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May 10, 2014 at 17:40 comment added Minimus Heximus btw, it reveals abeliannes
May 10, 2014 at 17:24 history edited Minimus Heximus CC BY-SA 3.0
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May 10, 2014 at 17:17 review Close votes
May 11, 2014 at 15:48
May 10, 2014 at 17:06 comment added Minimus Heximus 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).
May 10, 2014 at 16:59 comment added YCor 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.
May 10, 2014 at 15:58 comment added Minimus Heximus @YvesCornulier: If $G$ is abelian and $H$ is not; they are not isomorphic, nothing remains to ask. My limitations generalizes the question in title.
May 10, 2014 at 14:42 answer added shane.orourke timeline score: 4
May 10, 2014 at 13:44 comment added YCor 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)
May 10, 2014 at 12:24 comment added Minimus Heximus for finite abelian or simple they are isomorphic,
May 10, 2014 at 12:06 comment added Minimus Heximus yes, I have an answer for finite abelian or simple case already.
May 10, 2014 at 11:58 comment added YCor do you really mean "non-abelian non-simple"? this is a weird assumption.
May 10, 2014 at 11:50 history asked Minimus Heximus CC BY-SA 3.0