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I would like some help on the following problem from anyone who would like to help.

Let $f: H \to G$ be a group homomorphism. For $h \in H$, define $\rho(h) = \phi_{f(h)} \in Aut(G)$.

The situation being as described, prove that the semi-direct product $G\rtimes_{\rho} H$ is isomorphic to the direct product $G \times H$.

Help will be greatly appreciated!

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What is $\phi_{f(h)}$? Is it the inner automorphism induced by conjugating by $f(h)$? –  JSchlather Dec 1 '12 at 21:15
    
It is the conjugation by $f(h)$, i.e. $\phi_{f(h)}(g) = f(h)gf(h)^{-1}, \forall g \in G$ –  user44069 Dec 1 '12 at 21:18
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Look at the subgroup $(f(h),h^{-1})$ in the semidirect product. what does conjugation by $g\in G$ do? –  user641 Dec 1 '12 at 21:24
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@SteveD: Dear Steve, are the OP speaking about the internal semidirect product when he works on $G\time H$? Because as a fact D.Robinson noted in his book that we can find two subgroups like $N^*, H^*$ which $N^*\cong G$ and $H^*\cong H$ and $N^*\cap H^*=\{1\}$ such that $G\times_{\rho} H\cong N^*\times H^*$. Thanks. –  Babak S. Dec 2 '12 at 15:44
    
@Stefan: See this math.stackexchange.com/q/201710/8581. –  Babak S. Dec 2 '12 at 19:29

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