Timeline for Examples of groups such that order isomorphism of the subgroups of $G\times G$ and $H\times H$ does not imply isomorphism of $G$ and $H$
Current License: CC BY-SA 3.0
12 events
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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 |