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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 $f(h)$?2012-12-01
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    It is the conjugation by $f(h)$, i.e. $\phi_{f(h)}(g) = f(h)gf(h)^{-1}, \forall g \in G$2012-12-01
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    Look at the subgroup $(f(h),h^{-1})$ in the semidirect product. what does conjugation by $g\in G$ do?2012-12-01
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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.2012-12-02
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    @Stefan: See this http://math.stackexchange.com/q/201710/8581.2012-12-02
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    @BabakSorouh: It doesn't matter really. The subgroup $\lbrace f(h)h^{-1}\mid h\in H\rbrace$ is isomorphic to $H$, is centralized by $G$, and together they generate the whole group.2012-12-02
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    @AlexanderGruber: It is not via the "expected" isomorphism, so that $f$ does not need to map into $Z(G)$. See my comment above.2012-12-02
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    why don't you try $T(g, h) = (gf(h), h)$?2012-12-03

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