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\title{Group Theory: Math30038, Sheet 9 Solutions}
\date{GCS}
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\maketitle

\begin{enumerate}
\item Suppose that $G$ is a simple group of order $60$.
\begin{enumerate}
\item[(a)] {\em Show that $G$ has a subgroup $A$ of order 12.}\newline
\noindent {\bf Solution\ } Since $G$ is simple there must be more than
one Sylow 2-subgroup of $G$. Let $P$ and $Q$ be distinct subgroups of $G$
of size $4$. If $P \cap Q = T$ has order 2, then $T$ has index 2 in both
$P$ and $Q$ and so is normalized by each of them. Let $H = N_G(T)$ which
has order more than 4, divisible by 4 and dividing 60. Now $H \not = G$ else
$T \unlhd G$. Also $|G:H| \not = 3$ else a Poincar\'e argument would have
$60$ dividing $3! = 6$. Thus $|N_G(T)| = 12$. Let $A = N_G(T).$
\item[(b)] Show that $A$ has exactly 5 different conjugates. \newline
\noindent {\bf Solution\ } $|N_G(A)|$ divisible by 12 and divides 60.
However, $A$ is not normal in $G$ and so $N_G(A) = A$. Let $G$
act on
the conjugates of $A$ via conjugation. The size of the orbit
is $|G:N_G(A)| = 5$.
\item[(c)] {\it Show that there is an injective homomorphism from
$G$ to $S_5$.} \newline
\noindent {\bf Solution\ } 
Let the image of this map be $H$, a subgroup of index
2 in $S_5$. Let $\Omega$ be the set of conjugates of $A$, a set of
size 5. The conjugation action of $G$ on $\Omega$ induces a group
homomorphism $\theta: G \rightarrow \mbox{Sym}(\Omega).$ This
homomorphism has non-trivial image, and trivial kernel by
simplicity of $G$. This gives a monomorphism from $G$ to $\mbox{Sym}(\Omega)$.
Labelling the elements fo $\Omega$ using the first five natural numbers
we obtain a monomorphism from $G$ into $S_5$. The image $X$ is a copy
of $G$ and has index 2 in $S_5$.

\item[(d)] {\em Show that both $A_5$ and $H$ both contain every element
of $S_5$ of the form $g^2$ and therefore every $5$-cycle and
every $3$-cycle.}  \newline
\noindent {\bf Solution\ } Suppose that $Z$ is any subgroup of $S_5$
of index 2. Then $Z$ is normal in $S_5$ and the quoent group $S_5/Z$
is cyclic of order 2. Thus if $g \in G$, then $(gZ)2 = Z$ so
$g^2 \in Z$. Thus all squares are in $Z$. Now all $3$-cycles
and $5$-cycles are squares because $(a,b,c) = (a,c,b)2$ and
$(a,b,c,d,e) = (a, d. b, e, c)^2$.  
\item[(e)] {\em Show that $H = A_5$.} \newline \noindent {\bf Solution\ } There are $20$ three cycles
and $24$ 5-cycles in $S_5$ so $A_5 \cap H$ is a subgroup
of each of $A_5$ and $H$ and has size at least 44.
Therefore $A_5 \cap H$ has order 60 by Lagrange's theorem.
Thus $H = A_5$.
\item[(f)] {\em Deduce that any simple group of order 60 must be
isomorphic to $A_5$.} \newline
\noindent {\bf Solution\ } That is what we have shown.
\end{enumerate}
Incidentally, here is another way to show that $G$ is isomorphic to $A_5$,
and this method cuts out (d), (e) and (f):  Now let $Y = X \cap A_5$. Now
from an early sheet, $|A_5:Y| \leq |S_5:X| = 2$ so either
$|A_5:X|$ is 2 or 1. It cannot eb 2, else then $Y$ would be normal in
$A_5$ which would violate the simplicity of $A_5$. Therefore the
index is $1$ so $A_5 \leq Y$. Now each of $A_5$ and $Y$ has order 60
so $Y = A_5$.
\item Show that the following presentations all describe the trivial group.
\begin{enumerate}
\item[(a)] {\em $\langle x \mid x^2, x^3 \rangle,$} \newline
\noindent {\bf Solution\ }\ Let $x$ denote the image of $x$ in the
presented group $G$. Thus $x = x^3(x^2)^{-1} = 1$. Since $G$
is cyclic and generated by $x$, then $G$ is trivial.  
\item[(b)] {\em $\langle x, y \mid xy=yx, xyx = yxy, x^5, y^6 \rangle$} 
\newline {\bf Solution\ } As before we regard $x$ and $y$ as elements
of the presented group $G$. Now $G$ is abelian since $xy = yx$. Therefore
$y = x$ and $G$ is cyclic. Now $1 = x^5 = y^6 = x^6$ so
$x =1 =y$ and $G$ is trivial.
\item[(c)] {\em $\langle x, y \mid x^2, y^4, xyx = y^2, xyxyxy  \rangle.$}
\newline \noindent {\bf Solution\ } Working in the established way in
$G$ we discover that $(xyx)^2 = y^4 =1$. Thus $xyxxyx = 1$ so
$xyyx = 1$ so $yy = xx =1$. Thus $xyx = y^2 =1$ so $y = xx = 1$.
Now $xyxyxy = x^3 = 1 = x^2$ so $x =1$ in $G$ and $G$ is the trivial group.
\end{enumerate}
\item We discuss the group $G = \langle x, y \mid x^2 = y^2, (xy)^2 = x^2, 
x^4
\rangle$ in stages:
\begin{enumerate}
\item[(a)] Let $T = \langle x^2 \rangle \leq G$ so $|T| \leq 2$. Show that
$T \unlhd G$. \newline
\noindent {\bf Solution\ } $T = \{ 1, x^2\} = \{1,y^2\}.$
Thus in $G$ the element or elements of $T$ commute with both $x$
and $y$ and therefore with all elements of $G$. Thus $T$ is central
in $G$ and is therefore normal in $G$. 
\item[(b)] {\em Let $H = G/T$. Show that $xT$ and $yT$ commute so that
$H$ is abelian.}\newline
\noindent {\bf Solution } 
Now $yxT$ is the inverse of $xyT$
and so is $xyT$ so $xyT = yxT$. Thus $xT$ and $yT$ commute so $G/T$
is abelian.
\item[(c)] {\em Show that $|H| \leq 4$.} \newline
\noindent {\bf Solution\ } It is easy to verify
that the group generated by $xT$ and $yT$ is $\{ T, xT, yT, xyT\}$
and so $H$ has size at most $4$.
\item[(d)] Deduce that $|G| \leq 8$. \newline
\noindent {\bf Solution\ } $|G| = |G:T| \cdot |T| \leq 8.$
\item[(e)] Find a group of order $8$ containing elements $a, b$ such
that $a^2 = b2$, $(ab)^2 = a^2$ and $a^4 =1$. Apply von Dyck's theorem
to deduce that $G$ has a homomorphic image of order $8$, and by
part (d) deduce that this homomorphism is an isomorphism.\newline
\noindent {\bf Solution\ } The quaterion group will do the trick, in
the usual notation $i$ and $j$ satisfy the relations (for $x$ and $y$
respectively). By Von Dyck's theorem there is an epimorphism from
$G$ to a group of order $8$, so $G$ has size at least $8$. Thus
$G$ has order 8, and the homomorphism from $G$ onto the quaternion
is an isomorphism.
\end{enumerate}
\item {\em Discuss the (language) group $L$
presented by the small roman alphabet with
relations consisting of all homonyms. Thus we have relations
such as $too = two$, $practice = practise$, $sea = see$ and so on.
The group you get in my ideolect is $C_2$ (South London with a transparent
veneer of civilization), whereas in South African English you get the
trivial group. On the other hand, I understand that in German the spelling
of a word is determined by its pronunciation, so $L$ is a free group.
To get you started in $L$ for English, since $two = too$ in $L$, then
$t^{-1}twoo^{-1} = t^{-1}tooo^{-1}$ and thus $w = o$ in $L$. Also
$too = to$ so $o = 1$ and therefore $w = 1$. Here $1$ denotes the identity
element of $L$ for English.} \noindent
\newline {\bf Solution\ } Well, this is best done for fun. I reckon South
African Englsih can kill off the group with $felt = veldt$. 
\end{enumerate}
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