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Three finite groups with the same numbers of elements of each order

Suppose that we have two finite groups $G$ and $H$ such that for each $n\in\mathbb{N}$ $$\left|\left\{ g\in G:\text{ord}\left(g\right)=n\right\} \right|=\left|\left\{ h\in H:\text{ord}\left(h\right)=n\right\} \right|$$ meaning, the number of elements of order $n$ in each group is equal. Does that imply that they are isomorphic? (With abelian groups I think it is true and easily deduced from the fundamental theorem of finitely generated abelian groups)

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Ido
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    No: eg http://math.stackexchange.com/questions/62331/three-finite-groups-with-the-same-numbers-of-elements-of-each-order – Mark Bennet Aug 12 '12 at 12:16
  • What a disappointment. I always thought this was true and I just waited for a day when I knew enough group theory to be able to prove it! –  Aug 12 '12 at 12:34
  • Why is this a disappointment? It tells you that groups are too interesting to be captured by an invariant as boring as order profile! – Qiaochu Yuan Aug 12 '12 at 14:08
  • @Qiaochu: It isn't disappointment as a fact itself, I agree with your statement, but it is disappointing due to my expectation! –  Aug 12 '12 at 14:57
  • Interesting example. Thank you. – Ido Aug 12 '12 at 15:18

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