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Let $H$ be a normal subgroup in $G$. When is $G$ isomorphic to $H\times (G/H)$?

I think it's always true in the abelian case. Are there other rules?

  • It's not true in Abelian case! Hint: google "split short exact seuence". Moreover, in nonabelian cases, $G/H$ is not a group in general. – Peter Franek Jan 01 '16 at 22:17

1 Answers1

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It's not true in the abelian case. The smallest counterexample is $G = \mathbb{Z}_4, H = \mathbb{Z}_2$; the groups $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$ are not isomorphic.

In general you need $H$ to be normal for this question makes sense. Then $G$ is an extension of $G/H$ by $H$. The classification of these is difficult in general and usually there are interesting extensions other than the trivial extension $H \times G/H$. When all three groups are abelian the classification is in terms of the Ext group $\text{Ext}^1(G/H, H)$.

Qiaochu Yuan
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