# Prove that the number of elements of every conjugacy class of a finite group G divides the order of G.

Prove that the number of elements of every conjugacy class of a finite group $G$ divides the order of $G$.

I'm studying for my Group Theory exam and this was a question on a previous exam.

I know I have to use the counting orbit theory, but I have no idea how this would work. Furthermore I have some difficulties with understanding what conjugacy classes are. We use the book "Groups and Symmetry" by M.A. Armstrong which gives the following definition:

"Given elements x,y of a group G, we say that x is conjugate to y if $gxg^{-1}=y$ for some $g\in G$. The equivalence classes are called conjugacy classes."

But what does it mean that some elements are in the same conjugacy class? For me the deffiniton isn't clear enough. Could some one please try and explain some more? For example what the usage of conjugacy classes is.

• The conjugacy class of $x$ is $\{ gxg^{-1} : g \in G\}$, the set of all group elements that are conjugate to $x$. – Daniel Fischer Jan 1 '14 at 13:43

Regarding to Andreas's post including the very important equation in permutation groups, one of the usage can be:

If $G$ is a finite group and $H<G$ then $$G\neq\cup_{x\in G}x^{-1}Hx$$

The proof is routine and is based on defining an action of $G$ on $\Omega=\{Hx\mid x\in G\}$. The action is transitive and since $|\Omega|>1$ so $$\exists h\in G, |Fix(h)|=0$$. Now if we assume that every elements of $G$ is a conjugate element inside $H$ so $$\exists x\in G, \exists t\in H, h=x^{-1}tx$$ and so $(Hx)^h=Hx$ which is a contradiction.

$\newcommand{\Size}[1]{\lvert #1 \rvert}$Many questions in one...

You should have done some linear algebra. Then you should know that two $n \times n$ matrices $M, N$ represent the same linear map with respect to two different basis if and only if they are conjugate, that is, $g M g^{-1} = N$, where $g$ is the (invertible) matrix giving the change of bases.

As to the titular question, you should have done the orbit-stabilizer theorem, which tells you that if the finite group $G$ acts on the set $A$, and $a \in A$, then $$\Size{G} = \Size{a^G} \cdot \Size{G_a},$$ where $a^G$ is the orbit and $G_a$ is the stabilizer. In your case, let $G$ act on $A = G$ by conjugation, and you are done.

• Happy New Year my friend. ${+^{+_+}}$ – mrs Jan 1 '14 at 14:46
• Nicely written, Andreas. Happy New Year! – Namaste Jan 1 '14 at 15:07
• Thanks @B.S., a Happy New Year to you! – Andreas Caranti Jan 1 '14 at 17:56
• Thanks @amWhy, and a Happy New Year to you! – Andreas Caranti Jan 1 '14 at 17:57

The relation $\sim$ on the elements of $G$ defined by "$x\sim y$ iff there exist $g\in G$ so that $g^{-1}xg = y$" is an equivalence relation (this is a good exercise to prove), so $G$ can be partitioned into disjoint subsets of elements that are conjugates of one another (the equivalence classes under $\sim$).

These subsets are called the conjugacy classes, and the conjugacy class containing $x$ is called the conjugacy class of $x$. It can be constructed as the set of all conjugates of $x$, that is $\{g^{-1}xg\,|\,g\in G\}$.

Given any equivalence relation on a set $X$ you then have an equivalence class for each element $x\in X$. Simply put, the equivalence class of $x$ is the set of things related to (i.e. equivalent to) $x$.

Here you have a set (in fact a group) and you are told that there is an equivalence relation called conjugation. The sentence you quoted tells you the rule: We say $x$ and $y$ in $G$ are conjugate (i.e. "related" in some sense) if there exists $g\in G$ such that $gxg^{-1} = y$.

So clearly the equivalence class of $x$ under this equivalence relation is $\{gxg^{-1}\,|\,g\in G\}$, in other words the set of all conjugates of $x$. We give this the name of "conjugacy class" of $x$.

In fact $G$ viewed as a group acts on itself by conjugation. The orbit of an element $x\in G$ is then its conjugacy class, the stabilizer is something called the centralizer. Anyway the orbit-stabilizer theorem now tells you that the size of the conjugacy classes must divide $|G|$ when $G$ is a finite group.

As to why conjugates are useful, there are some nice applications mentioned above, you probably saw them all the way through linear algebra without knowing, especially in diagonalization etc. In fact they are a central part of group theory, geometrically they represent a kind of doing/undoing process. They are used in the mathematical solution of the Rubiks cube and facts about conjugation for symmetric groups were exploited in analysis of the Enigma machine.