How to classify finite groups ? : Part I

In these talks, we are going to study some basic tools which will enable us to classify some finite groups up to isomorphism. 

Aim: Classify finite groups up to isomorphism (We will explain what this means).

In this first talk we will introduce the basic tools we will use throughly without reference in the next talks.

-- Group Actions --

In this section, we define group actions and prove some important propositions which will is essential in all the next talks.

Definition: A left group action on a set $X$ is map $\theta: G \times X \rightarrow X$ satisfying 

1. Compatibility: $\theta(gh,x)= \theta(g,\theta(h,x))$
2. Identity: $\theta(e,x)=x$

The action can be considered as a group homomorphism $$\overline{\theta} : G \rightarrow \mathrm{Perm}(X) \  ; \ g \mapsto \theta(g,x)$$ With the action understood from the context, write $\theta(g,x) = g \cdot x$

Throughout $G$ acts on a finite set $X$

Definition: Let $X$ be a set, $x \in X$ and $\theta$ be an action of $G$ on $X$

• (Orbit) Define Orbit of $x$ by $\mathrm{Orbit}(x) = Gx =\{g\cdot x| g \in G \}$
• (Stabilizer) Define Stabilizer of $x$ by $\mathrm{Stab}(x) = \{g| g\cdot x = x\}$

Proposition (Orbit-stablizer formula): The map $\gamma : G/\mathrm{Stab}(x) \rightarrow Gx$ defined by 

$$g \mathrm{Stab}(x) \mapsto g\cdot x$$

is well-defined and bijective. As a result 

$$|G| = |\mathrm{Orbit}(x)| \cdot |\mathrm{Stab}(x)| = [G:\mathrm{Stab}(x)] \cdot |\mathrm{Stab}(x)|$$

proof. 

• $\gamma$ is well defined : $$g_1^{-1} g_2 \in S \implies (g_1^{-1} g_2)\cdot x = x \implies g_1 \cdot x = g_2 \cdot x$$
• $\gamma$ is bijective: let $$g_1 \cdot x = g_2 \cdot x \implies (g_2^{-1}g_1)\cdot x \implies (g_2^{-1}g_1) \in S \implies g_1 S=g_2 S$$ So $\gamma$ is injective. Surjectivity is immediate. 

$\square$

It is easy to see that the relation $E$ on $X$ defined by $$x E y \iff y = g \cdot x$$ is equivalence relation, and the equivalence class $[x]_E = \mathrm{Orbit}{(x)}$. So the set of all orbits is a partition of $X$ and is denoted by $G\backslash X$. Now if we choose a set representatives $R \subset X$ containing exactly a point from each orbit, we get the orbit equation

$$|X| = \sum_{x \in R} |\mathrm{Orbit}(x)| = \sum_{x \in R} [G:\mathrm{Stab}(x)]$$

Some terminology. An action $\theta$ of $G$ on $X$ is

1. Faithful if $\theta: G \rightarrow \mathrm{Perm}(G)$ is injective.
2. Free if for all $x \in X$, $\mathrm{Stab}(x) = \{e\}$
3. Transitive: If for some $x$ the orbit $Gx = X$. In fact this implies that for all $x\in X$, $Gx = X$. 

We want for sake of completeness to state some exercises and examples

Examples:

1. (Cayley theorem)  The action of $G$ on itself by left multiplication, defined by $g \cdot x = gx$ is faithful. As a result any group $G$ can be realized as a subgroup of $\mathrm{Perm}(G)$.
2. The group $O(n;\mathbb{R}) = \{A | AA^{T}=I \}$ of orthogonal $n \times n$ matrices acts on the unit $(n-1)$-sphere $S^{n-1} \subset \mathbb{R}^n$ by left multiplication, that's $A \cdot X = AX$. This action is transitive and $$\mathrm{Stab}(e_1) = \{ A \}$$ So the proposition above induces bijection $$\gamma : O(n)/O(n-1) \rightarrow S^{n-1}$$ . In fact if we endow $O(n)$ with its subspace topology, then $\gamma$ induces topological homomorphism from the quotient space $O(n)/O(n-1)$ to $S^{n-1}$.
3. The group $\Gamma = SL_2(\mathbb{\mathbb{R}})$ acts on the half plane $\mathbb{H}=\{z \in \mathbb{C} | \mathrm{Im}(z) > 0 \}$ as following $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \cdot z = \frac{az+b}{cz+d}$$ (Prove this is indeed an action). This action is transitive (prove this) and its isotropy subgroup is $SO(2)$ (prove this !)
4. For any permutation $\sigma \in S_n$, the cyclic group $\langle \sigma \rangle$ acts on $[n] =\{1,\cdots,n\}$, by the action $\sigma \cdot i = \sigma(i)$. The orbits form a partition $\langle \sigma \rangle \backslash [n] = \{O_1,\cdots,O_r \}$. We define $$\sigma_i(x) = \begin{cases}  \sigma(x)  &, \text{if} \ x \in O_i \\ x &, \text{if} \ x \notin O_i \end{cases}$$ It is clear that $\sigma_i \sigma_j = \sigma_j \sigma_i$ for $i \neq j$ and $$\sigma = \sigma_1 \cdots \sigma_r$$ Moreover, if $O_i = \mathrm{Orbit}(k)$ and $k_r = \sigma^r (k)$, then $$\sigma_i = (k_0 k_1  \cdots k_{|O_i|-1})$$ That's $\sigma_i$ is a cycle permutation. We conclude: Every permutation is composition of disjoint cycles.

Exercise. Suppose $G$ acts faithfully and transitively on $X$, where $|X| = p$ is prime and $\{e\} \neq H \trianglelefteq G$. Then the restriction of the action to $H$ is transitive

Exercise. (Burnside lemma) Let $G$ acts on a finite set $X$. Then $$|G\backslash X| = \frac{1}{|G|} \sum_{g \in G} |X^{g}|$$ For an application in combinatorics see wiki page.

The main group action we consider is the Conjugacy action where $G$ acts on itself by conjugation. In otherwords, $X=G$ and $\theta(g,x) = gxg^{-1}$. The orbit $\mathrm{Orbit}(x)$ is called conjugacy class and denoted by $C_x$, and the stablizer $\mathrm{Stab}(x)$ is called the centralizer of $x$ and denoted by $C_G(x)$. Clearly, the fixed point set $X^G$ is the center $$Z(G) = \{ x | gx =xg \ \ \text{for all} \  g \in G\} \trianglelefteq G$$ . The orbit equation is called the class equation

$$|G| = |Z(G)| + \sum_{C_g \text{is non-trivial conjugacy class}} [G:C_G(x)]$$

This can be generalized a bit to conjugacy action on subgroups, in which case the orbit $\mathrm{Orbit}(H)$ is the set of conjugate subgroups  $\{gHg^{-1}|g \in G\}$ and the stabilizer is the normalizer $N(H) = \{g H g^{-1} | g \in G \}$. Clearly $H < N(H)$, So we obtain

$$|H| \mid |N(H)| \mid |G|$$

-- Normal subgroups and Semi-direct product --

We begin with reviewing the basics of Normal Subgroups.

Definition. (Normal Subgroups) Let $N <G$, we say $N$ is normal subgroup of $G$ and write $N \trianglelefteq G$ if for all $g \in G$ we have $gN = Ng$.

The main points we want to recall are

• $G/N$ with the operation defined by $(aN) \cdot (bN) = (ab)N$ defines a group.
• $N \trianglelefteq G$ if and only if $N = \mathrm{ker}(\phi)$ for some homomorphism $\phi : G \rightarrow G'$.

Examples.

1. Any subgroup of an Abelian group is normal.
2. Let $D_{2n}$ be the Dihedral group. Recall that this is the group of reflection $r$ around $x$-axis and the counter clockwise rotation $r$ by angle $\frac{2\pi}{n}$ around the center. Under the operation of composition, these form a group with $2n$ elements. The following relations hold $$s^2 = 1 \ , \ r^n =1 \ , \ sr = r^{-1}s$$ One checks that $\langle r \rangle \trianglelefteq D_{2n}$
3. Let $A_4 < S_4$ be the subgroup of even permutations. Let $H$ be the set of all $$\{I,(12)(34),(13)(24),(14)(23)\}$$ $H \cong C_2 \times C_2$ is normal subgroup of $A_4$ (prove this) 

If we have a group $G$ and a non-trivial normal subgroup $N < G$, we can form a group $G/N$. of lower order. Suppose that we have an understanding of $G/N$. Sometimes we get $G=NK$, where $K \cong G/N$.

Every element of $G$ is written as $nk$ where $n \in N, k \in K$. Now let's study the multiplication of $G$ in terms of multiplications of $N$ and $K$. We get $$(n_1 k_1)(n_2 k_2) = ( n_1 k_1) (n_2 (k_1^{-1} k_1) k_2) = (n_1(k_1n_2k_1^{-1}))(k_1k_2)$$ where we note that $(k_1n_2k_1^{-1}) \in N$ by normality. 

So to understand the multiplication it suffices to understand the conjugation homomorphism $$C : K \rightarrow \mathrm{Aut}(N) \ ; \ (k \mapsto k^{-1}nk)$$

In other words $G$ is completely determined uniquely in terms of the subgroups $N$ and $K$, and the conjugation $C : K \rightarrow \mathrm{Aut}(N)$. Suppose we don't have $G$ how to reconstruct it (up to isomorphism) ? This motivates the following definition 

Definition and proposition (Semi-direct product) Let $N$ and $K$ be groups. Suppose $\phi: K \rightarrow \mathrm{Aut}(N)$ is a group homomorphism. Then the set $N \times K$ endowed with the operation $\star$ defined by $$(n_1,k_1) \star (n_2,k_2) = (n_1\phi(k_1)(n_2),k_1k_2)$$ is a group. We denote this group by $N \rtimes_\phi K$.

Note that: if $\phi$ is the trivial homomorphism, then $N \rtimes_\phi K$ is the direct product group.

Proposition. Let $N, K < G$. Suppose that  $N \trianglelefteq G$, $G = NK$ and $N \cap K = \{e\}$. Then $G \cong N \rtimes_C K$, where $C:K \rightarrow \mathrm{Aut}(N)$ is the conjugation homomorphism.

Example. Dihedral group $D_{2n} \cong C_n \rtimes_{\mathrm{inv}} C_2$, where $$\mathrm{inv}: \langle a \rangle =C_2 \rightarrow \mathrm{Aut}(C_n) \ ; \ a \mapsto (r \mapsto  r^{-1})$$

Moral of this section:

To understand a group $G$ , it is a good idea to study its proper subgroups hoping to "rebuild" $G$ from these subgroups.

In the next talk we will revisit this principle in depth, but now we will only content ourselves with the first sylow theorem which provides us with some proper subgroups of $G$ as intended.

-- First Sylow theorem --

Theorem. (Cauchy) If $p \mid |G|$, then there is an element $g \in G$ of order $p$

proof.

The proof consists of two parts:

First part: We prove the case $G$ is abelian.

The proof proceeds by induction on $|G|$. Pick an element $e \neq g$, if $\mathrm{ord}(g) =pk$, then $g^{k}$ has order $p$ and we are done, otherwise $p \nmid \mathrm{ord}(g)$ that's $p \mid [G:\langle g \rangle]$. We consider the factor group $G/ \langle g \rangle$. This is a group because $G$ is abelian, and $p \mid |G / \langle g \rangle | < |G|$.

By induction hypothesis there is an element $\overline{a} \in G / \langle g \rangle$ of order $p$, that means that $a^p= g^i$ for some $i$. It follows that $(a^{\mathrm{ord}(g)})^{p}= e$, But $\overline{a}^{\mathrm{ord}(g)}$ has order $p$, in particular $a^{\mathrm{ord}(g)} \neq e$. So, $\mathrm{ord}(a^{\mathrm{ord}(g)}) = p$

Second part: We prove the general case by induction on $|G|$ again.

 The case $|G| =1$ is trivial, now assume that the theorem holds for all groups $G'$ of order $\leq n$. Let $|G|=n$. In order to make use of the first part we have to consider the abelian subgroup $Z(G) \leq G$.  Consider the two cases

1. $p \mid |Z(G)|$ : Then the theorem follows from the first part.
2. $p \nmid |Z(G)|$: From the class equation taken modulo $p$ $$0 \equiv |G| = |Z(G)| + \sum_{C_g \text{is non-trivial conjugacy class}} \frac{|G|}{|C_G(g)|}$$ there is $g \in G$ such that $p \nmid \frac{|G|}{|C_G(g)|} > 1$. In particular $p | |C_G(g)| < |G|$, now the result follow by the inductive hypothesis.

$\square$

Exercise. (Tricky proof of Cauchy theorem) Let $G$ be a group of order divisible by $p$

1. Let $\theta$ be a group action of $K$ on a finite set $X$. Denote by $X^{K}$ the set of fixed points of the action. More precisely, $X^{K} = \{x : Gx = \{x\} \}$. Prove that $$|X| \equiv |X^{K}| \pmod{p}$$
2. Let $X = \{ (x_1,\cdots, x_p) \in G^p | x_1 \cdots x_p = e\}$. By using an appropriate action of the cyclic group $K=\mathbb{Z}/p\mathbb{Z}$ on $X$, use (1.) to prove Cauchy theorem.

Theorem. (First Sylow theorem (S.T)) If $p | |G|$ then there is a sylow $p$-subgroup of $G$.

Proof. 

The idea of the proof is the same as the idea of the second part of proof of Cauchy theorem. We induct on $|G|$. If $|G|=1$ then the statement is trivial. Now write $|G| = p^e m$ where $p \nmid m$. Assume $e >0$ otherwise the statement is trivial.

• $p \mid |Z(G)|$ : Then by Cauchy theorem there is an element $x \in Z(G)$ of order $p$. Since $x$ commutes with all elements of $G$, then $\langle x \rangle$ is normal in $G$. Now let $G' = G / \langle x \rangle$.  As $|G'| = p^{e-1}m$, by induction hypothesis there is a sylow p-subgroup $H' < G'$. By Third isomorphism theorem $H' = H/ \langle x \rangle$ for some subgroup $H < G$ containing $x$. Clearly $H$ is a Sylow p-subgroup of $G$.
• $p \nmid |Z(G)|$ : From class equation choose some $g \in G$ such that $$p \nmid | [G:C_G(g)] > 1 \iff p^e \mid |C_G(g)| < |G|$$ By induction hypothesis $C_G(g)$ which is a subgroup of $G$ has a Sylow p-subgroup $S$ of order $p^e$. So $S$ is a Sylow p-subgroup of $G$.

$\square$

In the next talk we will prove the second and third Sylow theorems and classify systematically groups of order 12. Now we will entertain you with the statements of Sylow theorems and some exercises

Theorem. (Second Sylow theorem) Let $H \leq G$ and $S$ be a Sylow p-subgroup of $G$, then then there is a conjugate $S'=gSg^{-1}$ (which is again a Sylow p-subgroup) such that $S' \cap H$ is a Sylow p-subgroup of $H$.

We conclude immediately the following corollaries,

1. A p-subgroup $H < G$ is contained in a Sylow p-subgroup of $G$
2. All Sylow p-subgroups of $G$ are conjugate.

Theorem. (Third Sylow theorem) The number $N_p$ of Sylow p-subgroups of $G$ satisfies:

• $N_p \equiv 1 \pmod{p}$ and $N_p \mid |G|$
• $N_p = [G:N_G(S)]$ where $S$ is any p-subgroup

Exercise. (Another proof of first S.T) Let $\mathcal{P}$ be the set of all $p^e$-subsets of $G$. Define the action 

1. Prove that for all $U \in \mathcal{P}$, $\mathrm{Stab}(U) \mid |U|$
2. Prove that $p \nmid |\mathcal{P}|$ and use the orbit equation and (1.) to prove First S.T.

Exercise. (A proof of first S.T assuming Second S.T)

1. Compute the order of $GL_n(\mathbb{F}_p)$
2. Find a Sylow p-subgroup of $GL_n(\mathbb{F}_p)$
3. Prove First S.T by embedding $G$ into $GL_{|G|}(\mathbb{F}_p)$ and using Second S.T.
4. (Bonus) Compute the number of Sylow p-subgroups of $GL_n(\mathbb{F}_p)$.

History: we suggest ???