Exact Sequence Splits Let
$$1\rightarrow N \rightarrow G\rightarrow Q\rightarrow 1$$
be an exact sequence of groups.

Question If it splits, then the extension group is a semi-direct product of the quotient by the kernel.


I have already known the proof about the ablien situation, i.e., module. However, I should deal with the nonablien situation now. Can someone help me with some details? Thank you!
 A: We are given a short exact sequence
$$1\rightarrow N \stackrel f \rightarrow G\stackrel g \rightarrow Q\rightarrow 1$$
We want to say that if there exists either an $u:G\rightarrow N$ or a $v:Q\rightarrow G$ such that $u\circ f=id_N$ or $g\circ v=id_N$, then $G$ is a semi-direct product of $N$ and $Q$. In particular, note that in this case $G\cong Q\ltimes N$, since $N$ must be normal in $G$ (as $f(N)=\ker g$). For simplicity, I will treat $N$ as identified with $f(N)\leq G$.
In the case we have a map $u:G\rightarrow N$, then $N$ and $\ker u$ are normal subgroups of $G$ such that $N\ker u=G$ and $\ker u\cap N=\{1\}$ and so $G\cong \ker u\times N$. Since $\ker u\cong G/N\cong Q$, we see that $G\cong N\times Q$, which is the trivial semi-direct product.
In the case $v:Q\rightarrow G$, we have a similar situation: $v(Q)\cap \ker g={1}$, $\ker g= N$, and $G=v(Q)N$. However, we have no guarantee that $v(Q)$ is normal in $G$; in that situation, we get a semidirect product $G\cong Q\ltimes_\phi N$ where for $q\in Q$, $\phi(q)\in {\rm Aut}(N)$ is the map defined by $\phi(q)(n)=v(q)nv(q^{-1})$.
