MIT 18.701 — Lecture 13

Conjugacy Classes, Centralizer, Center, Class Equation, Extensions, Simple Groups

There are two ways that a group $G$ can act on itself:

Let $G$ act on itself via group multiplication: $g \cdot x = gx$ for all actions $g \in G$ acting on $x \in G$ .
Let $G$ act on itself via conjugation: $g \cdot x = gxg^{-1}$ for all actions $g$ acting on $x \in G$ .

We've explored the former already in the proof of Cayley's Theorem. But conjugation is also worth exploring!

In $GL_n(\mathbb{R})$ , conjugation is the same as change of basis.
The identity $\Stab_G(gx) = g\Stab_G(x)g^{-1}$ holds.
Conjugation measures commutativity: $gx = yg$ for $y = gxg^{-1}$ .

§ Group Actions by Conjugation

Definition (Conjugacy Classes and Centralizers). Let $G$ act on itself via conjugation. Then:

The conjugacy class of $x \in G$ is the orbit of $x$ ; that is, the set $C(x) := \{g x g^{-1} : g \in G\}$ .
The centralizer of $x \in G$ is the stabilizer of $x$ ; that is, the set $Z(x) := \{g \in G: g xg^{-1} = x\} = \{g \in G: gx = xg\}$ .
The center of $G$ is $Z(G) := \bigcap_{x \in G} Z(x) = \{g \in G \mid C(g) = \{g\}\} = \{g \in G \mid Z(g) = G\}$ .

Some useful facts about conjugacy classes and centralizers:

Fact 1. Any group $G$ can be partitioned into disjoint conjugacy classes (i.e. disjoint orbits).
Fact 2. The Orbit-Stabilizer Theorem says that for all $x \in G$ , we have $|G| = |C(x)| \cdot |Z(x)|$ .
Fact 3. The center $Z(G)$ is a normal subgroup of $G$ , since $Z(G)$ commutes with everything.

Example. The conjugacy classes of $S_3$ are $\{1\}$ , $\{(1 \ 2), (1 \ 3), (2 \ 3)\}$ , and $\{(1 \ 2 \ 3), (1 \ 3 \ 2)\}$ .

Example. The conjugacy classes of $D_5$ are $\{1\}$ , $\{r, r^{-1}\}$ , $\{r^2, r^{-2}\}$ , and $\{\text{all five reflections}\}$ .

Example. The conjugacy classes of $D_4$ are $\{1\}$ , $\{r, r^{-1}\}$ , $\{r^2\}$ , $\{\text{diagonal reflections}\}$ , and $\{\text{midline reflections}\}$ .

Remark. The conjugacy classes capture the geometric distinctions between $D_4$ and $D_5$ . In particular, all reflections in $D_5$ look the same, whereas reflections in $D_4$ come in two different forms.

§ The Class Equation: Classifying Groups of $p^2$ Order

Definition (Class Equation). If finite $G$ has conjugacy classes $C_1, \dots, C_k$ , the class equation says $|G| = \sum_{i = 1}^k |C_i|$ .

Example. The class equations of $D_4$ and $D_5$ read $8 = 1 + 2 + 1 + 2 + 2$ and $10 = 1 + 2 + 2 + 5$ , respectively.

Definition ( $p$ -group). For primes $p$ , a group $G$ is a $p$ -group if $|G| = p^k$ for some $k \in \{0, 1, 2, \dots\}$ .

Example. All of the following are $p$ -groups:

\{1\} ~~ \text{ or } ~~ C_p ~~ \text{ or } ~~ C_p \times C_p ~~ \text{ or }~~ C_{p^2} ~~ \text{ or } ~~ \left \{ \left[\begin{smallmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{smallmatrix}\right]: x, y, z \in \mathbb{F}_p\right \}

The big goal of today's lecture will be to use the Class Equation to prove the following:

Theorem. If group $G$ has order $p^2$ for some prime $p$ , then either $G \cong C_p \times C_p$ or $G \cong C_{p^2}$ .

The proof of this theorem will come in three steps.

Theorem. (Claim #1: Nontrivial Center) Every nontrivial group of order $p^k$ has a nontrivial center.

Proof: Suppose the center has size $n$ . Then the Class Equation reads:

\underbrace{1 + \dots + 1}_{n \text{ times}} ~ + ~ |C_{n + 1}| + \dots + |C_\ell| = p^k.

The key observation is that, for all $C_i \in \{C_{n + 1}, \dots, C_{\ell}\}$ , we have $p \mid |C_i|$ by Orbit-Stabilizer. So the above equation modulo $p$ reads $n \equiv 0 \pmod{p}$ . But $n \neq 0$ since $\mathrm{id} \in Z(G)$ , so $n \geq p$ , done. $\blacksquare$

Note that the above claim holds for any nontrivial group $G$ of prime-power order! But when $|G| = p^2$ , specifically…

Theorem. (Claim #2: Abelian) Since $|G| = p^2$ , it must be that $G$ is abelian.

Proof: Equivalently, we wish to show $|Z(G)| = p^2$ . Since $Z(G)$ is a subgroup of $G$ , by Lagrange's Theorem, $|Z(G)|$ divides $p^2$ . Claim #1 then says either $|Z(G)| = p$ or $|Z(G)| = p^2$ . Argue by contradiction: suppose $|Z(G)| = p$ .

Recall that $Z(G)$ is normal. Thus, if $|Z(G)| = p$ , then $|G / Z(G)| = p$ , meaning $G / Z(G) \cong C_p$ . In particular, if we pick any $x \not \in Z(G)$ , then $G$ must be generated by $x$ and $Z(G)$ . Therefore,

G = \{x^k z \mid k \in \{0, 1, \dots, p - 1\} \text{ and } z \in Z(G)\}.

Since $Z(G)$ commutes with $x$ , the above means $G$ is abelian, and in particular $|Z(G)| = p^2$ (contradiction). $\blacksquare$

We're now ready to finish the proof.

Theorem. (Claim #3: Classification) If $|G| = p^2$ and $G$ is abelian, then $G \cong C_p \times C_p$ or $G \cong C_{p^2}$ .

Proof: If $G$ has an element of order $p^2$ , then $G \cong C_{p^2}$ . The only other case is that every $x \in G$ has order dividing $p$ .

In this latter case, we claim $G$ may be viewed as an $\mathbb{F}_p$ vector space. To confirm this is legal, we must check:

Does $G$ have abelian vector addition? Well, yes—this follows from Claim #2.
Does $G$ have well-defined scalar multiplication? Well, yes—just define $\lambda x = \overbrace{x + \dots + x}^{\lambda \text{ times}}$ .

(To check this is well-defined, we require that $px = 0$ for all $x \in G$ . But this is true because $x$ has order dividing $p$ .)

Recall that all $\mathbb{F}_p$ vector spaces are isomorphic to $\mathbb{F}_p^k$ (i.e. pick a basis). Thus, $G \cong \mathbb{F}_p^2$ , so $G \cong C_p \times C_p$ . $\blacksquare$

§ Definitions: Group Extensions

Definition (Extensions). For group $G$ and normal subgroup $N \unlhd G$ , say $G$ is an extension of $G/N$ by $N$ .

Suppose $G$ is an extension of $A$ by $B$ . Some warnings:

The extension $G$ need not be unique. For example, $C_6$ and $S_3$ are both extensions of $C_2$ by $C_3$ .
$G$ need not be an extension of $B$ by $A$ . For example, $S_3$ is an extension of $C_2$ by $C_3$ , but not of $C_3$ by $C_2$ .
$A$ need not be a subgroup of $G$ . For example, $\mathbb{Z}$ is an extension of $C_2$ by $\mathbb{Z}$ .

Definition (Split Extension). Say $G$ is a split extension of $G/N$ by $N$ , with $\pi: G \to G/N$ , if there exists a subgroup $H \subseteq G$ such that $\pi |_H: H \to G/N$ is an isomorphism.

In other words, $G$ is a split extension of $A$ by $B$ if the quotient $G / B \cong A$ is visibly embedded within $G$ .

Example. For any groups $G$ and $H$ , $G \times H$ is a split extension of $G$ by $H$ .

Example. For any $n$ , the isometry group $M_n$ is a split extension of $O_n(\mathbb{R})$ by $\mathbb{R}^n$ . In other words, the quotient $O_n(\mathbb{R}) \cong M_n / \mathbb{R}^n$ is visibly embedded within $M_n$ .

§ Definitions: We're a Finite Simple Group…

Definition (Simple Groups). A nontrivial group $G$ is simple if it has no normal subgroups besides $\{1\}$ and $G$ .

Example. For any prime $p$ , the group $C_p$ is simple. For any $n \geq 5$ , the group $A_n$ is simple—this is nonobvious!

The classification of all finite simple groups looks like:

Groups of prime order $C_p$ .
Groups of the form $A_n$ for some $n \geq 5$ .
Groups of Lie Type, constructed from matrix multiplication.
… and exactly $26$ others, called sporadic groups.

For further details, see here: https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups.

§ Group Actions by Conjugation

§ The Class Equation: Classifying Groups of p2p^2p2 Order

§ Definitions: Group Extensions

§ Definitions: We're a Finite Simple Group…

§ The Class Equation: Classifying Groups of $p^2$ Order