group generated by elements x,y satisfying x^p = y² = e and xy = yx⁻¹.
     Prove this using Sylow's Theorems. Hint: Show that there is a normal p-subgroup ⟨x⟩, and examine yxy⁻¹.
  6. Extra Credit: Consider the group G = GL_n(F) of invertible n×n matrices A = (a_ij) with entries a_ij in the p-element finite field F = Z_p: this is the symmetry group of the vector space Fⁿ. (This appeared in HW 1/18.) Let H be the subgroup of upper triangular matrices with 1's on the diagonal: that is a_ij = 0 for i > j, and a_ii = 1. Show that H is a Sylow p-subgroup of G.
  ⊞ Soln  

  1. Abelian p-groups: any products of cyclic groups C_p^k's, such as C₄ × C₂.

  Non-abelian p-groups: dihedral groups D_2^k such as D₈; quaternion group Q; and any product of an abelian p-group with a non-abelian one. Note: A symmetric group S_n is never a p-group, except for n = 2.

  2. Consider G = D₃ = ⟨r,a⟩ with r³ = a² = e and ra = ar⁻¹. The unique Sylow 3-subgroup ⟨r⟩ is normal. We have the three conjugate Sylow 2-subgroups ⟨a⟩, ⟨ar⟩ = r⟨a⟩r⁻¹, ⟨ar²⟩ = r²⟨a⟩r⁻², verifying Sylow Theorem II. Also, 3 divides |G| = 6 and 3 ≡ 1 mod 2, verifying Sylow Theorem III.

  For G = C₁₂ = ⟨r⟩ with r¹² = e, the unique Sylow 3-subgroup is ⟨r⁴⟩, and the unique Sylow 2-subgroup is ⟨r³⟩. We also have the 2-subgroup ⟨r⁶⟩ of order 2, as guaranteed by Sylow Theorem I.

  Consider G = D₆ = ⟨r,a⟩ with r⁶ = a² = e and ra = ar⁻¹; also the central element r³ with r³a = ar³. The unique Sylow 3-subgroup ⟨r²⟩ is normal. The Sylow 2-subgroups have order 2² = 4; there are three such subgroups, all conjugate:

  {e, r³, a, ar³}     {e, r³, ar, ar⁴}     {e, r³, ar², ar⁵}

  and indeed 3 divides |G| = 12, and 3 ≡ 1 mod 2. We also 2-subgroups of order 2, {e, ar^k} for any k = 0, . . . , 5 are all conjugate to each other, but not to {e, r³}. (Only the Sylow p-subgroups are guaranteed to be conjugate to each other by Sylow Theorem II.)

  3. G = S₄ has order |G| = 4! = 24 = 3·8. The Sylow 3-subgroups have order 3, which must be cyclic, generated by the 3-cycles. By Sylow III, the number s of these must divide 24, and s ≡ 1 mod 3, so it must be 1 or 4. In fact, we have:

  {(1), (123), (132)}     {(1), (124), (142)}     {(1), (134), (143)}     {(1), (234), (243)}.

  The Sylow 2-subgroups must have order 8, all conjugate to each other by Sylow II: this means the cycle structure of the elements is the same for each subgroup. By Sylow III, the number of these is an odd factor of 24, so it must be 1 or 3.

  There are two types of 4-subgroups, the three cyclic ones such as ⟨(1234)⟩ = {(1), (1234), (13)(24), (4321)}, all conjugate to each other; and the normal subgroup {(1), (12)(34), (13)(24), (14)(23)}. I turns out that adding one 2-cycle to a 4-cycle produces a subgroup of order 8:

  ⟨(13), (1234)⟩  =  {(1), (13), (24), (13)(24), (1234), (4321)},

  and similarly ⟨(14), (1243)⟩ and ⟨(12), (1324)⟩.

  4. Consider |G| = n ≤ 20. We know G must be abelian for n = 1,2,3,4. For even n = 2m ≥ 6, we always have the non-abelian dihedral group G = D_m. For odd n = p, or pq with p > q and q ∤ p− 1, our classification theorems guarantee G must be cyclic: that is, n = 5,7,11,13,15,17,19. The only uncertain case is n = 9, but we will show later that G is abelian for n = p² for prime p > 2.

  5. Proposition: If |G| = 2p for a prime p > 2, then either:

  • G ≅ C₂ × C_p ≅ C_2p, or
  • G ≅ D_2p= ⟨x,y⟩ satisfying x^p = y² = e and xy = yx⁻¹.

  Proof: By Sylow I there exists a Sylow p-subgroup H ⊂ G. Since |H| = p is prime, it must be a cyclic group: H = ⟨x⟩ with x^p = 1. Let s be the number of Sylow p-subgroups. By Sylow III, s divides 2p, so s = 1, 2, p, 2p, but also s = 1, p+1, 2p+1, . . . , which can only mean s = 1. Thus all the conugates of H are the same, and H is normal.

  Now take a Sylow 2-subgroup {e, y} with y² = e. Since H = ⟨x⟩ is normal, we have yxy⁻¹ = x^k for some k = 0, 1, . . . , p−1. We thus get:

  x  =  yyxy⁻¹y⁻¹  =  (x^k)^k  =  x^k2,

  and k² ≡ 1 mod p. In the field Z_p, the equation k²−1 = (k−1)(k+1) = 0 has only the roots k = ±1. If k = 1, then yx = xy, and xⁱy^j ↦ (i,j) gives an isomorphism from G to C_p × C₂. If k = −1, then yx = xy⁻¹ defines the multiplication of D_p.
• 3/15−22 Ch 9.4 Group actions, proof of Sylow I,II,III  ⊞ Notes  
  • Reading: Ch 9.4 pp. 304-310.
  • ⊞ Group actions: definitions  
    • An action of a group G on a set of objects 𝒮 = {objects S} is a way of making a group element g move each object S to another object S' = g∗S.
    • An action must be compatible with group multiplication, satisfying the identity and associative axioms: e∗S = S and g*(g'*S) = (gg')*S.
    • The G-orbit of S ∈ 𝒮 means the set of objects to which S is moved: G∗S = {g∗S for g ∈ G}.
    • The stabilizer of an object S is the set of group elements which move S to itself:
      Stab(S)  =  Stab_G(S)  =  {g ∈ G with g∗S = S}.

    • Orbit Size: The number of objects in an orbit is the size of the group divided by the size of the stabilizer:
      |G∗S|   =   ^|G|⁄_|Stab(S)| .

    • Example: Decorated shapes. Consider the symmetries of a hexagon X, the dihedral group
      G  =  D₆  =  Sym(X)  =  {e, r, r², r³, r⁴, r⁵, a, ar, ar², ar³, ar⁴, ar⁵}
      with   r⁶ = a² = e and ra = ar⁻¹. Also note r³a = ar³.
      Let 𝒮 = {X₁, . . . , X₁₅} be the set of hexagons decorated with two colored edges. Make each g ∈ G act by moving X, and thereby moving the decorated edges of X_i to some other g∗X_i = gX_i = X_j. We can picture this natural symmetry action by drawing arrows labeled with the generators r (clockwise rotation) and a (side-to-side reflection):
      
      The three G-orbits correspond to the adjacency types of edge pairs: adjacent and almost-adjacent edge pairs like X₁ and X₇ have |Stab(X_i)| = 2 and |G·X_i| = ¹²⁄₂ = 6; while the opposite edge pairs like X₁₃ have |Stab(X_i)| = 4 and |G·X_i| = ¹²⁄₄ = 3.
    • Example: Conjugation of subgroups. Taking the symmetry groups H = Sym(X_i), recall that Sym(gX_i) = g Sym(X_i) g⁻¹. This naturally defines the conjugation action of G = D₆ on the set of subgroups:
      g∗H  =  gHg⁻¹.
      We can picture this conjugation action as:
      
      The top orbits have only 3 objects because some X_i's have the same symmetry groups. Note that the bottom orbit contains the Sylow 2-subgroups: we might expect this because they are symmetry groups of the hexagon with p = 2 marked edges in maximally symmetric position.
  • ⊞ Conjugation action on G, classify groups of order p²  
    • Conjugation action on elements: Let any group G act on itself by conugation: 𝒮 = G with the action g∗x = gxg⁻¹ for x ∈ G. In this case, the orbits are called conjugacy classes, denoted G∗h = C(h).
    • The orbits of size one, G∗x = C(x) = {x} are precisely those with G = Stab(x), so that gxg⁻¹ = x for all g ∈ G; that is, gx = xg, meaning x is a central element of G, in the center subgroup:
      Z(G) = {x ∈ G with xg = gx for all g ∈ G}.
      That is,  |C(x)| = 1  ⇔  x ∈ Z(G).
    • Class formula: Let C(y₁), . . . , C(y_ℓ) be the distinct conjugacy classes of size |C(y_i)| > 1. The group splits into conjugacy classes as:
      G  =  ⋃_x∈Z(G){x}  ∪  C(y₁) ∪ ··· ∪ C(y_ℓ).
      Counting the size of each set gives:
      |G|   =   |Z(G)|  +  ^|G|⁄_Stab(y1)  +  ···  +  ^|G|⁄_Stab(yℓ),
      where Stab(y_i) ≠ |G|.
    • Theorem: Let G be a group of order |G| = p², for p a prime number. Then G is abelian, and G ≅ C_p² or G ≅ C_p × C_p.
      Proof: First, I claim that |Z(G)| > 1. Indeed, if we had |Z(G)| = 1, the class equation would give
      p² = 1 + ^|G|⁄_Stab(y1)  +  ···  +  ^|G|⁄_Stab(yℓ).
      Now, each Stab(y_i) is a subgroup, with size 1 or p, so the terms on the right are ^|G|⁄_Stab(yℓ) = p or p². This would mean p² = 1 + kp, which is impossible, so we must have |Z(G)| > 1.

      Therefore we may take a non-identity central element c ∈ Z(G). This generates a cyclic subgroup ⟨c⟩ whose order divides p². If |c| = p², we have G = ⟨c⟩ ≅ C_p².

      Otherwise, |c| = p, and we can take the quotient group of order |G/⟨c⟩| = p. This must be cyclic, generated by a coset [a] = a⟨c⟩ with [a]^p = [e] = ⟨c⟩, meaning a^p = cⁱ for some i = 0, 1, . . . , p−1. If i > 0, I claim |a| = p². Indeed, suppose a^j+pk = a^jc^ik = e for 0 ≤ j, k ≤ p−1. Since [a]^j = [e] only for p | j, we must have j = 0. We are left with c^ik = e, which means p | ik, so that p | k and k = 0, since 1 ≤ i ≤ p−1. Thus, a^m ≠ e for 0 < m = j+pk < p².

      Finally if i = 0, we have a^p = c^p = e with ac = ca, and we may define an isomorphism φ : C_p × C_p → G by φ(rⁱ, r^j) = aⁱc^j.

  • ⊞ Multiplication action on subsets, Sylow Theorem I  
    • Let G be a group of order |G| = p^em with p ∤ m. Let G act on the class of all subsets S of order p^e by left multiplication. That is:
      𝒮  =  {S ⊂ G  with  |S| = p^e}   with action   g·S = gS = {gs for s ∈ S}.

    • Lemma: p does not divide the number of subsets in 𝒮: that is, p ∤ |𝒮|.
      Proof: Recall that the binomial coefficient  (

      n

      k

      ) = ^n!⁄_k!(n−k)!  counts the number of subsets of {1, . . . ,n} of size k. We may show that p does not divide  |𝒮| = (

      pem

      pe

      )  by cancelling all powers of p in the numerator by corresponding powers in the denominator.
    • Sylow Theorem I: G has a Sylow p-subgroup, that is, there is some subgroup H ⊂ G of order |H| = p^e.
      Proof: We will find such a subgroup as the stabilizer Stab(S) for some S ∈ 𝒮. Since the size of 𝒮 is not divisible by p, and 𝒮 splits into orbits, there must be at least one orbit G·S whose size is not divisible by p:
      p  ∤  |G·S|  =  ^|G|⁄_|Stab(S)|  =  ^pem⁄_H,
      where we denote the stabilizer subgroup as H = Stab(S) ⊂ G. This means |H| cancels all the p-factors in the numerator, so p^e must divide |H|.

      Now, by definition, we have H·S = S, so we have a new action of the group H on the p^e elements g ∈ S. Thus S splits into H-orbits, which are actually the right cosets H·g = Hg of g ∈ S. Since the size of each coset is |Hg| = |H|, this means |H| divides |S| = p^e. Combining with the previous conclusion, we must have |H| = p^e, and we have found a Sylow p-subgroup.

  • ⊞ Multiplication on Sylow cosets, Sylow Theorem II  
    • Again take a group of order |G| = p^em, with a Sylow p-subgroup H ⊂ G of order |H| = p^e, which exists by Sylow I. Let G act on ℒ = G/H = {left cosets xH for x ∈ G} by multplication:  g·xH = gxH. Note that:
      g ∈ Stab_G(xH)   ⇔   gxH = xH   ⇔   x⁻¹gx ∈ H   ⇔   g ∈ xHx⁻¹.
      Thus Stab_G(xH) = xHx⁻¹. The size of ℒ is |ℒ|  =  ^|G|⁄_|H|  =  m, so there must be some orbit ℳ = G·yH whose size is not divisible by p.
    • Now let K ⊂ G be a subgroup, and define an action of K multiplying this G-orbit ℳ = G·yH, so that k·gyH = kgyH for k ∈ K. Since the size of ℳ is not divisible by p, there must be some K-orbit K·gyH whose size is not divisible by p.
    • Sylow Theorem II: Let H ⊂ G be a Sylow p-subgroup, so that |H| = p^e, and let K ⊂ G be a p-subgroup, meaning |K| = p^d. Then K lies inside a conjugate Sylow p-subgroup gHg⁻¹ for some g ∈ G. In particular, if K is another Sylow p-subgroup, then  K = gHg⁻¹, a conjugate subgroup of H.
      Proof: Above, we found a K-orbit K·gyH whose size is not divisible by p: that is,
      |K·gyH|  =  ^|K|⁄_|StabK(gyH)|  =  ^pd⁄_|StabK(gyH)|
      is not divisible by p, which must mean that the orbit is a single point, K·gyH = {gyH}, and Stab_K(gyH) = K. We conclude:
      K  =  Stab_K(gyH)  ⊂  Stab_G(gyH)  =  gyH(gy)⁻¹,
      so that K is indeed inside a conjugate of H.
  • ⊞ Conjugation action on Sylow subgroups, Sylow Theorem III  
    • Let G act on ℋ = {all Sylow p-subgroups H ⊂ G} by conjugation: g∗H = gHg⁻¹, and let s = |ℋ| be the number of Sylow p-subgroups. By Sylow II, all H's are conjugate to each other, so this is a single orbit:  ℋ = G∗H. Hence  s = |ℋ| = |G∗H| = ^|G|⁄_|Stab(H)|  is a divisor of |G|.
    • We have Stab_G(H) = {g ∈ G with gHg⁻¹ = H}, which is called the normalizer of H, denoted N_G(H); clearly H ⊂ N_G(H). By definition, this is the largest subgroup of G such that H is a normal subgroup of N(G).
    • Now consider a particular Sylow p-subgroup K ⊂ G, so that |K| = p^e, and restrict the conjugation action of G on ℋ to an action of K on ℋ; that is, we conjugate the Sylow subgroups H ∈ ℋ by k ∈ K. We again have the normalizer N_K(H) = Stab_K(H), and the size of the orbit K∗H is  ^|K|⁄_|NK(H)|  =  ^pe⁄_|NK(H)|.

      For H = K, we clearly have N_K(H) = K, so that |K∗K| = 1. However, suppose we had some other H ∈ ℋ with N_K(H) = K. Then H and K are both Sylow p-subgroups of N = N_G(H), so by Sylow II, they are conjugate: K = nHn⁻¹ for n ∈ N. However, H is normal in N, so K = nHn⁻¹ = H. This shows the orbit sizes:

      |K*H|  =

      {

      1   for H = K |K|⁄|NK(H)| = pj   with j ≥ 1 for H ≠ K.

      Now the K-orbit decomposition gives:

      s  =  |ℋ|  =  1 + p^j₁ + p^j₂ + ···  =  1 + kp   for j_i ≥ 1 and k ≥ 0.
      This proves Sylow Theorem III.
  ⊞ HW  
  In this HW, work out the above group actions in the case of the dihedral group, the symmetries of the triangle:
  G = D₃ = {e, r, r², a, ar, ar²}  with  r³ = a² = e and ra = ar⁻¹.
  This has order |G| = 2·3; we will concentrate on p = 2.
  1. Conjugation action of G on itself, 𝒮 = G.
     1. Draw the diagram of the conjugation action. That is, for x ∈ 𝒮 = G, draw arrows corresponding to the generators:
        x

        r

        →

        r∗x = rxr⁻¹   and   x

        a

        →

        a∗x = axa⁻¹.
        Circle the G-orbits, which are conjugacy classes: G∗x = C(x).
     2. Explicitly write the class equation |G| = |Z(G)| + ^|G|⁄_|Stab(x1)| + ···  in this case.
  2. Multiplication action on subsets. Let 𝒮 = {all subsets S ⊂ G with |S| = 2}, with the action on S = {u,v} ⊂ G given by g·S = gS = {gu, gv}.
     1. Compute the binomial coefficient (

        6

        2

        ) = ^6!⁄_2!4!, and show that it is not divisible by p = 2.
     2. List all the 2-element subsets S = {u,v} ⊂ G. and draw arrows for the G-action of r and a.
     3. Find the orbits not divisible by p = 2, and show that their stabilizers are Sylow 2-subgroups Stab(S) = H ⊂ G with |H| = 2. This illustrates our proof of Sylow Theorem I.
  3. Conjugation action on Sylow subgroups.
     1. Draw the diagram of G acting by conjugation on ℋ, the set of all Sylow 2-subgroups. Observe that this action has only one orbit, which is Sylow Theorem II. Explain why this means s = |ℋ|, the number of Sylow 2-subgroups, is a divisor of |G|.
     2. Choose the Sylow 2-subgroup K = {e,a}, and restrict the G-action to a K-action: this just means erasing the r-arrows from the above diagram.
     3. Show that one K-orbit has size 1, and the rest have order divisible by p = 2. This guarantees the Sylow Theorem III formula s = |ℋ| = 1 + pk.
  ⊞ Soln  

  1. The conjugacy classes (orbits of the conjugation action) are {e}, {r,r²}, {a,ar,ar²}. The only orbit of size 1 is {e}, so the center is Z(G) = {e}. The class equation thus reads: |G| = |Z(G)| + |C(r)| + |C(a)|, i.e. 6 = 1 + 2 + 3.

  2. There are 15 sets S ∈ 𝒮: {e,r}, {e,r²}, . . . , {ar,ar²}. The orbits are: G·{e,r} (6 elements), G·{e,a} (3 elements, stabilizer is Sylow 2-subgroup {e,a}); G·{e,ar} (3 elements, stabilizer is Sylow {e,ar}); G·{e,ar²} (3 elements, stabilizer is Sylow {e,ar²}).

  3a. G = D₃ acts by conjugation on its Sylow 2-subgroups as follows, with gnerator r indicated by black arrows, generator a by yellow arrows:

  

  All of ℋ is the orbit of a single K = {e,a}, so s = |ℋ| = |G∗K| = ^|G|⁄_|Stab(K)|, a divisor of |G| = 6. Since K ⊂ Stab(K), so that 2 = p^e divides|Stab(K)|, and s is not divisible by p = 2.

  3b,c. The yellow a-arrows give conjugacy orbits of K = {e,a}. Since the total number of elements s is not divisible by p = 2, there must be some odd-size K-orbit. The proof of Sylow III shows that only K∗K has size 1, with the other K-orbits divisible by p = 2, so that the number of orbits splits as s = 3 = 1 + 2·1 = 1 + pk.

• 3/27 Ch 5 Congruence-class arithmetic in F[x]  ⊞ Notes  
  • Reading: Ch 5, pp. 125-138.
  • An algebraic number α is a real or complex number which is the root of an equation with rational coefficients: that is, f(α) = 0 for some polynomial:
    f(x)  =  a₀ + a₁x + ··· + a_n−1xⁿ⁻¹ + a_nxⁿ  ∈  Q[x],
    where Q[x] is the ring of polynomial functions with rational-number coefficients a₀, a₁, . . . , a_n ∈ Q.
  • Examples
    • A rational number like α = ³⁄₅ is algebraic, since it is the root of f(x) = x − ³⁄₅ = 0. In fact, we can clear denominators to get integer coefficients:  5x − 3 = 0.
    • √2 is irrational but algebraic, the root of x² − 2 = 0.
    • Any rational numbers combined using +,−,×,÷ and root operations ⁿ√ give an algebraic number α: we say α can be written in radicals. For example, α = √2 + √3 is a root of the polynomial:
      f(x)  =  (x−√2−√3)(x−√2+√3)(x−+2−√3)(x+√2+√3)  =  x⁴ − 10x² + 1.

    • The complex square root of minus one, i = √(−1), is algebraic, since it is the root of x² + 1 = 0.
    • Not every algebraic number can be written in radicals: for example, the root of a higher-degree polynomial like x⁵ − x + 1 cannot be so written.
  • We compute with general algebraic numbers using abstract algebra as we do with simple algebraic numbers like √2 using high school algebra. To this end, we define the ring of polynomials in α,
    Q[α]  =  {g(α) for g(x) ∈ Q[x]}
    In fact, this is a field: every polynomial expression in α has a reciprocal which is a polynomial in α (no denominator).
  • Assuming f(α) = 0 for f(x) an irreducible polynomial (i.e. no non-trivial factorizations), we can model the ring generated by α as:
    Q[α]  ≅  ^Q[x]⁄_(f(x)) ,
    the quotient of the R = Q[x] by the ideal I = (f(x)) = {q(x)f(x) for q(x) ∈ Q[x]}. The elments of the quotient ring are cosets
    [a(x)]  =  a(x) + I  =  {a(x) + q(x)f(x) for q(x) ∈ Q[x]},
    so that [f(x)] = [0]. Writing the class of a constant [c] as just c, we get f([x]) = 0, and [x] is a solution of the original polynomial!
  • In fact, this ring is a field, with every class [a(x)] having a reciprocal [b(x)] with  [a(x)][b(x)] = [1], i.e.  a(x)b(x) ≡ 1 mod f(x), i.e.  a(x)b(x) = 1 + q(x)f(x)  for some polnomial q(x).
  • Example: Let α = √2, f(x) = x²−2, an irreducible polynomial in Q[x]. Then:
    Q[α] = {a + bα for a,b ∈ Q}  ≅  ^Q[x]⁄_(x²−2) .
    In fact, since α² = 2, α³ = 2α, etc., any polynomial expression g(α) can be written in the standard form above, a + bα. Two standard forms are multiplied as:
    (a+bα)(c+dα)  =  ac + (ad+bc)α + bdα²  =  (ac+2bd) + (ad+bd)α
    and divided by rationalizing the denominator:
    ^(a+bα)⁄_(c+dα)   =   ^(a+bα)⁄_(c+dα)   ^(c−dα)⁄_(c−dα)   =   ^(ac−2bd)⁄_(c²−2d²)  −  ^(bc−ad)⁄_(c²−2d²) α

  ⊞ HW  
  Consider the unique real solution x = α to the equation f(x) = x³ + x − 1 = 0. All our computations with α should be a consequence of the defining formula  α³ + α − 1 = 0.

  1. Using software, graph  y = x³ + x − 1  and show it has a unique real root x = α. Approximate α (perhaps with Newton's Method).
  2. Show that any polynomial g(α) for g(x) ∈ Q[x] can be reduced to the standard form a + bα + cα². Thus, Q[α] is a 3-dimensional vector space over Q with basis {1, α, α²}. Hint: Use polynomial long division to write g(x) = q(x)f(x) + r(x).
  3. Using α³ = 1 − α, write α⁴ in standard form. Then write a general product (a + bα + cα²)(d + eα + fα²) in standard form.
  4. Use long division of f(x) = x³ + x − 1 by bx + a to write q(x)(a+bx) + g(x)f(x) = 1 for some q(x) and g(x) depending on a,b. Use this to write ¹⁄_(a+bα) in standard form.
     Extra Credit: Write  ¹⁄_{(a+bα+cα²)}  in standard form. This shows Q[α] is a field.
  ⊞ Soln  

  1. Wolfram shows the unique real solution to  f(x) = x³ + x − 1 = 0  to be α ≈ 0.682. Since f '(x) = 3x² + 1 > 0 for all x, the function is increasing, and y = f(x) can only have one x-intercept.

  2. The long division of polynomials shows that any polynomial g(x) ∈ Q[x] can be divided by f(x) = x³ + x − 1 to give  g(x) = q(x)f(x) + r(x)  where the remainder has small degree:  deg r(x) < deg f(x) = 3; that is, r(x) = a + bx + cx². This allows us to write g(α) in standard form:

  g(α)  =  q(α)f(α) + r(α)  =  0 + a + bα + cα².

  3. One way to do this is to divide x⁴ by f(x), getting:  x⁴  =  x f(x) + x − x², so that α⁴  =  0 + α − α².

  4. Long division (using variable coefficients instead of numbers) gives:

  x³ + x − 1   =   (a+bx) (^x2⁄_b − ^ax⁄_b² + ^(a2+b2)⁄_b³)  −  ^(a3+ab2+b3)⁄_b³

  Dividing by the remainder and rearranging, we get:

  (a+bx) ^{(b2x2−abx+a2+b2)}⁄_{(a³+ab²+b³)}  −  ^{b3(x3+ x−1)}⁄_{(a³+ab²+b³)}   =   1

  Substituting x = α, the second term vanishes, and we have:

  (a+bα) ^{(b2α2 − abα + a2+b2)}⁄_{(a³ + ab² + b³)}   =   1

  This gives the reciprocal in standard form:

  ¹⁄_(a+bα)   =   ^(a2+b2)⁄_{(a³+ab²+b³)}  −  ^ab⁄_{(a³+ab²+b³)} α  +  ^b2⁄_{(a³+ab²+b³)} α² .

• 3/29 Ch 11.2 Simple extensions   ⊞ Notes  
  • Reading: Ch 11.2, pp. 376-381.
  • Recall that a field F is a number system in which we can perform +, −, ×, ÷ according to the usual rules of algebra. In terms of ring axioms, F is a commutative ring in which every nonzero element is invertible:  a ≠ 0 has a reciprocal a⁻¹ = ¹⁄_a ∈ F. In terms of group axioms:
    • (F, +) is a commutative group (5 axioms).
    • (F^×, ×) is a commutative group, where F^× = F−{0} (5 axioms).
    • Distributive axiom:  a(b+c) = ab + ac
  • We will usually fix a base field F, most often F = Q, which will be a subfield of all other fields we work with. Such a field K containing F is an extension field of F, and the pair F ⊂ K is an extension of fields.
  • We will use the polynomial ring:
    F[x]  =  { f(x) = a₀+a₁x+···+a_nxⁿ  for  n ≥ 0  and  a_i ∈ F }.
    The highest non-zero term a_nxⁿ defines the degree deg f(x) = n. In particular, deg f(x) = 0 means f(x) = a₀, a constant function.
  • We say α ∈ K is algebraic over the base field F if α satisfies (is a root of) a polynomial with coefficients in F: that is, f(α) = 0 for f(x) ∈ F[x].
  • A number which is not algebraic is transcendental. The familiar constants π ≈ 3.14 and e ≈ 2.71 are transcendental over F = Q. That is, π and e are not roots of any polynomial with rational (including integer) coefficients, so that f(π) and f(e) are never zero. Note that √2 is irrational, but algebraic, not transcendental.
  • Given an element of an extension field α ∈ K, we can form the smallest field containing both the base F and α:
    F(α)  =  { ^f(α)⁄_g(α)  for  f(x), g(x) ∈ F[x]  with  g(α) ≠ 0 }.
    We call this field the simple extension of F generated by α.
    We also have the smallest ring generated by F and α:
    F[α]  =  { f(α)  for  f(x) ∈ F[x] }.

  • Given an irreducible polynomial p(x) ∈ F[x], we can construct a simple extension field generated by a root α of p(x), whether or not we previously know a field with α ∈ K.
    • Inside the polynomial ring F[x], we construct the ideal (f(x)) = {q(x)p(x) for q(x) ∈ F[x]}, and take the quotient ring K  =  F[x]/(p(x)). By definition, K consists of polynomial classes [f(x)] modulo the relation [p(x)] = 0, so that [f(x)] = [f(x) + q(x)p(x)].
    • Then α = [x] is a root of p(x), since p(α) = p([x]) = [p(x)] = 0, and also f(α) = [f(x)], so we clearly have F[α] = K.
    • However, we know from Math 310 that (p(x)) is a prime ideal, so that K = F[x]/(p(x)) is a field: if g(α) = [g(x)] ≠ 0, then there is a polynomial reciprocal h(α) ∈ K with g(α) h(α) = 1. Therefore:
      K = F[α] = F(α).
      That is, every rational expression ^f(α)⁄_g(α) can be written as a polynomial f(α)h(α). We saw examples of this in the previous Notes and HW.
  • Minimal Polynomial Theorem:
    1. If α ∈ K is algebraic over F ⊂ K, then α is the root of a unique irreducible monic polyomial p(x) = a₀ + ··· + a_n−1xⁿ⁻¹ + xⁿ ∈ F[x]. Here:
       • irreducible means p(x) cannot be factored in F[x] except as p(x) = c · ^p(x)⁄_c for a constant c ∈ F;
       • monic means the leading coefficient is a_n = 1.
    2. Furthermore, α is a root of f(x) ∈ F[x] if and only if f(x) = q(x)p(x) for q(x) ∈ F[x].

    Proof: (i) Among all non-zero polynomials having α as a root, take one of minimal degree. Divide by its leading coefficient a_n to make it monic, and call the result p(x), with deg p(x) = n. If p(x) were reducible, we would have p(x) = a(x)b(x) with deg a(x), deg b(x) < n. But then p(α) = a(α)b(α) = 0, so either a(α) = 0 or b(α) = 0, which is impossible since we assumed α is not the root of any polynomial of degree less than n. Thus p(x) is irreducible, proving the first part except for uniqueness.

    (ii) If f(x) ∈ F[x] is any polynomial with f(α) = 0, perform long division to find f(x) = q(x)p(x) + r(x) with deg r(x) < n. Thus r(α) = f(α) − q(α)p(α) = 0; but as before, α cannot be the root of any non-zero polynomial of degree less than n; thus we must have r(x) = 0 and f(x) = q(x)p(x). This proves the second part.

    Finally, to prove the uniqueness of p(x), suppose p₁(x) and p₂(x) both have the required properties. By (ii), we get p₁(x) | p₂(x) and by the same reasoning p₂(x) | p₁(x), so p₁(x) = c p₂(x) for a constant c; but since both have leading term xⁿ, we must have c = 1. This completes the proof of (i).

  • Extension Degree Theorem: Consider the extension field F[α], where α has minimial polynomial p(x) of degree n. Then F[α] is an n-dimensional vector space over F, with basis {1, α, α², . . . , αⁿ⁻¹}. That is, any element of F[α] can be written in a standard form a₀ + a₁α + ···+ a_n−1αⁿ⁻¹ for unique a_i ∈ F.

    Proof: To show that the set spans, consider any element f(α) ∈ F[α]. Since f(x) = q(x)p(x) + r(x) for deg r(x) < n, we have:

    f(α)  =  q(α)p(α) + r(α)  =  r(α) = a₀1 + a₁α + ···+ a_n−1αⁿ⁻¹.
    Thus any element of F[α] is a linear combination of {1, α, α², . . . , αⁿ⁻¹}.

    To show linear independence, suppose a₀1 + a₁α + ··· + a_n−1αⁿ⁻¹ = 0, where not all a_i = 0. This means r(α) = 0 for a non-zero polynomial with deg r(x) < n, contradicting the assumption that deg p(x) = n is the minimal polynomial of α. Also, linear independence is equivalent to the uniqueness of the coefficients a_i .

  ⊞ HW  
  1. Prove that the following numbers are algebraic over Q by finding a polynomial satisfied by each.
     1. 1 + ³√2
     2. 3 + 5i
     3. √(1+√5)
     4. √2 + √5   Hint: Look at previous Notes.
  2. Show that if α² is algebraic over F, then so is α.
  3. Show that C = R[a+bi] for any a,b ∈ R with b ≠ 0.
  4. Show that if K is an extension field of Q of dimension 2, then K = Q[√d] for some d ∈ Q. Hint: Consider any α ∈ K−Q and show that K = Q[α]. Now consider the minimal polynomial of α.
  ⊞ Soln  

  1a. f(x) = (x−1)³ − 2 = x³ − 3x² - 1

  1b. f(x) = (x−3)² + 25

  1c. f(x) = (x²−1)² − 5

  1d. f(x) = (x−√2−√5)(x−√2+√5)(x+√2−√5)(x+√2+√5) = x⁴ − 14x² + 9

  2. If α² is the root of f(x), then α is the root of g(x) = f(x²).

  3. Clearly R[a+bi] ⊂ C. Also, for b ≠ 0, we have: i = ^{((a+bi) − a)}⁄_b = ^−a⁄_b + ¹⁄_b(a+bi), so i ∈ R[a+bi], and C = R[i] ⊂ R[a+bi].

  4. Suppose dim_Q(K) = 2, and take α ∈ K, α ∉ Q. Then {1, α} are linearly independent over Q, and make a basis of K. Thus α² = c + bα for b,c ∈ Q, and α is the root of x² − bx − c. The Quadratic Formula thus gives: α = ½(b ± √(b²+4ac)), so α ∈ Q[√d] for d = b² + 4ac. Therefore K = Q[α] ⊂ Q[√d]. But Q[α] = Q[√d] because both have dimension 2 over Q.

• 4/1 Ch 11.3 Algebraic extensions   ⊞ Notes  
  • Reading: Ch 11.3, pp. 382-386.
  • Dimension over a field: We know a simple extension field like Q ⊂ Q[√2] has elements a+b√2 for rational coefficients a,b ∈ Q. Instead of imagining a+b√2 on the real number line, we picture (a,b) as coordinates on two axes which are restricted to Q rather than R
    
    Thus Q[√2] is a vector space with scalars in the base field Q, with a basis {1,√2}. That is, as a vector space, Q[√2] ≅ Q², and dim_Q(Q[√2]) = 2.
  • Degree of an extension: We may think of any field extension F ⊂ K as a vector space over the base field F, that is, K ≅ Fⁿ. We call the dimension over F the degree of the extension, denoted as:
    [K : F]  =  dim_F(K)  =  n.
    We have seen that for a simple extension K = F[α], where α has the degree-n minimal polynomial p(x) ∈ F[x], the standard basis is {1, α, . . . , αⁿ⁻¹}. Thus the degree of the extension is equal to the degree of the minimal polynomial:
    [F[α] : F]  =  deg p(x)  =  n.

  • Degree product theorem: Suppose we have a chain of field extensions F ⊂ K ⊂ L. Then the degree of the full extension is the product of degrees of the two steps:
    [L : F]  =  [L : K] [K : F].
    
    Proof: Take an F-basis {α₁, . . . , α_n} of K, where n = [K : F], and a K-basis {β₁, . . . , β_m} of L, where m = [L : K]. Then we can write any element γ ∈ L uniquely as:  γ  =  ∑_j c_jβ_j   for   c_j ∈ K. We can also write each K-coefficient uniquely as  c_j = ∑_i a_ijα_i for a_ij ∈ F. Hence we have a unique F-linear combination:
    γ   =   ∑_i∑_j a_ijα_iβ_j ,
    which means {α_iβ_j} is a basis for  i = 1, . . . , n  and  j = 1, . . . , m. That is, dim_F(L) = nm, which proves the formula.
  • Finitely generated extensions: For an algebraic element α in some large ambient field (like the complex numbers), we have seen how to adjoin it to a field F to form the simple extension F[α]. We may iterate this to adjoin more algebraic elements. For example:
    F[α,β]  =  F[α][β]  =  {f(α,β) for f(x,y) ∈ F[x,y]},
    the ring (in fact a field) of all polynomial expressions in α and β with coefficients in F.
    In general, we may adjoin any finite list of algebraic elements to form F[α₁, . . . , α_k].
  • Algebraic extensions: We say F ⊂ K is an algebraic extension when every element γ ∈ K is algebraic over F, satisfying its own minimal polynomial. Note that for a simple extension F ⊂ F[α] by an algebraic α, it is not clear that every element γ = f(α) is also algebraic. However, the next proposition shows this is true.
  • Finite Extension Theorem: An extension F ⊂ K is finite, meaning its degree [K : F] = n is finite, if and only if K is a finitely generated algebraic extension.
    Proof: (⇐) If we adjoin finitely many elements to F, each of finite degree, the Degree Product Theorem shows the resulting extension has finite degree.
    (⇒) Suppose [K : F] = n, so that K ≅ Fⁿ is an n-dimensional vector space over F. Then for any element γ ∈ K, the n+1 elements {1, γ, γ², . . . , γⁿ} cannot be linearly independent, and so must obey a linear dependence relation:
    a₀1 + a₁γ + ··· + a_nγⁿ  =  0   for   a_i ∈ F.
    But then  f(x) = a₀ + a₁x + ··· + a_nxⁿ  is a polynomial satisfied by γ, so that γ is algebraic. Therefore K is an algebraic extension. Finally, adjoining an outside element increases the degree of a field, so that finitely many elements successively adjoined to F must produce K.
  • Example: Consider the multiple extension
    F = Q   ⊂   K = Q[³√2]   ⊂   L = Q[³√2,√5].
    Then α = ³√2 and β = √5 have the respective minimal polynomials x³ − 2 and x² − 5 ∈ Q[x]. If either were reducible in Q[x], it would have a rational root, but it does not by the Rational Root Test. This shows that [K : F] = 3 by the Extension Degree Theorem in the previous Notes, and it follows from the Degree Product Theorem that [L : F] = [L : K] [K : F] = 3[L : K]. Thus [L : F] = 6, provided we can show [L : K] = 2.

    Now, we could have [L : K] < 2 if x² − 5 becomes reducible in K[x]; that is, if √5 ∈ K = Q[³√2]: could we write  √5 = a + b³√2 + c³√4  for rational a,b,c? However, K' = Q[√5] clearly has [K' : F] = 2, so [L : F] = [L : K'] [K' : F] = 2[L : K']. Thus [L : F] is divisible by 2 and 3, and so [L : F] = 6. (That is, √5 ∉ K.) This illustrates how considering fields can easily resolve difficult questions about individual numbers.

    Furthermore, for a number like γ = ³√2 + √5, it is not clear that it is the root of a polynomial in Q. However the Finite Extension Theorem tells us it is: since the ambient field L has dimension 6 over F = Q, the 7 powers {1, γ, γ², . . . , γ⁶} have a linear dependence over Q, and γ satisfies a polynomial of order at most 6. In fact, by the Degree Product Theorem, the degree of the minimal polynomial must be a divisor or 6, namely 1,2,3, or 6.

  ⊞ HW  
  1. Let K = Q[√2,√3], with basis {1, √2, √3, √6} over the base field Q.  Let γ = √2 + √3.
     1. Compute 1, γ, . . . , γ⁴ by expanding in terms of the basis. The coefficients are the coordinates of each vector.
     2. Write a 4×4 matrix whose column vectors are the basis coordinates of {1, γ, γ², γ³}. Show this matrix is invertible: for example compute its determinant, or do row and column reduction, or rearrange rows and columns to make the matrix as upper-triangular as possible. Conclude that this is a basis of K, and that K = Q[γ], a simple extension generated by γ alone.
     3. We know the 5 vectors {1, γ, γ², γ³, γ⁴} must be linearly dependent over Q. Find a linear dependency relation:
        a₀1 + a₁γ + a₂γ² + a₃γ³+ a₄γ⁴  =  0
        by writing each vector in coordinates, reducing to a system of linear equations with unknown a_i's, and solving by row reduction.
     4. Extra Credit: Repeat the above for K = Q[³√2,√5] and γ = ³√2 + √5.
  2. Show that if F ⊂ K is a finite field extension, and D is a ring with F ⊂ D ⊂ K, then D is a field, containing the reciprocals of all its non-zero elements.
  ⊞ Soln  

  1a,b. Wolfram Alpha is useful to compute powers of γ. The matrix of {1, γ, γ², γ³, γ⁴} with respect to the basis {1, √2, √3, √6} is:

  1   0   5   0  49 0 1 0 11 0 0 1 0 9 0 0 0 2 0 20

  For example, the second column means γ = 0(1) + 1√2 + 1√3 + 0√6, the third column γ² = 5(1) + 0√2 + 0√3 + 2√6, etc.
  Taking the first 4 columns corresponding to {1, γ, γ², γ³}, and rearranging rows and columns to put the non-zero elements near the diagonal, we get:

  1   5   0   0    0 2 0 0   0 0 1 11   0 0 1 9

  Since the determinant of a block-diagonal matrix is equal to the product of the the determinants of its 2×2 blocks, we find the determinant of the original matrix is ±(2)(−2) = ±4 ≠ 0, and the matrix is non-singular (invertible). This is equivalent to the columns forming a basis.

  1c. To find a linear depenency among {1, γ, γ², γ³, γ⁴}, we must solve the linear system given by the matrix equation:

  1   0   5   0  49 0 1 0 11 0 0 1 0 9 0 0 0 2 0 20

  a0   a1   a2   a3   a4

  =

  0   0   0   0

  The general solution is (a₀, . . . , a₄)  =  Q(1, 0, −10, 0, 1), which means:

  1 − 10γ² + γ⁴  =  0.

  Since {1, γ, γ², γ³} is linearly independent, this is the minimal polynomial of γ, which shows [Q[γ] : Q] = 4 so Q[γ] fills all of K.

  2. We are given F ⊂ D ⊂ K with D a ring and F ⊂ K a finite extension. For any δ ∈ D, we have F[δ] ⊂ D, since F ⊂ D, and D is closed under addition and multiplication. Now, δ ⊂ K is algebraic by the Finite Extension Theorem, so we have the simple extension F[δ] = F(δ), meaning the reciprocal ¹⁄_δ is a polynomial in δ. Thus ¹⁄_δ ∈ D, which is a field.

• 4/5 Ch 11.4 Splitting fields and field symmetries, quadratic extensions   ⊞ Notes  
  • Reading: Ch 11.4, pp. 388-392.
  • Symmetry of Field Extensions: Given an extension F ⊂ K, we define a symmetry or automorphism to be a bijective mapping φ : K → K which is:
    • a field isomorphism with structure equations  φ(a+b) = φ(a) + φ(b),  φ(ab) = φ(a)φ(b);
    • equal to the identity mapping on F, meaning φ(a) = a for a ∈ F.
  • Complex conjugation: This is the most familiar field automorphism, the symmetry of the complex plane C defined by φ(a+bi) = a−bi, often denoted by φ(z) = z: geometrically, this is reflection across the real axis. It is easy to check the structure equations using the usual rules of algebra, and φ is the identity on the base field R. The reason this works is that C = R[i], and i and φ(i) = −i are both square roots of −1, both roots of the polynomial x²+1 = (x−i)(x+i), which is irreducible in R[x].
  • Quadratic extensions: For any K = Q[√r], where r ∈ Q with √r ∉ Q, we similarly have the symmetry φ(a+b√r) = a−√r. Again, this works because x²−r = (x−√r)(x+√r), and the two square roots of r are algebraically equivalent. This means that in any formula involving √r, we can replace it with −√r and it is still correct. For example:
    ¹⁄_(a+b√r)  =  ^(a−b√r)⁄_(a²−b²r)     ⇔     ¹⁄_(a−b√r)  =  ^(a+b√r)⁄_(a²−b²r)
    Note that r = (√r)² = (−√r)² is the same on both sides.
  ⊞ HW  
  1. Consider the field:
     L  =  Q[³√2]  =  { a + b³√2 + c³√4   for   a,b,c ∈ Q }.
     1. Compute the general product (a+b³√2+c³√4)(a'+b' ³√2+c' ³√4) in standard form.
     2. Consider the mappings φ : L → L defined by:
        φ(a + b ³√2 + c ³√4)  =  a ± b ³√2 ± c ³√4
        for any given choice of + and − signs. Show that φ respects Q-linear combinations:
        φ(aα + bβ)  =  a φ(α) + b φ(β)  for  a,b ∈ Q,  α, β ∈ L.
        This means φ is a vector space isomorphism.
     3. Show that φ does not respect multiplication, by giving a counterexample with φ(αβ) ≠ φ(α) φ(β). (This is for any choice of signs, except when φ is the identity.) Thus, L does not have any obvious symmetries.
        Extra Credit: Show L has no symmetries at all, by showing φ(³√2) must be another cube-root of 2
  2. Consider the following field obtained by adjoining three complex numbers to Q:
     K  =  Q[α, ζα, ζ²α]
     where α = ³√2 = 2^1/3 is the real cube root of 2, and ζ = ⁻¹⁄₂ + ^√3⁄₂ i ∈ C (Greek leter zeta) is a complex cube root of 1.
     1. Using the usual rules of algebra, show that ζ³ = (ζ²)³ = 1, so that ζ, ζ² are the complex cube roots of unity. Also, the minimal polynomial of ζ is x² + x + 1. Then it is easy to show α, ζα, ζ²α are the three complex cube roots of 2.
     2. Show that K = Q[³√2, ζ]. Hint: The main thing to show is ζ ∈ L.
     3. Show that the minimal polynomial p(x) = x³ − 2 of α splits into linear factors in K[x]:
        x³ − 2   =   (x − α) (x − ζα) (x − ζ²α).
        We say that K is the splitting field of p(x) over the base field Q.
     4. Find a basis of K in terms of α and ζ, and determine the degree [K : Q]. Hint: Keep in mind ζ² + ζ + 1 = 0, so ζ² = −1−ζ; and α³ = 2.
     5. Show that the complex conjugate mapping σ : K → K with σ(z) = z is a symmetry of K. Hint: We have shown above that σ respects the operations; you only need to show that σ takes K to K.
     6. Show that K has a symmetry ρ defined by ρ(α) = ζα and ρ(ζ) = ζ. This clearly can be extended to respect Q-linear combinations, and check that it respects multiplication on the basis elements of K.
     7. Consider σ and ρ as permutations of the three roots α₁ = α , α₂ = ζα , α₃ = ζ²α. Using this, show that we can compose these symmetries to get a symmetry group isomorphic to S₃ ≅ D₃.
  ⊞ Soln  

  1a. W|A

  1c. If φ is to be multiplicative, we must have φ(³√4) = φ(³√2)² = (±³√2)² = ³√4. But then 1 = φ(1) = φ(³√2·³√4) = ±³√2 ³√4 = ±1; and hence φ(³√2) = ³√2. Thus φ must be the identity mapping.

  2a. Use the Binomial Theorem (a+b)³ = a³ + 3²b + 3ab² + b³. Once you know ζ³ = 1, it follows that (ζ²)³ = (ζ³)² = 1, and (ζ^{j 3}√2)³ = 2.

  2b. Clearly K ⊂ Q[³√2, ζ]. Also ζ = (³√2)² (ζ ³√2) ∈ L, so Q[³√2, ζ] ⊂ K.

  2c. Each ζ^{j 3}√2 is clearly a root of the polyonomial, and this means (x − ζ^{j 3}√2) is a linear factor; but the degree 3 polyomial p(x) cannot be any more factors than these three.

  2d. K has Q-basis given by the powers of α = ³√2 up to 2 (since its minimal polynomial p(x) has degree 3); and the powers of ζ up to 1 (since its minimal polynomial has degree 2). Thus a basis is 1, α, α², ζ, ζα, ζα², and [K : Q] = 6.

  2e. It is clear that the conjugate field K is equal to K, i.e. K is closed under conjugation, and this is the only property missing to show the conjugation mapping σ is a symmetry of K.

  2f. We have ρ(α^jζ^k) = ζ^j α^j ζ^k = α^j ζ^j+k. Now check ρ(ab) = ρ(a) ρ(b) for each pair of basis elements, or do it on a product of standard forms with W|A.

  2g. In cycle notation for S₃, we have σ = (23) and ρ = (123). These two elements clearly generate all permutations in S₃. Alternatively, one could check the defining relations of D₃, namely σ² = ρ³ = 1 (the identity symmetry) and ρσ = σρ⁻¹.

• 4/10 & 4/12 Ch 12 Galois Theory, Q[³√2, ζ₃], Q[√2,√3]  ⊞ Notes  
  • Reading: Ch 12.1−12.2, pp. 407-421.
  • Splitting field: This means a field K = F[α₁, . . . , α_n] formed by adjoining to F all the roots of a polynomial over F:
    f(x)   =   (x−α₁) ··· (x−α_n)  ∈  F[x].
    We say that f(x) splits over K, that is, it splits into linear factors x−α_i  ∈  K[x].
  • Galois group: The algebraic symmetry group (or automorphism group) of the extension F ⊂ K is:  G  =  Gal(K/F)  =  {field isomorphisms φ : K → K with φ(a) = a for all a ∈ F}.
    We construct splitting fields as above because the roots α_i are symmetric to each other. In fact, any φ ∈ G is equivalent to a permutation of these roots, so we may consider G ⊂ S_n , and for any two roots, α_i, α_j, there is at least one φ ∈ G with φ(α_i) = α_j .
  • Fundamental Theorems of Galois Theory
    1. Let K = F[α₁, . . . , α_n] be the splitting field of a polynomial of degree n, with n distinct roots. Then the order of the Galois group is equal to the degree of the extension: |Gal(K/F)| = [K : F] = N.
    2. There is a bijective correspondence between subgroups H ⊂ G and intermediate fields F ⊂ L ⊂ K, given by the two inverse mappings:

       H	→	L = KH = {α ∈ K with φ(α) = α for all φ ∈ H}
       Gal(K/L) = H	←	L

       That is: Gal(K/K^H) = H  and  K^Gal(K/L) = L.
    3. The above gives a bijective correspondence between normal subgroups H and splitting field extensions F ⊂ L, whose Galois group is the quotient group: Gal(L/F) = G/N.
    The following results are useful in the proof of these theorems.
  • Primitive Element Theorem: If F ⊂ K is an extension of finite degree [K : F] = N, then there is some element γ ∈ K with K = F[γ], so that K is a simple extension. In fact, γ can be chosen to be a generic element of K, i.e. a random combination of basis elements will probably work.
  • Splitting Field Theorem: Any polynomial f(x) ∈ F[x] of degree n has a unique splitting field K of degree [K : F] = N ≤ n!. If g(x) ∈ F[x] is any other polynomial having a root β ∈ K, then all the roots of g(x) lie in K:
    g(β) = 0  for  β ∈ K       ⇒       g(x)  =  (x−β₁) ··· (x−β_m)   for   β₁, . . . , β_m ∈ K.
    That is, g(x) splits over K as soon as it has one root in K.
  • Example: We work out these theorems for the case considered in the previous HW, the splitting field of f(x) = x³−2 ∈ Q[x], that is:
    K  =  Q[α₁, α₂, α₃]  =  Q[α,ζ]     where     α = ³√2 , ζ³ = 1, and α₁ = α , α₂ = ζα , α₃ = ζ²α.
    • We saw the Galois group G ≅ S₃ ≅ D₃ is generated by ρ ≅ (123), defined by ρ(α) = ζα and ρ(ζ) = ζ; and by the complex conjugation mapping σ = (23) defined by σ(α) = α and σ(ζ) = ζ = ζ². Thus:
      G  =  ⟨ρ,σ⟩  =  {e, ρ, ρ², σ, σρ, σρ²}.

    • A primitive element with K = Q[γ] is given by almost any combination of the α_i's, for example:
      γ  =  α₁−α₂  =  (1−ζ) α.
      To find the minimal polyomial g(x) of γ, we can (as in HW 4/1 #1) expand the 7 powers 1, γ, . . . , γ⁶ in terms of a basis of the 6-dimensional K, and solve a system of linear equations to find a linear dependence c₀ + c₁γ + ··· + c_nγⁿ = 0 with c_i ∈ Q. The result is γ⁶ + 108, so g(x) = x⁶ + 108 ∈ Q[x], and it turns out that γ = ⁶√(−108).
    • As stated in the Splitting Field Theorem, all the roots γ_i of g(x) must be contained in K. Since we already know the Galois group G, we can find the other roots of g(x) as φ(γ) for symmetries φ ∈ G:
      • γ₁ = γ = α₁−α₂ = (1−ζ)α
      • γ₂ = ρ(γ) = α₂−α₃ = (ζ−ζ²)α
      • γ₃ = ρ²(γ) = α₃−α₁ = (ζ²−1)α
      • γ₄ = σ(γ) = α₁−α₃ = (1−ζ²)α
      • γ₅ = σρ(γ) = α₃−α₂ = (ζ²−ζ)α
      • γ₆ = σρ²(γ) = α₂−α₁ = (ζ−1)α
      To check, we multiply out:  g(x)  =  (x−γ₁) ··· (x−γ₆)  =  x⁶ + 108  ∈  Q[x]. This must be irreducible, since the minimal polynomial of γ has a root at every φ(γ), so g(x) divides any polynomial with h(γ) = 0.
    • Now we see why to expect Fundamental Theorem I. Any φ ∈ G is determined by φ(γ), which must be one of the conjugates γ_i , and there is a unique symmetry φ_i defined by φ_i(γ) = γ_i. Thus, the number of conjugates is N = deg g(x) = [K : F] = 6, but each conjugate corresponds to an element of G. Thus, |G| = [K : F].
    • Fundamental Theorem II says that each subgroup H ⊂ G corresponds to an intermediate field Q ⊂ L ⊂ K, namely the fixed field L = K^H. Fundamental Theorem III says that normal subgroups H ⊂ G correspond to splitting field extensions F ⊂ L, with Galois group G/H.
      • H = {e} fixes all elements, so L = K, a splitting field with Gal(K/Q) = G/{e} = G
      • H = {e, σ} has σ(α) = α, and L = Q[α] = Q[α₁].
      • H = {e, σρ} corresponds to L = Q[α₂]
      • H = {e, σρ²} corresponds to L = Q[α₃]
      • H = {e, ρ, ρ²} corresponds to L = Q[ζ], the splitting field of x²+x+1 ∈ Q[x] with Gal(L/Q) = G/⟨ρ⟩ = {[e], [σ]}
      • H = G fixes only the base field, so L = Q, a trivial extension with Gal(Q/Q) = G/G = {[e]}.
  ⊞ HW  
  Work out the consequences of the above results for the field K = Q[√2, √3] considered in HW 4/1 #1.

  1. Find a (reducible) polynomial in Q[x] such that K is its splitting field.
  2. Describe the Galois group G = Gal(K/Q), and verify Fundamental Theorem I.
  3. Show that γ = √2 + √3 is a primitive element, so that K = Q[γ]. Find the mimimal polyomial of γ by taking its conjugates γ_i = φ(γ) for φ ∈ G, and multiplying out the product g(x) = (x−γ₁) ··· (x−γ₄). Why must g(x) be irreducible?
  4. Apply Fundamental Theorems II, III: find all the subgroups H ⊂ G, their corresponding fixed fields L = K^H, and the Galois groups Gal(L/Q) = G/H if H is normal.
  ⊞ Soln  

  1. The smallest polynomial whose roots include both √2 and √3 is f(x) = (x²−2)(x²−3), and K = Q[±√2, ±√3]. Note f(x) does not need to be irreducible to produce a splitting field.

  2. G = {e, σ, τ, στ = τσ} ≅ C₂ × C₂, where σ(√2) = −√2, σ(√3) = √3; and τ(√2) = √2, τ(√3) = −√3. As we saw in HW 4/1, [K : F] = 4, and this is equal to |G| as stated by Fundamental Theorem I.

  3. As in HW 4/1, you can check that 1, γ, γ², γ³ are linearly independent in the 4-dimensional space K ≅ Q⁴: that is, the give an invertible 4×4 matrix with respect to the basis {1, √2, √4, √6}. This shows that γ generates K.

  As in Notes 3/27, the minimal polynomial is g(x)  =  (x−√2−√3)(x−√2+√3)(x−+2−√3)(x+√2+√3)  =  x⁴ − 10x² + 1. This must be minimal polynomial (and hence irreducible) since every polynomial h(x) ∈ Q[x] with γ as a root must also have all the conjugates φ(γ) as roots, so that each x−φ(γ) is a linear factor, and so g(x) divides h(x).

  4. Since G is abelian, all subgroups are normal, so Fundmental Theorems II and III both apply.

  • H = {e},  L = K,  G/H = G
  • H = {e, σ} fixes √3, so L = Q[√3], and Gal(L/Q) = G/H = {[e], [τ]}.
    Here [τ] = τH is a coset, giving the symmetry [τ](√3) = τ(√3) = −√3.
  • H = {e, τ} fixes √3, so L = Q[√2],  G/H = {[e], [σ]}.
  • H = G,  L = Q,  G/H = G/G = {[e]}

• 4/15 Galois Theory for a cubic x³−4x−1, discriminant  ⊞ HW  
  Let K = Q[α,β,γ] be the splitting field of the irreducible cubic (degree 3) polynomial:

  f(x)  =  x³ − 4x − 1  =  (x−α)(x−β)(x−γ)  ∈  Q[x]    with real roots   α  >  β  >  γ.

  1. Graph y = f(x) and approximate the values of the three roots.
  2. Use Newton's Method to approximate α to 5 decimal places. Hint: Start with an initial estimate x₀ from the previous problem, and iterate using the formula  x_n+1  =  x_n − ^f(x_n)⁄_{f '(xn)} . Continue until the digits stabilize. A spreadsheet is ideal for such a recursive algorithm: copy down.
  3. Expand the factored form  f(x) = (x−α)(x−β)(x−γ), and equate with x³ + 0x² − 4x − 1  to find γ in terms of α and β. The real numbers α,β,γ are algebraic conjugates, intimately related because they are roots of the same minimal polynomial with rational coefficients: when we multiply out, all the irrationalities cancel, leaving only the coefficients 0, −4, −1.
  4. The polynomial f(x) has no roots in Q, but by definition it has a root x = α in the extension field L = Q[α]. Divide f(x) by the linear factor x−α to get a quadratic factor g(x) ∈ L[x]; this is the minimal polynomial of β and γ in L[x].
  5. Show that K = Q[α,β]. Find a Q-basis for K by using the relations f(α) = 0 and g(β) = 0.
  6. Conisider the symmetry ρ ∈ G = Gal(K/Q) defined by ρ(α) = β, ρ(β) = γ, ρ(γ) = α. Write this as a Q-linear mapping in terms of the basis from the previous problem: that is, as a 6×6 matrix with Q entries. Do the same for the symmetry σ with  σ(α) = α,  σ(β) = γ,  σ(γ) = β.
  7. We have the Galois group G = {e, ρ, ρ², σ, σρ, σρ²} ≅ D₃ ≅ S₃. Draw the upside-down picture of inclusion of subgroups H ⊂ G, and the corresponding picture of subfields L = K^H. At the top, the trivial subgroup H = {e} fixes all of K, so K^H = K. At the bottom, H = G corresponds to the field of elements fixed by every element of G, namely the base field: K^G = Q.
  8. The fixed field E = K^H corresponding to the normal subgroup H = ⟨ρ⟩ ⊂ G is the set of elements λ ∈ K, having ρ(λ) = λ. Show that if θ is any element of K, then λ = θ + ρ(θ) + ρ²(θ) is fixed by ρ. Apply this to the basis elements to find λ ∈ E with λ ∉ Q; that is, σ(λ) ≠ λ.
  9. Show that if λ ∈ K, then δ = λ − σ(λ) has σ(δ) = −δ, and therefore σ(δ²) = δ². Apply this to the above λ ∈ E to get an element μ ∈ M with δ² = d ∈ Q; that is, δ = √d and E = Q[√d]. The number d is called the discriminant of cubic f(x): it is positive when f(x) has three distinct real roots, negative when it has a real root and two complex roots, and zero when it has a double or triple root.
     Hint: To reduce a polynomial g(α, β) to standard form in terms of the basis of K, use the PolynomialRemainder[f(x), p(x), x] function of Wolfram Alpha, which gives the remainder of the division f(x) = q(x)p(x) + r(x), where x is the variable. First the remainder of g(α, β) modulo β² + αβ + α² − 4 with variable β, regarding α as a constant; then the remainder of the result modulo α³ − 4α − 1 with variable α, regarding β as a constant.
  10. Extend the fields by a cube root of unity ζ, so that ζ³ = 1. That is, let:
      F' = Q[ζ]  ⊂  E' = E[ζ]  ⊂  K' = K[ζ].
      Then all of the above computations are valid, but coefficients in the base field are allowed to contain ζ. Show that
      μ  =  α + ζ ρ(α) + ζ²ρ²(α),         ν  =  α + ζ⁻¹ρ(α) + ζ⁻²ρ²(α)
      satisfy  ρ(μ) = ζ⁻¹μ,  ρ(ν) = ζν,  and  ρ(μ³) = μ³, ρ(ν³) = ν³. Thus, μ³, ν³ ∈ E' = (K')^ρ = F'[δ] = Q[ζ,δ]. Write α in terms of μ, ν and hence in terms of square and cube roots of rational numbers and ζ. This is a special case of the Cubic Formula.
  ⊞ Soln  

  1a. Plot

  2. Newton's Method obtains 5 correct decimal places after only 3 steps.

  3.

  f(x)   =   (x−α)(x−β)(x−γ)   =   x³ − (α+β+γ)x² + (αβ+αγ+βγ)x − αβγ   =   x³ − 4x − 1

  Equating coefficients and solving gives γ  =  −α−β  =  ¹⁄_αβ . Note also that αβ+αγ+βγ  =  −4 .

  4. Long division gives f(x) = (x − α)(x² + αx + α² − 4) + (α³−4 α−1), where the remainder vanishes. Thus β² + αβ + α² − 4

  5. The relation β² = −αβ − α² + 4 allows us to reduce powers βⁿ for n ≥ 2; then α³ = 4α + 1 allows us to reduce all powers αⁿ with n ≥ 3. Thus a Q-basis is {1, α, α², β, βα, βα²}.

  6. We reduce using the relations:

  α³  =  4α + 1,     β²  =  4 − α² − αβ,     γ  =  −α − β.

  Then σ acts on the basis as:

  σ(1) = 1,     σ(α) = α,     σ(α²) = σ(α)² = α²,
  σ(β)  =  γ  =  −α−β,     σ(αβ)  =  σ(α)σ(β)  =  −α²−αβ,
  σ(α²β)  =  α²(−α−β)  =  −α³−α²β  =  −1−4α−α²β.

  Writing the matrix by rows:  σ  =  {{1,0,0,0,0,-1}, {0,1,0,-1,0,-4}, {0,0,1,0,-1,0}, {0,0,0,-1,0,0}, {0,0,0,0,-1,0}, {0,0,0,0,0,-1}}.     Check σ² = I (identity matrix).

  We get ρ similarly. The most complicated computation uses αβ + αγ + βγ = −4.

  ρ(αβ)  =  βγ  =  −4−αβ−αγ  =  −4−αβ−α(−α−β)  =  −4+α².

  Writing the matrix by rows:  ρ  =  {{1,0,4,0,-4,0}, {0,0,0,-1,0,0}, {0,0,-1,0,1,0}, {0,1,0,-1,0,-4}, {0,0,-1,0,0,0}, {0,0,0,0,0,1}}.     Check ρ³ = I.

  7. The diagram has H = {e} at the top, under it {e, σ}, {e, σρ}, {e, σρ²}, and {e, ρ, ρ²}; then H = G at the bottom. The corresponding diagram of fixed fields has K at the top, three fields Q[α], Q[β], Q[γ] with degree 3 over Q; one field E with degree 2 over Q; and the base field F = Q at the bottom.

  8. To obtain a generator λ of the mysterious field E above with [E : Q] = 2, consider the element:

  λ  =  α²β + ρ(α²β) + ρ²(α²β)  =  (0,0,0,0,0,1) + (0,0,0,-4,0,-1) + (0,4,0,0,0,1)
   =  (0,4,0,-4,0,3)  =  4α−4β+3α²β.

  For all other basis elements θ, the elements θ + ρ(θ) + ρ²(θ) are also invariant under σ, so they are in Q and not useful for understanding E.

  9. An especially nice element of E is:

  δ  =  λ − σ(λ)  =  (0,4,0,-4,0,3) + (3,4,0,-4,0,3)  =  (3,8,0,-8,0,6)  =  3+8α−8β+6α²β.

  To find the discriminant d = δ², first expand δ², then reduce powers of β, then reduce powers of α: all α's and β's cancel, and we get d = 229, so that δ = √229 = 3+8α−8β+6α²β. Thus E = Q[√229].

• Hand-In HW: Problems: Classify |G| = 21 with Sylow; C₂₁ or non-abelian with F₂-linear rep on F₈.
• 4/24 Review: Geometric symmetries, triangular prism rotations D₃. See old notes.   ⊞ HW  
  Consider the following solid X, an equilateral triangular prism like a cheese wedge with the axes like toothpicks.
  
  This is defined as the set of vector points v = (x,y,z) satisfying the dot product inequalities:
  −1  ≤  v · (0,0,1)  ≤  1,     v · (1,0,0)  ≤  1,     v · (⁻¹⁄₂ ,^√3⁄₂, 0)  ≤  1,     v · (⁻¹⁄₂ ,^−√3⁄₂, 0)  ≤  1.
  For example, the first inequalities mean −1 ≤ z ≤ 1. Each vector in the right factor of a dot product is an axis of symmetry of X (for example the z-axis R(0,0,1) for the first inequalities). We let G = Sym(X) be the rotation symmetry group of X (excluding reflection symmetries).
  1. G is generated by a ¹⁄₂ rotation around the x-axis, denoted S, and a ¹⁄₃ right-handed rotation around the z-axis, denoted R. Find the matrices of S and R. Remember that the matrix of a linear mapping L has the three column vectors [L(1,0,0) | L(0,1,0) | L(0,0,1)].
  2. Multiply the matrices R and S repeatedly to generate all 6 elements of G. In particular, generate a rotation S' around the axis w = (⁻¹⁄₂ ,^√3⁄₂, 0) by rotating w to the x-axis by R⁻¹, then rotating around the x-axis, then rotating the x-axis back to w by R: that is, S' = RSR⁻¹
  3. Find the eigenvectors and eigenvalues of the matrix S' above, and verify its geometric description as a ¹⁄₂ rotation around w.
  4. Show that G cannot have more than 6 elements. Consider the images of the vertices {v₁, . . . , v₆} under an arbitrary symmetry A ∈ G: count the possible choices for A(v₁), and show that this determines all the other A(v_i)'s.
  5. For each subgroup H ⊂ G, add some decorations to X so that the resulting X⁺ has H = Sym(X⁺). Verify that if we move the decorations to get gX⁺, the symmetry group becomes the conjugate subgroup gHg⁻¹. What is special about the decoration for the normal subgroup ⟨R⟩?
  6. For each subgroup H ⊂ G, decorate X by its set of fixed points X^H = {x ∈ X with h(x) = x for all h ∈ H}, so that X_H⁺ = X ∪ X^H. Note that H = Sym(X_H⁺); that the inclusion of subgroups H ⊂ H' corresponds to a reverse inclusion X_H'⁺ ⊂ X_H⁺; and that for the normal subgroup H = ⟨R⟩, we have G/H = Sym(X^H). What Theorem does this situation remind you of?
  ⊞ Soln  

  1. Tracing the rotation on the picure, we see S(1,0,0) = (1,0,0), S(0,1,0) = (0,−1,0), etc., giving the matrices:

  S   =

  1 0 0 0 −1 0 0 0 −1

  R   =

  −1⁄2 −√3⁄2   0 √3⁄2 −1⁄2 0 0 0 1

  The upper left submatrix of R is a 2-dimensional rotation matrix by angle θ = 120°.

  2 & 3. Wolfram gives:

  S'   =   RSR−1   =

  −1⁄2 −√3⁄2   0 −√3⁄2 1⁄2 0 0  0  −1

  .

  The characteristic polynomial is −(λ−1)(λ+1)², which is consistent with a 180° rotation about the λ = 1 eigenvector as axis, which is indeed (⁻¹⁄₂ ,^√3⁄₂, 0) as expected. (Remember that the eigenvector is not unique, but can be multiplied by any scalar.)

  The full group of symmetries is the dihedral group G = {I,R,R²,S,SR,SR²} ≅ D₃.

  4. The symmetry A must take corners to corners, so A(v₁) = v_i for i = 1, . . . , 6. Then the vertex counterclockwise from v₁ must go to the vertex counterclockwise from v_i, the vertex vertical from v₁ must go to the vertex vertical from v_i, etc., so there is no further choice needed to define A. Hence there are at most 6 symmetries.

  5. The trivial subgroup {I} can be obtained by adding any asymmetric decoration to X, for example marking one of the corners. Each of the order 2 subgroups ⟨S⟩, ⟨SR⟩, ⟨SR²⟩ can obtained by marking one of the vertical faces, or two vertices switched by the subgroup. The normal subgroup ⟨R⟩ can be obtained by marking one of the triangular faces, or three vertices around a triangular face.

  6. The fixed set of the trivial subgroup {I} is all of X; the fixed sets of the three two-element subgroups are their respective horizontal axes of rotation; the fixed set of the normal 3-element subgroup is its vertical axis of rotation; and the fixed set of all of G is the origin {0}. We clearly have Sym(X ∪ X^H) = H, except for the case of H = {I}, since Sym(X ∪ X^I) = Sym(X) = G ≠ H. Also, for the three normal subgroups, we have: G/{I} ≅ G = Sym(X^I), since X^I = X; G/⟨R⟩ ≅ {I, S}, = Sym(X^R), since the only non-trivial geometric linear symmetry mapping of the z-axis is a vertical flip; G/G ≅ {I} = Sym(X^G) = Sym(0), since a single point has no non-trivial symmetry mappings.

  This setup exactly parallels the Main Theorems of Galois Theory, in which G = Gal(K/F), and each subgroup H ⊂ G corresponds to its fixed field K^H with Gal(K/K^H) = H. Also, for normal subgroups, we have G/H ≅ Gal(K^H/F).

  We can make the analogy perfect by defining the relative symmetry group of a pair of objects Y ⊂ X to be symmetries of X which fix every point of Y: that is, Sym(X,Y) = {symmetries σ : X → X such that σ(y) = y for all y ∈ Y}. Then we have Sym(X,X^H) = H for all subgroups H ⊂ G.