Understanding The Automorphisms to make a Semidirect Product I have used semi-direct products for a while now, and frankly, they still don't make much sense to me. Especially where we define the homomorphism $\phi = K \to \text{Aut}(H)$. The trivial homomorphism makes sense to me, but non-trivial homomorphisms don't. I understand the definition,  but it seems very hard to work with. How do you actually construct the non-trivial homomorphisms, and just as importantly, how do you actually write them? Most of the examples I have seen are written similarly to this: $H \rtimes_{\phi} K$ with $\phi(x) = x^{m}$, but this makes no sense to me.
When are the different homomorphisms in the semidirect products isomorphic, and when are they not?
Once you do construct a non-trivial homomorphism, how do you find the presentation of the resulting semidirect product?
Thank you!
 A: Take for example $H=C_4$ and $K=C_2$, you can think of $C_4$ as $\langle i\rangle$, where $i^2=-1$. I like to think of the homomorphism $K\rightarrow Aut(H)$ as an action of $K$ on $H$. In this case the action of $C_2=\{1,\sigma\}$ on $C_4=\{1,i,-1,-i\}$ is given my multiplication by $-1$, i.e. $\varphi(\sigma): i\mapsto -i$, $1\mapsto -1$. This is a nontrivial homomorphism and you can easily check that  $C_4\rtimes_{\varphi} C_2\simeq Q_8$ with this action. Now write $C_4=\{1,\rho,\rho^2,\rho^3\}$, and consider the action $\psi(\sigma):1\mapsto 1, \rho\mapsto \rho^3,\rho^2\mapsto \rho^2$. Now you can check that $C_4\rtimes_{\psi}C_2\simeq D_8$.
I also like to think of the semidirect product as a split solution to the extension problem
$$ 1 \rightarrow K \rightarrow G \rightarrow H\rightarrow 1,$$
where the above is a short exact sequence of groups, and split means that there is a section $H\rightarrow G$ for the map $G\rightarrow H$. As far as I know the solutions to this problem are captured by the Ext functor and the classes of nonisomorphic solutions are given by $Ext^1$.
As to how to construct nontrivial homomorphisms, I suppose that it depends much on the particular groups at hand. But hopefully seeing them as permutation representations of groups might make things more clear.
