Cayley's Theorem

If \(G\) has a left action on a set \(X\), given \(g\in G\) let \(\ell_g:X\to X\) be defined by \(\ell_g(x)=g\cdot x\). Then

1. \( \ell_g \circ \ell_h= \ell_{gh} \quad \forall g,h\in G\)
2. \(\ell_e=1\) (identity function)
3. \(\ell_g\) is bijective for all \(g\in G\).

Proof. Straight forward.

Every left action of \(G\) on \(X\) gives a homomorphism \(\phi:G\to S_X:g\mapsto \ell_g\) with \(\phi(g)(x)=\ell_g(x)=g\cdot x\).

Proof. Straight forward.

Note. \(\phi:G\to S_X\) is known as a permutation representation of \(G\) on \(X\).

Note. If \(|X|=n\), then \(S_X\cong S_n\). So, \(\phi\) also gives a homomorphism to \(S_n\).
For example, consider a diheral group \(D_{2n}\). It acts on \(n\) vertices of \(n\)-gon, so there is a homomorphism \(D_{2n}\to S_n\). More precisely, if we let \(X=\{v_0,v_1,\dots, v_{n-1}\}\) be vertices of \(n\)-gon, and define a map \(v_i=i+1\) for \(i+1\in \{1,2,\dots, n\}\). Then we can write the elements of \(S_X\) as the elements of \(S_n\) as follow:

Let \(\phi:D_{2n}\to S_n\) be the permutation representation given by the action of \(D_{2n}\) on \(X\) where
• \(\phi(s)\) (\(s\in D_{2n}\) is a rotation)

\(s\cdot v_0=v_1, s\cdot v_1=v_2,\dots, s\cdot v_{n-1}=v_0\)
\[\implies \phi(s)=(1 2 \dots n)\]

• \(\phi(r)\) (\(r\in D_{2n}\) is a reflection)

\(r\cdot v_0=v_0, r\cdot v_1=v_{n-1}, \dots, r\cdot v_{i}= v_{n-i}\)
\[\implies \phi(r)=\begin{cases} (2 n)(3 n-1)\cdots(\frac{n+1}{2} \frac{n+3}{2}) & \text{if n is odd}\\ (2 n)(3 n-1)\cdots (\frac{n}{2} \frac{n+2}{2}) & \text{if n is even}\end{cases}\]

If \(\phi:G\to S_X\) is a homomorphism, then \(g\cdot x=\phi(g)(x)\) defines a group action of \(G\) on \(X\).

Proof. Straight forward.

Note. We say that the action is faithful if \(\ker(\phi)=\{e\}\)

The action of \(G\) on \(X\) is faithful \(\iff\) for all \(g\in G\backslash \{e\}\), there exists \(x\in X\) such that \(g\cdot x\neq x\).

Proof. \(\exists x\in X \text{ such that } g\cdot x=\ell_g(x)\neq x \iff \ell_g\neq \ell_e\).

(Cayley's Theorem) The left regular action of \(G\) on \(G\) is faithful. Consequently, \(G\) is isomorphic to a subgroups of \(S_G\). In particular, if \(|G|=n \lt +\infty\), then \(G\) is isomorphic to a subgroup of \(S_n\).

Proof. For \(g\in G\backslash \{e\}, g\cdot e=g\neq e\). So the left regular action is faithful. It implies that \(\phi:G\to S_G\) is injective. Thus, \(G = G/\ker(\phi)\cong Im(\phi) \leq S_G\) by the first isomorphism theorem. If \(|G|=n\lt +\infty, S_G\cong S_n\).

Sylow's Theorem

If \(G\) acts on \(X\), then \( \sim_G \) defined by \(x\sim y \iff \exists g\in G \text{ such that } g\cdot x= y\) is an equivalence relation on \(X\).

Proof. Straight forward.

Note. For \(x\in X\), \( [x]_{\sim}=\{y\in X: g\cdot x=y \; \forall g\in G\}= \mathcal{O}_x \) where \(\mathcal{O}_x\) is the \(G\)-orbit of \(x\in X\).

If \(G\) acts on \(X\), then orbits of \(G\) form a partition of \(X\). In particular, the action is transitive if and only if there is one orbit.

Proof. Straight forward.

Note. Suppose \(G\) acts on \(X\). Let \(S\subseteq X\) be the set of representatives for \(\sim_G\). Then naturally we have \[ |X|=\sum_{x\in S} |\mathcal{O}_x| \]

If \(G\) acts on \(X\), then for \(x\in X\), \(G_x\leq G\).

Proof.

• \(e\in G_x\)
• For \(g,h\in G_x\), \(gh\cdot x=g\cdot h\cdot x=g\cdot x=x \implies gh\in G_x\)
• \(g\in G_x\), \(g^{-1}\cdot x=g^{-1}\cdot g\cdot x=e\cdot x= x \implies g^{-1}\in G_x\)

(Orbit-Stabilizer Theorem) If \(G\) acts on \(X\), then for \(x\in X\), there eixsts a bijection \(G\backslash G_x \to \mathcal{O}_x\) defined by \(gG_x\mapsto g\cdot x\) where \(G_x\) is the stabilizer of \(x\) and \(\mathcal{O}_x\) is the \(G\)-orbit of \(x\).

Proof.
(well defined) Suppose \(gG_x=hG_x\). Then \(g^{-1}h\in G_x\). So, \(g^{-1}h\cdot x= x \implies h\cdot x =g\cdot x\).
(injective & surjective) Straight forward

Cor. If \(G\) acts on \(X\), then for \(x\in X\), \[ |\mathcal{O}_x| = [ G: G_x ] \]

Fun application Let \(G=S_n\) and \(X=\{1,\dots,n\}\). Note that the action of \(G\) on \(X\) is transitive (\(\sigma(i)=j, \tau(j)=k \implies \tau \circ sigma(i)=k\)). (Intuitively, for \(1\leq i\neq j \leq n\), there exists \(\pi\in S_n\) such that \(\pi(i)=j\) (e.g. \(\pi=(i j)\)). So \(\mathcal{O}_x=\{g\cdot x: g\in G\}=X \implies \) the action is transitive.) Thus, \(\mathcal{O}_i=X\) for all \(i=1,2,\dots, n\). So, \(|\mathcal{O}_x|=n=[G:G_i]=\frac{|G|}{|G_i|}=\frac{n!}{|G_i|}\implies |G_i|=(n-1)!\). So we can infer that the stabilizer of \(i\in X\); \(G_i=\{\pi\in S_n: \pi(i)=i\}\) is given such that \(G_i \cong S_{n-1}\).

If \(G\) is finite and \(H\leq G\) with \([G:H]=p\) where \(p\) is the smallest prime dividing \(|G|\), then \(H\unlhd G\).

Proof. Let \(K\) be the kernel of action of \(G\) on \(G/H\). Then it is easy to see that \(K\leq G_{eH}\). Let \(k=[G_{eH}:K]=\frac{|G_{eH}|}{K}\). Since \([G:K]=\frac{|G|}{|K|}=\frac{|G|}{|G_{eH}|}\cdot \frac{|G_{eH}}{|K|}=p\cdot k\), we have \(G/K\cong \text{ subgroup of } S_p\). So, \(|G/K|=kp \mid |S_p|=p! \implies k\mid (p-1)!\). On the other hand, \(k\mid |G|\). Since \(p\) is the smallest prime dividing \(|G|\), we must have \(k=1\). Thus, \(|G_{eH}|=|K| \implies G_{eH}=K\).

Note. If \(H\leq G\), then \(gHg^{-1}\leq G\) for all \(g\in G\).
Proof. For \(g\in G\), \(\phi:G\to G: h\mapsto ghg^{-1}\) is an automorphism. So, \(gHg^{-1}\) preserves the structure of \(H\).

(Sylow's Theorem) Let \(|G|=p^km\) where \(p\) is a prime, \(k\geq 1\) and \(\gcd(p,m)=1\). Then:

1. \(Sylow_p(G)\neq \emptyset\), and
2. If \(Q\) is a p-subgroup of \(G\) and \(P\) is a Sylow p-subgroup, then there is \(g\in G\) such that \(Q\leq gPg^{-1}\). (This implies that every Sylow p-subgroups are conjugate!)
3. \(n_p(G):=|Sylow_p(G)| \equiv 1 \mod p\) and \(n_p=[G:N_G(P)]\) for any \(P\in Sylow_p(G)\).

Some notes on conjugation action before the proof of Sylow's Theorem