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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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  • $\begingroup$ For the history of the question: tea.mathoverflow.net/discussion/647/collaborative-math-news-tab/… $\endgroup$
    – Unknown
    Sep 4, 2010 at 17:04
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    $\begingroup$ A much stronger question was asked and answered (negatively!) here: mathoverflow.net/questions/35455/… $\endgroup$ Sep 4, 2010 at 19:37
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    $\begingroup$ I just noticed that you yourself posted this link in the other question. Why doesn't this answer your question completely? $\endgroup$ Sep 4, 2010 at 20:05
  • $\begingroup$ I think in my question there is no requirement for bijection. So, this question generalizes the other. $\endgroup$
    – Unknown
    Sep 5, 2010 at 7:06
  • $\begingroup$ 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. $\endgroup$ Sep 5, 2010 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.

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  • $\begingroup$ ...and the same number of subgroups(which is always two). $\endgroup$
    – Unknown
    Sep 4, 2010 at 18:31
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    $\begingroup$ 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. $\endgroup$
    – user6976
    Sep 4, 2010 at 20:10
  • $\begingroup$ So, is it true that there are two non-isomorphic groups with equal order and with equal number of subgroups? $\endgroup$
    – Unknown
    Sep 5, 2010 at 7:22
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    $\begingroup$ @Elohemahab Solomon: They have an equal number of subgroups, because that is the number of vertices in the subgroup lattice. See en.wikipedia.org/wiki/Lattice_of_subgroups $\endgroup$
    – S. Carnahan
    Sep 5, 2010 at 9:15

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