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Given a finite group $G$, let $\{(1,1),(m_1,n_1),\ldots,(m_r,n_r)\}$ be the list of pairs $(m,n)$ in which $m$ is the order of some element, and $n$ is the number of elements with this order. The order of $G$ is thus $1+n_1+\cdots+n_r$, and the pair $(1,1)$ accounts for the neutral element.

Let $G,G'$ be two finite groups, with the same list. Is it true or not (I bet not) that $G$ and $G'$ are isomorphic ? If not, please provide a counter-exemple.

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For what it's worth, it's true for finite abelian groups. –  Gerry Myerson Sep 24 '10 at 12:40

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up vote 20 down vote accepted

There are easy examples that are $p$-groups. For instance, the mod 3 Heisenberg group is the nilpotent group with presentation $\left < a,b,c \;\bigg |\, [a,b] = c, [a,c] = [b,c] = a^3 = b^3 = c^3 = 1 \right >$ has order 27, and all but the trivial element of order 3. This has the same order portrait as $C_3^3$ where $C_3 = \mathbb Z / 3\mathbb Z$ is the cyclic group of order 3.

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Thanks a lot, Bill. –  Denis Serre Sep 24 '10 at 12:33

See this question and the first answer:

Order information enough to guarantee 1-isomorphism?

A fortiori, any counterexample given to that question will work for your question as well.

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