Conjugation Action, Class Equation and Cauchy's Theorem

The conjugation action was briefly covered on the previous post, namely it is an action of G on itself defined by $g*a = gag^{-1}$. The prbots of this action are called the conjugacy classes and if g and h are the same conjugacy class, we say g and h are conjugated.

Examples: $\{1\}$ is always a conjugacy class. If G is abelian, every conjugacy class is a singleton (more generally, the conjugacy class of a is a singleton if and only if $a \in Z(G)$. In $D_8$ the conjugacy classes are $\{1\}, \{r, r^3\}, \{r^2\}, \{s, r^2s\}, \{rs, r^3s\}$. In $S_3 : \{1\}, \{(1 2 3), (1 3 2)\}, \{(1 2), (2 3), (1 3)\}$. In $S_n$ the conjugacy class of $\sigma$ is all permutations whose cycle decomposition has the same shape (ie the lenghts of the cycles are the same).

Theorem (class equation): let g be a finite group, let $g_1,...,g_k \in G\backslash Z(G)$ such that if $g_i$ and $g_j$ are conjugated then i = j. If $g \in G\backslash Z(G)$ then there exists an i such that g and $g_i$ are conjugated. Then $\vert G \vert = \vert Z(G) \vert + \sum[G:C_G(g_i)]$

Proof: let $O_g$ denote the conjugacy class of g. Then $G = \bigcup_{g \in G} O_g = \bigcup_{g \in G, \vert O_g \vert = 1} O_g \cup \bigcup_{g \in G, \vert O_g \vert > 1} O_g = \bigcup_i O_{g_i}$ Hence $\vert G \vert = \vert Z(G) \vert + \sum \vert O_g \vert = \vert G \vert = \vert Z(G) \vert + [G: S_G(g_i)]$ where $S_G(g_i)$ is the stabilizer for conjugation action and equal to $C_G(g_i)$

Theorem: Let $\vert G \vert = p^n, n \in \mathbb{Z}_{>0}$. Then $Z(G) \neq \{1\}$.

Proof: $\vert G \vert = \vert Z(G) \vert + \sum [G: C_G(g_i)]$ note that the last part is divisible by p. So $0 = \vert Z(G) \vert + \sum 0$ mod p so p divides $\vert Z(G) \vert \geq 1. \vert Z^i(G) \vert > p > 1$ and $Z(G) \neq \{1\}$.

Corollary: if $\vert G \vert = p^2$ then either $g \cong \mathbb{Z}/p^2\mathbb{Z}$ or $G \cong (\mathbb{Z}/p\mathbb{Z})^2$

Theorem (Cauchy’s theorem): Let G ve a finite group and p prime dividing $\vert G \vert$ then there exists a $g \in G$ of order p.

Proof: let’s prove the theorem when G is abelian. We proceed by generalized induction. Pick an element $x \in G$. If p divides $\vert x \vert = pd$ then $x^d$ is order p. If p does not divide $\vert x \vert$ then $\vert G/<x> = \frac{\vert G \vert}{\vert x \vert}, \vert G/<x> \vert < \vert G \vert$. By induction there exists $y \in G$ such that $y<x>$ is order p. Since $y \not\in <x> and (y<x>)^p = <x>$ then $y^p \in <x>$ and $<y^p> < <y>, [<y> : <y^p>] = gcd(\vert y \vert , p) \neq 1$ so p divides $\vert y \vert$. We are done by the cyclic group case. General case: (we also proceed by generalized induction). By the class equation $\vert G \vert = \vert Z(G) \vert + \sum \vert [G: C_G(g_i)]$. Assume p divides all $[G: C_G(g_i)]$ then p divides $\vert Z(G) \vert$ so we are done by the abelian case. Otherwise, i does not divide $[G: C_G(g_i)]$ for some i. $\vert G \vert = [G: C_G(g_i)]\cdot \vert C_G(g_i) \vert$ (p divides $\vert G \vert$ and $\vert C_G(g_i)\vert$) $\vert C_G(g_i) \vert < \vert G \vert$ so by induction we’re done.