This is another question from Artin's book Algebra (2nd edition). The question is as follows:
Let $H$ be a subgroup of a group $G$, let $\varphi:G\to H$ be a homomorphism whose restriction to $H$ is the identity map, and let $N$ be its kernel. What can one say about the product map $H\times N\to G$?
I guess for the product map, Artin is referring to $f(h,n)=hn$.
If so, I can show that $N\cap H=\{1\}$ and $HN=G$ and thus $H\times N\to G$ is bijective.
I'm trying to prove that $H$ is normal so that $H\times N\to G$ is an isomorphism. May I ask if that is possible? or if $H\times N\to G$ may not be a homomorphism?
Any hint is appreciated, thanks in advance!
We have a short exact sequence $1\to N\to G\stackrel{\varphi}\to H\to1$. Now since the restriction $\varphi\restriction_H=\rm{id}_H$, there is a morphism, namely the inclusion, going from $H\hookrightarrow G$ with $\varphi\circ i=\varphi\restriction_H=\rm{id}_H$. That's the sequence is split exact, and we have $G\cong N\rtimes H$.
$H$ is not necessarily normal, that is when the action of $H$ on $N$ is nontrivial. For instance, consider $S_n\cong A_n\rtimes\mathbb Z_2$.
So we can at least say that the product map is surjective. It doesn't have to be a homomorphism. Again, for instance, if the multiplication map $A_n\times \mathbb Z_2\to S_n$ were a homomorphism, it would be an isomorphism.