# Is a finite group determined by the family of all its 2-generated subgroups?

At the last week I meet my old coauthor, Oleg Verbitsky who proposed me the following question. I think that here should be an easy counterexample, but I am not a pure group theorist and I am usually interested in infinite groups, so I decided to forward the question here. So, here it is:

For a group $G$ and a positive integer $k$ let $Sub_k(G)$ be the family of all subgroups of $G$ generated by its subsets of size at most $k$ and indexed by these subsets, that is $$Sub_k(G)=\{\langle K\rangle_K: K\subset G, |K|\le k\}.$$ Moreover, we shall call two indexed families $\{G_i:i\in I\}$ and $\{G’_{i’}:i’\in I’\}$ of groups isomorphic, if there exists a bijection $\delta:I\to I'$ such that for each $i\in I$ the groups $G_i$ and $G_{\delta(i)}$ are isomorphic.

Now suppose that we have two finite groups $G_1$ and $G_2$ of the equal size $n$. If I remember Oleg’s words right, the groups $G_1$ and $G_2$ are not necessarily isomorphic provided the families $Sub_1(G_1)$ and $Sub_1(G_2)$ are isomorphic, and there is a counterexample of two subgroups of small order. But the isomorphism of the families $Sub_1(G_1)$ and $Sub_1(G_2)$ implies the isomorphism of the groups $G_1$ and $G_2$, provided the groups are abelian. So Oleg asked me: are groups $G_1$ and $G_2$ isomorphic provided the families $Sub_2(G_1)$ and $Sub_2(G_2)$ are isomorphic? He proposed this question to a group theorist, who expects that the answer is negative in general, but positive for the abelian groups.

Thanks.

• Note that the result is false for infinite groups even in abelian case, since ${\bf Z}^2$ and ${\bf Z}^3$ have isomorphic $Sub_2$ by combinatorial considerations. The result for finite abelian groups should follow from fundamental theorem on finitely generated abelian groups and, again, by combinatorial considerations. Note that also in case of $Sub_2$, abelianness of one group implies that of the other. Commented Jul 16, 2013 at 13:54
• Now that I think of it, I think this question is closely related to finite model theory. The model-theoretical version of this question would be "when are two groups satisfying the same formulas with at most $k$ quantifiers isomorphic?". Of course, this seems weaker than $Sub_k$ being isomorphic, as you require a bijection, but related, and I think probably well-examined, but I've only ever had a glance at finite model theory. Commented Jul 16, 2013 at 14:02
• For the $Sub_1$ case, consider two distinct $p$-groups of same order, of exponent $p$ (with $p$ odd). For example, let $G_1$ be the elementary abelian group of order $p^3$ and $G_2$ the Heisenberg group of order $p^3$. So if $Sub_1(G_1)$ and $Sub_1(G_2)$ are isomorphic, it does not necessarily follow that $G_1$ and $G_2$ are isomorphic. Commented Jul 16, 2013 at 14:18

No. The groups \begin{align} G_{44} &= \langle a,b,c : a^3 = b^9 = c^9 = 1, [b,a]=b^3 c^3, [c,a]=b^3, [c,b]=1 \rangle \\ G_{45} &= \langle a,b,c : a^3 = b^9 = c^9 = 1, [b,a]=c^3 c^3, [c,a]=b^3, [c,b]=1 \rangle \end{align} have isomorphic 2-generated subgroups, but are not themselves isomorphic.
$p$-groups are just a sea of continuous change, so it is not at all surprising to find examples there. Conveniently, it is very easy to find minimal generating sets of $p$-groups, so the condition is more easily checked.
• How can you check that they have the same $2$-generated subgroups (and are not isomorphic)? Commented Jul 16, 2013 at 14:41
• I just had GAP check. I don't believe there is any deep reason. No sane numerical invariants serve to classify $p$-groups up to isomorphism. Commented Jul 16, 2013 at 14:45