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.
Best Answer
$\newcommand{\Size}[1]{\lvert #1 \rvert}$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.