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\title{Group Theory: Math30038, Sheet 7}
\date{Solutions --  GCS}
\author{}
\begin{document}
\maketitle

\begin{enumerate}
\item {\em Suppose that $n \geq 3$. Let $g \in G = S_n$ be an $n$-cycle.
Recall that all $n$-cycles are conjugate in $S_n$.}
\begin{enumerate}
\item[(a)] {\em Prove that $C_G(g) = \langle g \rangle.$} \newline
\noindent {\bf Solution }\ Since the conjugacy class of $g$ has size
$(n-1)!$ it follows that $C_G(g)$ has order $n$ and so is
$\langle g \rangle$. 
\item[(b)] {\em Deduce that $Z(G) = 1$.} \newline
\noindent {\bf Solution }\ 
Let $g_1 = (1,2,3,\ldots,n)$ and $g_2 = (2,1,3, \ldots n)$
(so $g_1$ and $g_2$ only differ in their effects on $1$ and $2$).
Now $Z(G) \leq C_G(g_1) \cap C_G(g_2)
= \langle g_1 \rangle \cap \langle g_2 \rangle = 1$.
\item[(c)] {\em Show that $\mbox{Aut}(G)$ has a subgroup which
is isomorphic to $G$.} \newline
\noindent {\bf Solution }\ Let $\tau: G \longrightarrow \mbox{Aut}(G)$
be the homomorphism which sends each group element $x$ to the automorphism
$\tau_x$ which is ``conjugation by $x$''. Now $\mbox{Ker }\tau = Z(G) =1$
so $\tau$ is  a monomorphism and $\mbox{Im }\tau$ is a subgroup
of $\mbox{Aut}(G)$ which is isomorphic to $G$ as required.
\end{enumerate}
\item {\em Suppose that $G$ is a group and that $\mbox{Aut}(G)$ is the trivial group.}
\begin{enumerate}
\item[(a)] {\em Show that $G$ must be abelian.} \newline
\noindent {\bf Solution }\ If $x \in G - Z(G)$, then
$\tau_x$ (conjugation by $x$) will be a non-trivial automorphism of $G$.
Thus $G - Z(G) = \emptyset$ so $G = Z(G)$ is abelian.
\item[(b)] {\em By 
considering the inversion map, deduce that $g^2=1$ for all $g \in G$.}\newline
\noindent {\bf Solution}\ The inversion map is a non-trivial automorphism
unless $g = g^{-1}$ for every $g \in G$, so $g^2 =1$ for all $g \in G$.
\item[(c)] {\em Write $G$ additively, so that $G$ becomes a vector space 
over the field
$\mathbb Z_2$ of ``integers modulo 2''. You may assume that every vector space has a basis.
Deduce that $G$ must either 
be cyclic of order 2 or the trivial group.}\newline
\noindent {\bf Solution }\ Pick a basis $B$ of the vector space $G$.
If dim $G > 1$, then we can choose two distinct elements $u, v$ of $B$.
Now the map which swaps $u$ and $v$ and fixes all other elements of $B$
extends uniquely to a non-trivial linear isomorphism from $G$ to $G$.
This linear isomorphism will also be a non-trivial group automorphism
of $G$. We deduce that $G$ has dimension 0 or 1, and so is
either the trivial group or the cyclic group of order 2. Finally
we observe that these two groups do indeed have trivial groups of
automorphisms.
\end{enumerate}
\item {\em Suppose that $G$ is a group and that $\mbox{Aut}(G)$ is finite.
Prove that $|G:Z(G)|$ must be finite.} \newline
\noindent {\bf Solution } $G/Z(G)$ is isomorphic to a subgroup
of Aut $G$, and so is finite. Thus $|G:Z(G)|$ is finite.
\item {\em (Challenge) Suppose that $G$ is a group and that $\mbox{Aut}(G)$ is
cyclic. Prove that $G$ must be abelian.}\newline
\noindent {\bf Solution} Since a subgroup of a cyclic group
is cyclic, it follows that $G/Z(G)$ is a cyclic group. Say
$G/Z(G)$ is generated by $tZ(G)$. Thus every element of $G$
lies in a coset $t^kZ(G)$ for some integer $k$, and therefore
$G = \langle \{t\} \cup Z(G) \rangle.$ Thus $G$ is abelian.
{\em \item Prove that for a finite 
group, the notions of isomorphism, monomorphism
and epimorphism all co-incide.} \newline
\noindent {\bf Solution }\ This was poorly worded. I was thinking
of homomorphisms from a finite group $G$ to itself (so called 
endomorphism of $G$). Since $G$ is finite, the notions of 
injection, surjection and bijection all co-incide.
\item In each case give an example of a group $G$ and a map $\zeta: G
\rightarrow G$ which satisfies the specified property.
\begin{enumerate}
\item[(a)] {\em An epimorphism (i.e. a surjective homomorphism) but
not an isomorphism (suggestion for $G$: fix a prime number $p$ and
consider those complex numbers $z$ such that $z^{p^n} =1$ for some
natural number $n$ which may vary as $z$ varies).} \newline
\noindent {\bf Solution }\ The map whic sends each element to its $p$-th
power is a homomorphism, and is surjective but not injective.
\item[(b)] {\em A monomorphism (i.e. an injective homomorphism)  but
not an isomorphism (suggestion for $G$: $\mathbb Z$ under addition).}\newline
\noindent {\bf Solution }\ The map which doubles every integer
is an injective homomorphism, but is clearly not surjective.
\end{enumerate}
\item {\em Suppose that $G$ is a finitely generated group (i.e. there is
a finite subset $X$ of $G$ such that $\langle X \rangle = G$) and that
every conjugacy class of $G$ is finite. 
Prove that $|G:Z(G)|$ is finite.}\newline
\noindent {\bf Solution }\ Since every conjugacy class is finite,
each centralizer of an element is of finite index in $G$.
Now an element is in the centre of $G$ if and only if it
commutes with every element of a generating set. 
Thus
\[ Z(G) = \cap_{x \in X} C_G(x).\]
Now, we have proved that the intersection of two (and therefore
finitely many) subgroups of finite index is of finite index.
Since $X$ is finite it follows that $|G:Z(G)| < \infty.$ 
\item {\em Let $V$ be a vector space of dimension $n$ over a field $F$ which
has $q$ elements. Let $\mbox{GL}(V)$ denote the group of linear
isomorphisms from $V$ to $V$ (a group under composition of maps).
Determine $|\mbox{GL}(V)|$.} \newline
\noindent {\bf Solution }\ Pick and fix a basis for use both in domain
and codomain. The invertible linear maps on $V$ are in bijective correspondence
with the invertible $n$ by $n$ matrices. we count such matrices row by row.
The first row (viewed as an element of $F^n$) can be any non-zero vector,
of which there are $q^n -1$. The second row must not be a linear multiple of
the first, yielding $q^n - q$ choices. The first two rows are now linearly independent,
and their span must have size $q^2$. The third row can be any vector not in the
span of the first two rows, giving us $q^n - q^2$ possibilities.

Continuing in this fashion, we deduce that
\[ |GL(V)| = (q^n-1)(q^n-q)(q^n - q^2) \cdots (q^n - q^{n-1}).\]
\item
\begin{enumerate}
\item[(a)] {\em Suppose that $G$ is a group acting on a set $\Omega$.
Define a map $\eta: G \longrightarrow \mbox{Sym}(\Omega)$
by $(g)\eta : \omega \mapsto \omega \cdot g$ for all $\omega \in
\Omega$ and for all $g \in G$. Prove that $\eta$ is a
homomorphism of groups.} \newline
\noindent {\bf Solution }\ Suppose that $u, v \in G$, then
$(uv)\eta : \omega \mapsto \omega \cdot (uv) = (\omega \cdot u) \cdot v$.
On the other hand $(u) \eta \circ (v) \eta : \omega \mapsto (\omega \cdot u) \cdot v$.
Now for each $g \in G$ we have $(g)\eta \circ (g^{-1}) \eta =
(g^{-1}) \eta \circ (g)\eta  = (1)\eta$ which is the identity map.
Therefore each $(g)\eta$ has a two-sided inverse and so is an
element of $\mbox{Sym}(\Omega).$ We have already established the
multiplicativity property which ensures that $\eta$ is a group homomorphism.

\item[(b)] {\em Suppose that $\eta: G \longrightarrow \mbox{Sym}(\Omega)$
is a group homomorphism. Show that one may define a group action of
$G$ on the set $\Omega$ by defining $\omega \cdot g$ to
be $(\omega)((g)\eta)$ for all $\omega \in \Omega$ and for all
$g \in G$. Show that this is indeed a group action.} \newline
\noindent {\bf Solution }\
Now suppose that $\eta : G \rightarrow \mbox{Sym}(\Omega)$
is a group homomorphism. We define an action of $G$ on $\Omega$ by
$\omega \cdot g = (\omega)(g)\eta.$ We check that this is an action:
$\omega \cdot 1 = (\omega)(1)\eta= (\omega)\mbox{Id}_{\Omega} 
= \omega$ for every
$\omega \in \Omega$. Also if $u, v \in G$, then
$ \omega \cdot (uv) = (\omega)(uv)\eta= (\omega)((u)\eta (v)\eta )
= ((\omega)(u)(\eta)) (v)(\eta) = (\omega \cdot u) \cdot v.$ Thus
we have an action.
\item[(c)] {\em Show that the procedures outlined in parts (a) and (b)
are mutually inverse (i.e. if you perform one, then the other, then
you recover the situation in which you started).} \newline
\noindent {\bf Solution }\ 
Suppose that we have an action denoted by a dot which gives rise to the group
homomorphism $\eta : G \rightarrow \mbox{Sym}(\Omega)$ outlined in part (a).
Now we use part (b) to define an action of $G$ on $\Omega$ denoted $\star$.
The definition of $\star$ is that if $\omega \in \Omega$ and $g \in G$,
then $\omega \star g = (\omega)((g)\eta) = \omega \cdot g$. Thus $\star$
is the same action as the original one.
Conversely suppose that you have a homomorphism $\eta:  G \rightarrow \mbox{Sym}(\Omega)$.
This gives rise to an action as outlined in part (b). According to part
(a) this gives rise to a homomorphism $\xi : G \rightarrow \mbox{Sym}(\Omega)$
defined by $(g)\xi : \omega \mapsto \omega \cdot g = (\omega)(g)\eta$.
Thus the maps $(g)\xi$ and $(g)\eta$ have the same domains and codomains, and agree
at all arguments. Thus $(g)\xi = (g)\eta$ for every $g$, and so
$\xi = \eta$.
\end{enumerate}
\end{enumerate}
\end{document}