How the existence of a special homomorphism of a group can help me constract a special normal subgroup to represent uniquely the group?

Question

The problem is like this:

Let $H$ be a subgroup of $G$, and let $f$ be a homomorphism: $G\to H$ which has $f(h)=h$ for all $h$ belongs to $H$. Then how to construct (or prove the existence of) a normal subgroup $K$ of $G$ such that any element of $G$ can be presented uniquely as $hk$ with $h$ from $H$ and $k$ from $K$?

Thank you, and the why is because $f(g)^{-1}$ is in $H$.

Answer 1

Put $\;K:=\ker f\;$ . Clearly $\;K\lhd G\;$, and for all $\;g\in G\;$ we can put $\;g=f(g)\left(f(g)^{-1}g\right)\;$. Trivialy $\;f(g)\in H\;$ , and

$$f\left(f(g)^{-1}g\right)=f(f(g)^{-1})f(g)\stackrel{\text{Why?}}=f(g)^{-1}f(g)=1\implies f(g)^{-1}g\in K=\ker f$$

The above expression is unique because if:

$$g=f(g)\cdot f(g)^{-1}g=hk\,,\,\,k\in H\,,\,\,k\in K\implies h^{-1}f(g)=kg^{-1}f(g)\in H\cap K$$

But

$$x\in H\cap K\implies x\stackrel{x\in H}=f(x)\stackrel{x\in K=\ker F}=1$$

Finish now the argument.