For 2 groups to be isomorphic, they clearly have to have the same class equation (i.e., they must have the same number of conjugacy classes of each size), but I'm wondering when the converse is not true. Someone pointed out to me that different abelian groups of the same order form a counterexample, since in an abelian group, every element is in its own conjugacy class.

Are there less trivial counterexamples?


Look at the non-isomorphic non-abelian groups of order 8: the dihedral group $D_4$ and the quaternion group $Q$.

  • 3
    $\begingroup$ The first time you can also get the orders of the class reps to match is order 16. $\endgroup$ – Jack Schmidt Dec 30 '13 at 22:03
  • $\begingroup$ Do not know if the OP means 2-groups or two groups ... $\endgroup$ – Nicky Hekster Dec 30 '13 at 22:45
  • $\begingroup$ I meant two groups. I guess I have to be careful when I have a prime number of groups. $\endgroup$ – Nishant Jan 1 '14 at 2:23

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