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\noindent
{Student Name:\underline{\quad \sc group theory review\quad}\hfill\mbox{}}
\\[1em]
\noindent
{\bf MTH 418H \hfill Fall 2004}
\\[-1.2em]
\centerline
{\Large \bf Quiz 11/5}
\vspace{-.2em}

\noindent
If $H$ is a subgroup of $G$, the {\it index}  $[G\sh:H]$  is the number of distinct right cosets $gH$, where $g\in G$.
Note that this is also the number of left cosets $Hg$, since the bijection $G\stackrel{\sim}\to G$,\ $g\mapsto g^{-1}$ takes each right coset $gH$ to a left coset $Hg^{-1}$.
\\[.5em]
{\bf 1.}  Is it possible that $G$ is an infinite group, but $[G\sh:H]$ is finite?  
\\[.5em]
{\sc solution:} Yes, it is possible.  For example, take $G=\Z^+$, the additive group of integers, and $H=n\Z$.  Then the distinct cosets are:
$$n\Z,\ 1\sh+n\Z,\ 2\sh+n\Z,\ldots,\  n\sh-1\sh+n\Z\,$$
and $G/H=\Z/n\Z\cong \Z_n^+$, 
the additive group of clock arithmetic mod $n$.
\\[1em]
Another example:  $G=\R^\times$ (non-zero reals under multiplication),  $H=\R_{>0}^\times$ ,
$[G\sh:H]=2$.
\\[1em]
{\bf 2.} Show that if $[G\sh:H]=2$, then $H$ is a normal subgroup of $G$.
\\[.5em]
{\sc proof:} 
If $[G\sh:H]=2$, then $G$ partitions into right and left cosets as  
$G=H\cup gH=H\cup Hg$, where $g$ is any element not in $H$.  Thus:
$$
gH = G\smallsetminus H = Hg\,,
$$
or equivalently $gHg^{-1}=H$, which is the definition of a normal subgroup.  (Here $G\smallsetminus H$ means $G$ with the elements of $H$ removed.)
\\[1em]
{\bf 3.} Show that if $n,m>0$ with $\gcd(n,m)=1$, then 
the product of the cyclic groups $C_n$ and $C_m$ is isomorphic to 
the cyclic group $C_{nm}$ :
$$C_n\times C_m\cong C_{nm}\,.$$
\\[-1em]
{\sc proof:} Let $C_n=\langle x\rangle$ and $C_m=\langle y\rangle$, so that $C_n\times C_m = \{(x^i,y^j)\mid i,j\in \Z\}$.    Of course, these are not all distinct elements, since $x^k = 1$ whenever $n|k$ , and $y^k=1$ whenever $m|k$ .  Now, let $z:=(x,y)$.  

I claim $z$ has order $nm$. Clearly $z^{nm}=(x^{nm},y^{nm})=(1,1)=1$, so $\ord(z)\leq nm$.  Now, since $\gcd(n,m)=1$, we can write $an+bm=1$ for integers $a,b$.  Thus $z^k=1$ means $(x^k,y^k)=(1,1)$, i.e., $n|k$ and $m|k$, so that:
$$
nm\,|\, (ank+bmk) = k\,.
$$
That is, \,  $z^k=1\Longrightarrow nm|k$ , \, and $\ord(z)=nm$.

Therefore $C_n\times C_m=\{z,z^2,\ldots,z^{nm}=1\}$, 
which is clearly a cyclic group $C_{nm}$. 
\\[1em]
{\sc note:}
Let us rewrite this as: 
$C_n\times C_m\cong\Z_n^+\times\Z_m^+$, so that $z\leftrightarrow(1\mod n,1\mod m)$ and $z^k\leftrightarrow
k(1,1)=(k\mod n, k\mod m)$.  Thus, we have the isomorphism:
$$\begin{array}{rcl}
\Z_{nm}^+&\rightarrow&\Z_n^+\times\Z_m^+\\[.3em]
k\mod nm&\mapsto&(k\mod n\,,\, k\mod m)\,.
\end{array}$$
Since this is an bijection, we get the remarkable fact:\\[.5em]
{\it Chinese Remainder Theorem}: Suppose $n,m$ are relatively prime. Then for any $i\mod n$ and $j\mod m$, there is a unique $k\mod nm$ such that $k\equiv i\mod n$ and $k\equiv j\mod m$. 

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\noindent
{\bf Proposition:}  If $G$ is a group with 6 elements, then $G$ is isomorphic to the cyclic group $C_6$ or the dihedral group $D_3$.
\\[1em]
{\bf Proof:}  {\sc case (1)} \ Suppose $G$ has an element $x$ of order 6.  Then the cyclic subgroup $\langle x\rangle
=\{1,x,x^2,\ldots,x^5\}$ has 6 elements and is all of $G$,
so that $G$ is cyclic.
\\[1em]
{\sc case (2)} \ Suppose $G$ has an element $x$ of order 3, but none of order 6.  Taking  some element $y\not\in\langle
x\rangle=\{1,x,x^2\}$ , we have:
$$
G=\langle x\rangle \cup y\langle x\rangle
=\left\{\begin{array}{ccc}
1,&x,&x^2\\
y,&yx,&yx^2
\end{array}\right\}\,.$$
{\sc question}:  Which of these 6 elements is $y^2$ ?  
\begin{itemize}
\item Since the index $[G\sh:\langle x\rangle]=2$, the subgroup $\langle x\rangle$ is normal by Quiz Question 2. Thus we have a quotient group $G/\langle x\rangle
=\{\overline 1,\overline y\}$, and clearly $\overline y^2 = \overline 1$ , i.e., $y^2\in \langle x\rangle$ .
\item
If $y^2=x$, what is the order of $y$ ?  
We have $y^6=x^3=1$, so $\ord(y)$ divides $6$, and $\ord(y)\neq 1,2$. If $\ord(y)=3$, then $1=y^3=xy$ and $y=x^{-1}=x^2$, which is false.  Thus $\ord(y)=6$, contrary to our assumption.  Hence $y^2=x$ is impossible.
\item
We can show $y^2=x^2$ is impossible by an exactly similar argument.  For example, if $\ord(y)=3$, then $1=y^3=x^2y$, so that $y=x^{-2}=x$, which is false.
\item
The only remaining possiblility is $y^2=1$ .  
\end{itemize}
{\sc question}: What is $yxy^{-1}$ ?  
\begin{itemize}
\item Since as noted 
$\langle x\rangle$ is normal, we have
\ $y\langle x\rangle y^{-1}=\langle x\rangle$ \ and \ $yxy^{-1}\in\langle x\rangle$ .
\item Since conjugating does not change the order of an element, we have $\ord(yxy^{-1})=\ord(x)=3$.  Thus $yxy^{-1}=x$ or $x^2$.
\item
If $yxy^{-1}=x$ , then $yx=xy$ and:
$$
\langle xy\rangle=
\{1,xy,x^2y^2,x^3y^3, x^4y^4,x^5y^5\}
=
\{1,xy,x^2,y,x,x^2y\}\,,
$$
so that $\ord(xy)=6$, contrary to assumption.
(In other words: $C_2\times C_3\cong C_6$.)
\item
The only remaining possibility is: $yxy^{-1}=x^2$.
\end{itemize}
{\sc summary}:  $G$ is generated by elements $x,y$ with $x^3=y^2=1$ and $yx=x^2y$ .  But we know that this defines the multiplication table of $D_3$ , and we have $G\cong D_3$ .
\\[1em]
{\sc case (3)} \ Suppose $G$ has only elements of order 1 and 2.  Then for any $x\in G$, we have $x^{-1}=x$.  For any $x,y\in G$, we have $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$, so $G$ is abelian.  

Now consider two distinct elements $x,y\neq 1$, which clearly generate the subgroup:
$$
H:=\langle x,y\rangle = \{1,x,y,xy\}\cong C_2\times C_2\,.
$$
But then $G$, with 6 elements, could not possibly be partitioned into disjoint cosets of $H$, each with 4 elements.  (Indeed, for any subgroup $H\subset G$, we have $\#H\,|\,\#G$ for this same reason.) \   Thus this case is impossible.

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{\sc lemma:} If $p$ is prime, and $G$ is any group of order $p$, then $G\cong C_p$ , a cyclic group.
\\[.3em]
{\sc proof:} Let $1\neq x\in G$.  Then $