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.

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  • 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.

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