I'm trying to consolidate my notes on inner semidirect products, and I would like to verify that the following statements are precisely correct. This is my attempt to clarify my understanding after studying various sources. I'm hoping to get some help verifying that these statements are correct and also complete in their characterization of an inner semidirect product.
Let $H,K<G$ and $f\ :\ H\times K\rightarrow G$ where $f(h,k)=hk$
Suppose $H\cap K=\{e\}$. Then $f$ is a bijection, but not necessarily a homomorphism. That is the same as saying: $HK$ might not be a subgroup of $G$.
Suppose that $H<N_G(K)$, without the assumption that $H\cap K=\{e\}$. Then $f$ is a homomorphism and $HK$ is a subgroup, but $f$ is not necessarily a bijection.
Now suppose that $H\cap K=\{e\}$ and $H<N_G(K)$. Then we have $H\ltimes K=HK$.
1) Not a bijection, just an injection. It is a bijection from $H\times K$ to its image $HK$, not $G$.
Indeed $HK$ may not be a subgroup. The simplest example is $G=\langle a,b\rangle$, the free group on two generators, and $H=<a>,K=<b>$. Then $HK=\{a^k b^l:k,l\in\mathbb{Z}\}$ does not contain the inverse of $ab$.
2) Yes, $HK$ is a subgroup. This is not hard to see. But is $f$ a homomorphism? If $f(hh',kk')=f(h,k)f(h',k')$ for all elements, then $hh'kk'=hkh'k'$ so choosing $h=k'=e$ we get $h'k=kh'$ for all $h',k$ in the respective groups, which is much stronger than one being in the normalizer of the other: one must centralize the other.
And again, when you say it is not a bijection, you mean from $H\times K$ to $HK$, and yes, it is true it does not have to be a bijection.