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I was studying some properties of semidirect products, and I noticed this sentence on Wiki

Suppose $G$ is a semidirect product of the normal subgroup $N$ and the subgroup $H$. If $H$ is also normal in $G$, or equivalently, if there exists a homomorphism $G\to N$ which is the identity on $N$, then $G$ is the direct product of $N$ and $H$.

So I tried proving myself that having such homomorphism from $\varrho \colon G\to N$ which is the identity on $N$ was sufficient to imply $H$ normal (because I think this is the point) and so having two normal subsets $H,N$ with $H\cap N=1$ it's easy to see that $G\cong N \times H$.

The problem is that I'm not able to prove that $\ker \varrho=H$ (either the two implications).

Being $\varrho$ surjective I tried considering the iso $G/\ker \varrho \to N$, but I do not know how to prove that $H \subset \ker \varrho$. In fact I proved that, for every $g \in G$, exists $n \in N$ such that $[g]=[n]$, but from this I do not know how to proceed further, because I'd have $h w=n$ for $w$ element in the kernel. But to apply the fact that the intersection is trivial I need to know that the kernel is contained in $H$, which I cannot prove.

Can someone help me please? because this fact seems a triviality but it's bugging me

Brian M. Scott
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Luigi M
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  • Show that $\ker\rho\cap N=1$, from this $[\ker\rho,N]=1$ and then you're on your way to show that $G=\ker\rho\times N$. – Myself Jan 19 '15 at 22:55
  • I should add that if what you are trying to show is that if (1) $G = A\times B$ and (2) there is a map $\varrho:A\to G$ which is the identity on $A$, then this implies that $\ker\varrho=B$ then you will fail because it is wrong: there may in general be many such maps and only one will correspond to this particular decomposition. Play around with $\mathbf C_2\times\mathbf C_2$ to see what I mean: it has 3 subgroups $A$, $B$ and $C$ and it is the direct product of any two of them. In other words the 'decomposition' (1) must correspond to the map (2) for this to hold. – Myself Jan 19 '15 at 23:02
  • @Myself I'll work on your hint right now! by [G,H] you mean the commutator between G and H? – Luigi M Jan 19 '15 at 23:08
  • That's right! (here's a hint/solution for that part.) – Myself Jan 19 '15 at 23:46
  • @Myself if $G=N \ltimes H$ (can't remember if this is the right notation) can we guarantee that the homomorphism satisfies that $\ker \rho = H$? – hjhjhj57 Jan 20 '15 at 05:29
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    @hjhjhj57 I've proved that decomposing $G=\ker \rho N = HN$ and then using uniqueness of both decompositions – Luigi M Jan 20 '15 at 08:19
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    @hjhjhj57 Since a kernel of a homomorphism is always a normal subgroup, there is no such homomorphism unless $N$ and $H$ are both normal. – Myself Jan 20 '15 at 08:45
  • @Myself got it from both your comments. – hjhjhj57 Jan 20 '15 at 09:56

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