## If a set of conjugacy class representatives of a finite group commute pairwise, the group is abelian

Let $g_1,g_2,\ldots,g_n$ be representatives of the conjugacy classes of a finite group $G$ and suppose that these elements commute pairwise. Prove that $G$ is abelian.

Let $G$ act on itself by conjugation. Note that, in particular, $g_1,g_2,\ldots,g_n \in \mathsf{stab}(g_k)$ for all $k$ under this action. Let $x \in G$; then $x = ag_ia^{-1}$ for some $a \in G$ and $g_i$. Thus $x \in a \mathsf{stab}(g_k) a^{-1}$ for each $k$ (since $g_i$ stabilizes each $g_k$), and more generally, $x \in \bigcup_{a \in G} a \mathsf{stab}(g_k)a^{-1}$ for all $k$. So $G = \bigcup_{a \in G} a \mathsf{stab}(g_k)a^{-1}$ for each $k$. Since $G$ is finite, by this previous exercise, $\mathsf{stab}(g_k)$ must not be a proper subgroup; that is, $\mathsf{stab}(g_k) = G$ for each $g_k$.

Now let $a,b \in G$, with $a = xg_ax^{-1}$ and $b = yg_by^{-1}$. Now $ab = xg_ax^{-1}yg_by^{-1}$ $= xx^{-1}g_ag_byy^{-1}$ $= g_bg_a = yy^{-1}g_bg_axx^{-1}$ $= yg_by^{-1}xg_ax^{-1}$ $= ba$. Thus $G$ is abelian.

• j2kun  On October 7, 2011 at 11:05 pm

What is $A$? representatives of the the non-trivial orbits of the action of conjugation?

• nbloomf  On October 10, 2011 at 8:38 am

Do you mean little $a$? This is just an element of $G$ which conjugates our arbitrary $x$ into one of the $g_i$.

• nbloomf  On December 15, 2011 at 10:10 am

Ah- I understand the question now. $A$ should be $G$. I’m a little slow. 🙂

• Gobi Ree  On December 12, 2011 at 1:52 am

This is trivial, but isn’t the notation $a=xg_ax^{-1}$ awkward? it will be better $a=xg_ix^{-1}$.

• nbloomf  On December 15, 2011 at 10:15 am

I was probably thinking of $g$ as a function that takes group elements to their conjugacy class representatives.