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\title{Group Theory: Math30038, Sheet 9}
\date{GCS}
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\begin{enumerate}
\item Suppose that $G$ is a simple group of order $60$.
\begin{enumerate}
\item[(a)] Show that $G$ has a subgroup $A$ of order 12.
\item[(b)] Show that $A$ has exactly 5 different conjugates.
\item[(c)] Show that there is an injective homomorphism from
$G$ to $S_5$. Let the image of this map be $H$, a subgroup of index
2 in $S_5$.
\item[(d)] 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.
\item[(e)] Show that $H = A_5$.
\item[(f)] Deduce that any simple group of order 60 must be 
isomorphic to $A_5$.
\end{enumerate}
\item Show that the following presentations all describe the trivial group.
\begin{enumerate}
\item[(a)] $\langle x \mid x^2, x^3 \rangle,$
\item[(b)] $\langle x, y \mid xy=yx, xyx = yxy, x^5, y^6 \rangle$ and 
\item[(c)] $\langle x, y \mid x^2, y^4, xyx = y^2, xyxyxy  \rangle.$ 
\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
$T \unlhd G$.
\item[(b)] Let $H = G/T$. Show that $xT$ and $yT$ commute so that
$H$ is abelian.
\item[(c)] Show that $|H| \leq 4$.
\item[(d)] Deduce that $|G| \leq 8$.
\item[(e)] Find a group of order $8$ containing elements $a, b$ such
that $a^2 = b^2$, $(ab)^2 = a^2$ 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.
\end{enumerate} 
\item 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.
\end{enumerate} 
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