# Abstract Nonsense

## Review of Group Theory: Conjugacy Classes of The Product of Two Groups

Point of Post: In this post we show how one can completely classify the conjugacy classes of the product of two groups $G$ and $H$ given the conjugacy classes of each.

Motivation

It’s a natural question whether given two groups $G$ and $H$ and their set of conjugacy classes can we tell anything about the conjugacy classes of the product of the two groups. In fact, (as may or may not be obvious) there is an extremely easy way to find the set of all conjugacy classes of $G\times H$. Namely, if $\mathcal{C}_G$ denotes the set of conjugacy classes of $G$ and $\mathcal{C}_H$ those of $H$ then $\mathcal{C}_{G\times H}$ is equal to $\left\{A\times B:A\in\mathcal{C}_G\text{ and }B\in\mathcal{C}_H\right\}$. Thus, in particular $\#\left(\mathcal{C}_{G\times H}\right)=\#\left(\mathcal{C}_G\right)\#\left(\mathcal{C}_H\right)$.

Classifying the Conjugacy Classes of  The Product Group

We now aim to classify the conjugacy classes of $G\times H$ given those of $G$ and $H$. Indeed:

Theorem: Let $G$ and $H$ be groups which have the sets of conjugacy classes $\mathcal{C}_G$ and $\mathcal{C}_H$ respectively. Then, if $\mathcal{C}_{G\times H}$ denotes the set of conjugacy classes of $G\times H$ then

$\text{ }$

$\mathcal{C}_{G\times H}=\left\{A\times B:A\in\mathcal{C}_G\text{ and }B\in\mathcal{C}_{H}\right\}$

$\text{ }$

Proof: Let $A\in\mathcal{C}_G$ and $B\in\mathcal{C}_H$ with $g\in A$ and $h\in B$, we claim that $A\times B=\mathcal{C}_{(g,h)}$. Indeed, let $(r,s)\in \mathcal{C}_{(g,h)}$ then there exists some $(x,y)\in G\times H$ such that $(r,s)=(x,y)(g,h)(x,y)^{-1}=(xgx^{-1},hyh^{-1})$ and so $r=xgx^{-1}$ and $s=yhy^{-1}$ so that $(r,s)\in A\times B$. Conversely, let $(r,s)\in A\times B$ then $r=xgx^{-1}$ and $s=yhy^{-1}$ for some $x\in G$ and $y\in H$ and so $(r,s)=(xgx^{-1},yhy^{-1})=(x,y)(g,h)(x,y)^{-1}$ and so $(r,s)\in\mathcal{C}_{(g,h)}$. Thus $A\times B=\mathcal{C}_{(g,h)}$ as claimed and so $A\times B\in\mathcal{C}_{G\times H}$.

Conversely, let $\mathcal{C}_{(g,h)}=C\in\mathcal{C}_{G\times H}$ and let $A=\pi_1(C)$ and $B=\pi_2(C)$ where $\pi_k$ is the $k^{\text{th}}$ canonical projection onto the $k^{\text{th}}$ coordinate group. Our first claim is that $C=A\times B$. Indeed, evidently $C\subseteq A\times B$, so to show the reverse inclusion let $(a,b)\in A\times B$. We evidently have then that there is some $a'\in G$ and $b'\in H$ so that $(a,b')\in C$ and $(a',b)\in C$. Thus, by definition there must exists $(x,y),(u,v)\in G\times H$ so that $(a,b')=(x,y)(g,h)(x,y)^{-1}=(xgx^{-1},yhy^{-1})$ and similarly $(a',b)=(ugu^{-1},vhv^{-1})$ and so in particular $a=xgx^{-1}$ and $b=vhv^{-1}$ and so $(a,b)=(x,v)(g,h)(x,v)^{-1}$ and so $(a,b)\in C$ as desired. What we now claim then is that $A=\mathcal{C}_g$ and $B=\mathcal{C}_h$. Indeed, let $a\in A$ then $(a,h)\in A\times B=\mathcal{C}_{(g,h)}$ and so $(a,h)=(x,y)(g,h)(x,y)^{-1}$ for some $(x,y)\in G\times H$ and so in particular $a=xgx^{-1}$ so that $a\in\mathcal{C}_g$. If $xgx^{-1}\in\mathcal{C}_g$ then $xgx^{-1}\in A$ since $(xgx^{-1},h)=(x,e)(g,h)(x,e)^{-1}$. Thus, $A=\mathcal{C}_g$, and similarly one can prove that $B=\mathcal{C}_h$. Thus, $C=\mathcal{C}_g\times\mathcal{C}_h$ as required.

The conclusion follows. $\blacksquare$

References:

1. Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.