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.