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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!

  • What is $\phi_{f(h)}$? Is it the inner automorphism induced by conjugating by $f(h)$? – JSchlather Dec 01 '12 at 21:15
  • It is the conjugation by $f(h)$, i.e. $\phi_{f(h)}(g) = f(h)gf(h)^{-1}, \forall g \in G$ –  Dec 01 '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? –  Dec 01 '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. – Mikasa Dec 02 '12 at 15:44
  • @Stefan: See this http://math.stackexchange.com/q/201710/8581. – Mikasa Dec 02 '12 at 19:29
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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. –  Dec 02 '12 at 19:32
  • @AlexanderGruber: It is not via the "expected" isomorphism, so that $f$ does not need to map into $Z(G)$. See my comment above. –  Dec 02 '12 at 19:35
  • why don't you try $T(g, h) = (gf(h), h)$? – Shankara Pailoor Dec 03 '12 at 23:38

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