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Let $H\trianglelefteq G$, $N\trianglelefteq G$ such that $N\cap H=\{e\}$ and $HN=G$. Then obviously $G=N\rtimes H$, but also $G=N\ltimes H$. This implies $G=N\times H$.

How do I obtain this last implication? At first I have tried constructing an isomorphism $\phi: N\rtimes H\to N\times H, (h,n)\mapsto (h,n)$ but I failed. Then I have tried to construct an isomorphism $\psi: HN\to H\times N, hn\mapsto (h,n)$ but I have faild at that too. For example, I don't see how $\psi(g_1g_2)=\psi(g_1)\psi(g_2)$ unless I know that $G$ is abelian when having the listed properties.

Buh
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1 Answers1

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Consider

$$f:N\times H\to G$$ $$f(h, n)=hn$$

  1. Since $G=HN$ then $f$ is "onto".
  2. Since $H\cap N=\{e\}$ then $f$ is "1-1".
  3. $f$ is a homomorphism.

Proof of 3. For that realize that if $h\in H$ and $n\in N$ then

$$hnh^{-1}n^{-1}\in H$$ $$hnh^{-1}n^{-1}\in N$$

Both follow from the fact that $H$ and $N$ are normal. Thus

$$hnh^{-1}n^{-1}\in H\cap N=\{e\}$$

and thus

$$hn=nh$$

Note that it doesn't mean that $G$ is abelian (and indeed it isn't if one of $H$, $N$ isn't). It only means that elements from $H$ commute with elements from $N$.

Anyway this will be enough to show that $f$ is a homomorphism:

$$f\big((h,n)(h',n')\big)=f(hh', nn')=hh'nn'=hnh'n'=f(h,n)f(h',n')$$

freakish
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