# semidirect product, split extension

I working on my thesis on semidirect products and splitting. I am trying to prove that if you assume that $G$ is a split extension of $N$ and $H$ then you can show that $G$ is a semidirect product of $N$ and $H$.

Let $0\xrightarrow{}N\xrightarrow{\alpha}G\xrightarrow{\beta}H\xrightarrow{}0$ be exact.

Define $\gamma :H\to G$ with $\beta\circ \gamma =id_{H}$

$N_{0}:=\alpha(N)$, $H_{0}:=\gamma(H)$

Show that $G=N_{0}H_{0}$ and that $N_{0}\cap H_{0}=1$

This is what I´ve done right now

I define $g\in N_{0}\cap H_{0}$ and with $H_{0}=\gamma(H)$ I have that $g=\gamma(h)$ since $g\in H_{0}$

At the same time $g\in N_{0}$ so that gives me $\beta(g)=id_{H}$.

So now I have that $\beta(\gamma(h))=id_{H}$ which is $\beta \circ \gamma$. That shows that only element in $N\cap H$ is the identity element.

Now to show that $g=nh$:

For $g\in G$ I can define $h:=\gamma(\beta(g))\in H_{0}$ and $n:=gh^{-1}$.

That gives me that $g=nh$. Now I have to show that $n\in N_{0}$.

With $n=g\gamma^{-1}(\beta(g))$ and $\beta(n)=id$ assuming $n\in N_{0}$

I can make the following equation:

$id=\beta(n)=\beta(g\gamma^{-1}(\beta(g)))$ and using that $\beta$ is homomorphic I get

$id=\beta(n)=\beta(g\gamma^{-1}(\beta(g)))$ = $\beta(g)\beta(\gamma^{-1}(\beta(g)))$

This is where I am now ... I dont know how to make the last step which gets me $\beta(g)\beta^{-1}(g)$ on the right side.

I appreciate any help and comments of my calculations. Thank you.

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As you prove you can write every $g \in G$ as a product $$g = n\gamma(\beta(g))$$ for some $h \in H$, now if you applies $\beta$ to $g$ what you get is $$\beta(g) = \beta(n \gamma(\beta(g))) = \beta(n)\beta(\gamma(\beta(g)))=\beta(n)\beta(g)$$ now a simple multiplication by $\beta(g)^{-1}$ on both sides proves that $$\text{id} = \beta(n)$$ which exactly states that $n \in N_0$.
well the problem is how do i get $β(γ^{-1}(β(g)))$ to become $\beta^{-1}(g)$. –  grendizer Dec 8 '12 at 11:44
oh, i think i know now. is it from what i stated earlier that $g=\gamma(h)$? –  grendizer Dec 8 '12 at 12:54
@grendizer yes, that or more easily from the fact that $\beta \circ \gamma = \text{id}$, by splitting property. –  Giorgio Mossa Dec 8 '12 at 18:29