Group actions and Sylow theorems

If a positive integer $m$ divides the order of a finite group $G$, does $G$ have a subgroup of order $m$? The answer is true for finite abelian groups, but it is not true for arbitrary groups. Sylow theorems discuss this situation when $m$ is a prime power.

Before discussing Sylow theorems, we first discuss group actions.

\begin{definition} An action of a group $G$ on a set $S$ is a function $G \times S \to S$ denoted $(g,x) \mapsto gx$ such that $\forall x \in S, \forall g_1, g_2 \in G, ex = x$ and $(g_1 g_2)x = g_1 (g_2 x)$. When such an action is given, we say that $G$ acts on the set $S$. \end{definition}

Let $G$ is a group and $H \leq G$, the action of group $H$ on the set $G$ where $(h,x) \mapsto hx$ is the product on $G$ is called a left translation. The action of $H$ on $G$ where $(h,x) \mapsto h x h^{-1}$ is called conjugation by $h$ and the element $h x h^{-1}$ is said to be a conjugate of $x$. If $K$ is any subgroup of $G$ and $h \in H$, $hKh^{-1} \cong K$. Thus $H$ acts on the set $S$ of all subgroups of $G$ by conjugation $(h,K) \mapsto hKh^{-1}$. The group $hKh^{-1}$ is said to be conjugate to $K$.

\begin{lemma} Let $G$ be a group acting on a set $S$ \begin{enumerate}[label=(\roman*)] \item The relation $\sim$ on $S$ defined by $x \sim x' \iff \exists g \in G, g x = x'$ is an equivalence relation. \item $\forall x \in S, G_x = \{g \in G \mid g x = x\}$ is a subgroup of $G$. \end{enumerate} \end{lemma}

The equivalence classes are called the orbits of $G$ on $S$, denoted by $\bar{x}$ for $x \in S$. The group $G_x$ is called the stabilizer of $x$. If $G$ acts on itself by conjugation, the orbits are called conjugacy classes. If a subgroup $H$ acts on $G$ by conjugation, $H_x = \{h \in H \mid h x h^{-1} = x\}$ is called the centralizer of $x$ in $H$ and is denoted $C_H(x)$. $C_G(x)$ is simply called the centralizer of $x$. If $H$ acts by conjugation on the set $S$ of subgroups of $G$, the subgroup of $H$ fixing $k \in S$, $\{h \in H \mid h K h^{-1} = K \}$ is called the normalizer of $K$ and is denoted $N_H(K)$. The group $N_G(K)$ is simply the normalizer of $K$. Every subgroup $K$ is normal in $N_G(K)$ and $K$ is normal iff $N_G(K) = G$.

\begin{theorem} If a group $G$ acts on a set $S$, the cardinal number of the orbit of $x \in S$, is the index $[G: G_x]$. \end{theorem}

Proof. Let $g, h \in G$. Since $gx = hx \iff g^{-1} h x = x \iff h G_x = g G_x$ it follows that $g G_x \mapsto g x$ is a well-defined bijection of the set of cosets of $G_x$ in $G$ onto $\bar{x}$. Hence $[G: G_x] = |\bar{x}|$. ◻

\begin{corollary} Let $G$ be a finite group and $K \leq G$. \begin{enumerate}[label=(\roman*)] \item The number of elements in the conjugacy class of $x \in G$ is $[G: C_G(x)]$ which divides $|G|$ \item If $\bar{x}_1, \cdots, \bar{x}_n$ are the distinct conjugacy classes of $G$, then $$|G| = \sum_{i=1}^n [G: C_G(x_i)]$$ \item The number of subgroups of $G$ conjugate to $K$ is $[G: N_G(K)]$ which divides $|G|$. \end{enumerate} \end{corollary}

Proof. (i) and (iii) follow from the previous theorem and Lagrange’s theorem. Since conjugacy is an equivalence relation, (ii) follows from (i). ◻

\begin{theorem} If a group $G$ acts on a set $S$, this induces a homomorphism $G \to A(S)$, where $A(S)$ is the group of permutations of $S$. \end{theorem}

Proof. If $g \in G$, define $\tau_g: S \to S$ by $\tau_g(x) = g x$. Since $x = g(g^{-1} x)$, $\tau_g$ is surjective. Similarly, $g x = gy$ implies $x = y$ whence $\tau_g$ is injective. Since $\tau_{g g'} = \tau_g \tau_{g'}$, the map $G \to A(S)$ given by $g \mapsto \tau_g$ is a homomorphism. ◻

\begin{corollary} If $G$ is a group, there is a monomorphism $G \to A(G)$. Hence every group is isomorphic to a group of permutations. In particular, every finite group $G$ is isomorphic to a subgroup of $S_n$ with $n = |G|$. \end{corollary}

Proof. Let $G$ act on itself by left translation and obtain $\tau: G \to A(G)$. If $\tau(g) = \text{id}_G$, then $\forall x \in G, gx = x$. In particular, $ge = e$ whence $g = e$ and $\tau$ is a monomorphism. Note if $|G| = n$, $A(G) \cong S_n$. ◻

If $G$ is a group, $\operatorname{Aut} G$, the set of all automorphisms of $G$ is a group under composition.

\begin{corollary} Let $G$ be a group. \begin{enumerate}[label=(\roman*)] \item $\forall g \in G$, conjugation by $g$ induces an automorphism of $G$. \item There is a homomorphism $G \to \operatorname{Aut} G$ whose kernel is $C(G) = \{g \in G \mid \forall x \in G, gx = xg\}$. \end{enumerate} \end{corollary}

Proof. (i) If $G$ acts on itself by conjugation, $\tau_g: G \to G$ given by $\tau_g (x) = g x g^{-1}$ is a bijection. $\tau_g$ is also a homomorphism and hence an automorphism. (ii) Let $G$ act on itself by conjugation. The homomorphism $\tau: G \to A(G)$ has image contained in $\operatorname{Aut} G$. Clearly $$g \in \operatorname{ker} \tau \iff \tau_g = \text{id}_G \iff \forall x \in G, g x g^{-1} = x$$ whence $\operatorname{ker} \tau = C(G)$. ◻

The automorphism $\tau_g$ is called the inner automorphism induced by $g$. $C(G)$ is called the center of $G$. An element $g \in C(G)$ iff the conjugacy class of $g$ consists of $g$ alone. Thus if $x \in C(G)$, then $[G:C_G(x)] = 1$. Thus if $G$ is finite, then $$|G| = |C(G)| + \sum_{i=1}^m [G: C_G(x_i)]$$ where $\bar{x}_1, \bar{x}_2, \cdots, \bar{x}_m$ are distinct conjugacy classes of $G$ and each $[G: C_G(x_i)] > 1$. The above equation is called the class equation.

\begin{proposition} Let $H$ be a subgroup of $G$ and $G$ act on $S$ the set of all left cosets of $H$ in $G$ by left translation. The kernel of the induced homomorphism $G \to A(S)$ is contained in $H$. \end{proposition}

Proof. The induced homomorphism $\tau: G \to A(S)$ is given by $g \mapsto \tau_g$ where $\tau_g: S \to S$ and $\tau_g (xH) = gxH$. If $g \in \operatorname{ker} \tau, \tau_g = \text{id}_S$ and $\forall x \in G, gx H = xH$. In particular, $geH=eH$ implying $g \in H$. ◻

\begin{corollary} If $H$ is a subgroup of index $n$ in a group $G$ and no nontrivial normal subgroup of $G$ is contained in $H$, then $G$ is isomorphic to a subgroup of $S_n$. \end{corollary}

Proof. Apply the proposition. The kernel of the induced homomorphism $G \to A(S)$ is a normal subgroup of $G$ contained in $H$ and thus must be $\langle e \rangle$. Hence $G \to A(S)$ is a monomorphism. ◻