Hi I am trying to solve the following problem:
Let $g_1,g_2,\ldots,g_r$ representatives of conjugacy classes of the finite group $G$ such that they all commute with each other. Show $G$ is abelian. Hint: some $g_i's$ determine $Z(G)$.
I have found a solution here. But I do not understand why the following statement:
In particular, $g_1,g_2,\ldots,g_r \in \mathsf{stab}(g_i)$ for all $i$ under the conjugation of $G$.
Here is my thought process so far...
We want to show that $g_i\cdot a=a$ for $a\in G$ in order to show that an arbitrary $g_i\in\mathsf{stab}(g_i)$. We also know that $g_ig_j=g_jg_i$ for all representatives of the conjugacy classes.
Notice $g_1\cdot a = g_1(a)g_1^{-1}$. And this is where I get stuck. Any help would be much appreciated. Thanks!
In general, if $g, h \in G$ commute, then $g \in \mathrm{stab}(h)$ if $G$ acts on itself by conjugation (and vice versa, i.e. $h \in \mathrm{stab}(g)$). To see this, note $g \cdot h = ghg^{-1} = (gh)g^{-1} = (hg)g^{-1} = h(gg^{-1}) = h$. In fact, $\mathrm{stab}(h)$ consists exactly of elements of $G$ which commute with $h$. Indeed, we have already shown that the element of $G$ which commute with $h$ belong to $\mathrm{stab}(h)$. Now suppose $g \in \mathrm{stab}(h)$. Then $g \cdot h = ghg^{-1} = h$, so $ghg^{-1}g = hg$, i.e. $gh = hg$.