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Let $G$ and $H$ be groups, $\operatorname{Sub}(G\times G)$ be the set of all subgroups of $G\times G$ and $\operatorname{Sub}(H\times H)$ be the set of all subgroups of $H\times H$. Assume there exists a bijection $\phi :\operatorname{Sub}(G\times G) \to \operatorname{Sub}(H\times H)$ such that always $$A\le B~~\leftrightarrow~~\phi(A)\le\phi(B)$$

1) If $G$ and $H$ are infinite abelian groups, are they isomorphic?

2) If $G$ and $H$ are finite non-abelian non-simple groups, are they isomorphic?

Edit:

As shane.orourke's answer shows below and also by an example by Schmidt in 

R. Schmidt. Der Untergruppenverband des direkten Produktes zweier isomorpher Gruppen. J. Algebra 73 (1981), 264–272.

The first question and has a negative answer. (Still I'm not sure if Schmidt's example is abelian)

The second question remains unanswered.

Minimus Heximus
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  • do you really mean "non-abelian non-simple"? this is a weird assumption. – YCor May 10 '14 at 11:58
  • yes, I have an answer for finite abelian or simple case already. – Minimus Heximus May 10 '14 at 12:06
  • for finite abelian or simple they are isomorphic, – Minimus Heximus May 10 '14 at 12:24
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    But it's weird to exclude them (for instance the question makes sense if $G$ is abelian and not $H$). You could just mention that the answer is yes in a few particular cases (e.g. when $G,H$ are both finite abelian) – YCor May 10 '14 at 13:44
  • @YvesCornulier: If $G$ is abelian and $H$ is not; they are not isomorphic, nothing remains to ask. My limitations generalizes the question in title. – Minimus Heximus May 10 '14 at 15:58
  • I don't agree with "nothing remains to ask": in principle you might have $G$ abelian, $H$ not abelian, but $G^2$ and $H^2$ (or $G$ and $H$) having isomorphic lattice of subgroups. – YCor May 10 '14 at 16:59
  • yes whether the sturucture of $\operatorname{Sub}(G\times G)$ reveals abelian-ness or simple-ness, can be interesting. But my question is about whether it reveals the structure of $G$ completely (and not partially). – Minimus Heximus May 10 '14 at 17:06
  • btw, it reveals abeliannes – Minimus Heximus May 10 '14 at 17:40

1 Answers1

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The answer to the first question is no. A. L. S. Corner showed that (quoting from the mathscinet review): `given a positive integer $q$, there exist standard abelian $p$-groups $G$ and $H$ with no elements of infinite height such that $G^n\cong H^n$ if and only if $q$ divides $n$'.

So if $q=2$ we get $G\times G\cong H\times H$ -- and certainly a bijection as required by the question -- even though $G\not\cong H$.

This result is in the paper

A. L. S. Corner On endomorphism rings of primary abelian groups. Quart. J. Math. Oxford Ser. (2) 20 1969 277–296.

(This is basically a duplicate of my answer to If $G \times G \cong H \times H$, then is $G \cong H$?; this question in turn was a duplicate of when is A isomorphic to A^3?.)

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shane.orourke
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    Pardon my ignorance, but what is a standard group? – Emil Jeřábek May 10 '14 at 14:59
  • @EmilJeřábek If I'm not mistaken, a standard abelian $p$-group $G$ is one that coincides with one of its `standard' subgroups, namely $G_p$ which consists of elements of order dividing $p$, or $G^p$ which consists of $p$th powers. – shane.orourke May 10 '14 at 20:51
  • Thank you, however, this does not seem to work. If $G=G_p$, $G$ is of exponent $p$ and thus a direct sum of $\kappa$ copies of the $p$-element cyclic group for some $\kappa$, and if $G=G^p$, it is divisible and thus a direct sum of $\kappa$ copies of the Prüfer $p$-group. In either case, the isomorphism type is uniquely determined by $\kappa$, which easily implies $G^n\simeq H^n\iff G\simeq H$. Could it be that the meaning is $G=G_{p^k}$ for some $k$, i.e., that $G$ has a finite exponent? – Emil Jeřábek May 10 '14 at 21:42
  • @EmilJeřábek You could be right, I don't have the paper to hand. – shane.orourke May 10 '14 at 22:09