# Isomorphism and number of subgroups

This question arose while I was trying to work out examples for the second question of this thread: Reconstruction Conjecture: Group theoretic formulation?

In the beginning, I considered some computable properties of groups and wondered whether two groups of the same order having equal value for that computable property would necessarily be isomorphic. For instance, take centers of groups and it is not difficult to find many specific examples where two groups have the same order and isomorphic centers but then the two groups are not necessarily isomorphic. Considering lattice of groups, Scott Carnahan has already given a counterexample there.

Are there any two finite groups of the same order that have the same number of subgroups?

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For the history of the question: tea.mathoverflow.net/discussion/647/collaborative-math-news-tab/… –  Unknown Sep 4 '10 at 17:04
A much stronger question was asked and answered (negatively!) here: mathoverflow.net/questions/35455/… –  Cam McLeman Sep 4 '10 at 19:37
I just noticed that you yourself posted this link in the other question. Why doesn't this answer your question completely? –  Cam McLeman Sep 4 '10 at 20:05
I think in my question there is no requirement for bijection. So, this question generalizes the other. –  Unknown Sep 5 '10 at 7:06
You have the logic backwards. If there can be non-isomorphic groups with a structure-preserving bijection between their subgroups, certainly there can be non-isomorphic groups with the same number of subgroups. –  Cam McLeman Sep 5 '10 at 23:45
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There are pairs G,H of nonisomorphic p-groups with isomorphic subgroup lattices (and therefore of the same order). The book Subgroup Lattices of Groups", by R. Schmidt, is an excellent reference on this subject.
No, they are not cyclic. If two p-groups (same $p$) have isomorphic lattices of subgroups, then they have the same length longest chains of subgroups. For a p-group of order $p^n$, the length of this chain is exactly $n$. Hence the groups have the same order. –  Mark Sapir Sep 4 '10 at 20:10