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Let's denote all subgroups of a group $G$ by $\text{sub}(G)$.

Let $G$ and $H$ be groups. Can $\text{sub}(G\times H)$ be determined in terms of $\text{sub}(G)$ and $\text{sub}(H)$?

$G\times H$ is the direct product.

SE Anarchist
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  • Take the subgroup generated by $(1,1)$ in $Z \times Z$.... – N. S. Sep 12 '13 at 03:47
  • ${\rm sub}(G\times H)={\rm sub}(G)\times{\rm sub}(H)$ fails badly in general. There is a tautological sense in which it is possible though. Take the maximal element of ${\rm sub}(G)$ and ${\rm sub}(H)$ (namely $G$ and $H$ themselves), form their direct product and then apply ${\rm sub}(\cdot)$... – anon Sep 12 '13 at 03:49
  • well I don't mean ${\rm sub}(G\times H)={\rm sub}(G)\times{\rm sub}(H)$. But I think altho finding all subgroups of an arbitrary group is not possible, but finding all subgroups of $\text{sub}(G\times H)$ is possible in terms of subgroups of $H$ and $G$. – SE Anarchist Sep 12 '13 at 07:15
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    For more information on this, see my answer http://math.stackexchange.com/questions/485512/subgroups-of-a-direct-product/488222#488222 As I also mention in a comment there, all subgroups being of the form mentioned be Andreas is equivalent to the two groups not having any subquotient in common. – Tobias Kildetoft Sep 12 '13 at 07:37
  • Can you say exactly what you mean by "determined in terms of $sub(G)$ and $sub(H)$"? – Derek Holt Sep 12 '13 at 07:42
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    Maybe it should be noted though, that it is not quite enough to know the subgroups of $G$ and $H$, but one also needs to know which ones are normal in each other. – Tobias Kildetoft Sep 12 '13 at 07:44
  • @TobiasKildetoft: thanks. @ DerekHolt: well in the same way the set {x+y,xy,z} is determined in terms of the sets {x+1,y+1} and {z} in real numbers. – SE Anarchist Sep 12 '13 at 08:24
  • @CrossChris: The point is that $G$ and $H$ themselves are among the subgroups of $G$ and $H$, and obviously the subgroups of $G \times H$ are determined by $G$ and $H$. So I am genuinely unsure what you are asking (although everyone else seems to know). – Derek Holt Sep 12 '13 at 08:52
  • @DerekHolt: well similar ambiguities is common in mathematics. for example there is not a general formula for roots of a polynomial of order 6. is completely ambiguous but everyone thinks it's completely understandable or integration of $\frac{1}{cos(x)}$ is vague. btw, I wanted to obtain more information about how having subgroups of H and G helps in finding subgroups of $G\times H$ and I think it's clear enough. – SE Anarchist Sep 12 '13 at 09:53

1 Answers1

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See Goursat's lemma (http://en.wikipedia.org/wiki/Goursat's_lemma).

ronno
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Boris Novikov
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