\documentclass[12 pt]{article} \usepackage{amssymb} \usepackage{latexsym} \usepackage{epsf} \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}