Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
For what it's worth, it's true for finite abelian groups. –  Gerry Myerson Sep 24 '10 at 12:40
add comment

2 Answers 2

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.

share|improve this answer
    
Thanks a lot, Bill. –  Denis Serre Sep 24 '10 at 12:33
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.