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If $G, H$ are groups and $f : G \rightarrow H, g : H \rightarrow G$ are homomorphisms such that $fg = id_H$ then

$$G \cong Ker(f) \rtimes_\varphi H$$

where $\varphi: H \rightarrow Aut(Ker(f))$ is given by $\varphi(b)(a) := g(b)ag(b)^{-1}$ for $a \in Ker(f), b \in H$.

I have tried to defined an isomorphism from $G$ to $Ker(f) \rtimes_\varphi H$, but could not. Any help is appreciated.

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  • $\begingroup$ Hint: There is a map $\ker(f) \rtimes_\varphi H \rightarrow G$ induced by the inclusion of $\ker(f)$ into $G$ and $g: H \rightarrow G$ that could be of use to you. $\endgroup$ – Matthias Klupsch Apr 30 '15 at 6:33
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You can always define a function $\psi$ :

$$\psi :Ker(f)\rtimes_{\varphi}H\rightarrow G $$

$$(k,h)\mapsto kg(h) $$

Using the existence of both $f$ and $g$ you can show that your function $\psi$ is a bijection (I leave this to you).

Furthermore :

$$\psi((k_1,h_1)(k_2,h_2))=\psi(k_1\varphi(h_1)(k_2),h_1h_2)=k_1\varphi(h_1)(k_2)g(k_1)g(k_2)=k_1g(h_1)k_2g(h_1)^{-1}g(h_1)g(h_2)=\psi(h_1,k_1)\psi(h_2,k_2) $$

Hence $\psi$ is also a group morphism.

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