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Let $1\rightarrow G{\rightarrow}^\varphi E\rightarrow^\psi H\rightarrow 1$ be short exact sequence of groups and homomorphisms where $E$ is a semidirect product $E=G \rtimes H$.

I am struggling to see why does this short exact sequence split? That is, why is there a homomorphism $\gamma : H\rightarrow E$, such that $\psi \gamma = 1_H$?

edit: fixed $1_H$.

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    $\begingroup$ If there is no further information on the maps, the statement is false. (It is even false for direct products of abelian groups) $\endgroup$ – MooS Mar 13 '15 at 9:48
  • $\begingroup$ Can you provide counterexample? $\endgroup$ – Burt Named Mar 13 '15 at 10:23
  • $\begingroup$ There is no further information on the maps. $\endgroup$ – Burt Named Mar 13 '15 at 10:32
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    $\begingroup$ Consider the the exact sequence $$0 \to \mathbb Z \to \mathbb Z \times \prod_{n \in \mathbb N} \mathbb Z/p\mathbb Z \to \prod_{n \in \mathbb N} \mathbb Z/p\mathbb Z \to 0$$ where the first map is multiplication by $p$ on the first factor and the second map is $(n,x) \mapsto (n \mod p, x)$. $\endgroup$ – MooS Mar 13 '15 at 11:31
  • $\begingroup$ Thanks, that's a good example! $\endgroup$ – Burt Named Mar 13 '15 at 12:08

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