Proving the existence of a homomorphism $\overline f:G/H\rightarrow G'$ such that $\overline f \pi = f.$

Question

I'm working on a problem where I'm given that $G$ is a group, $H$ is a normal subgroup of $G,$ $f:G\rightarrow G'$ is a homomorphism, and $H\subseteq \ker(f).$ I need to show that there exists a homomorphism $\overline f: G/H\rightarrow G'$ such that $\overline f \pi = f.$

I have an answer which I believe to be too simple... Define $\overline f:G/H\rightarrow G'$ as $\overline f(gH) = f(g).$ Then $\forall g\in G,$ $\overline f(\pi(g)) = f(g).$ But I didn't even use the fact that $H\subseteq \ker(f),$ so this can't be right. Any advice?

Answer 1

In general if you define maps in terms of representatives of their arguments, you are defining the same function value multiple times; this is only meaningful if all these definitions coincide. In the current case $g$ is just a representative of its coset $gH$, so $gH=g'H$ holds for any other element $g'$ of its coset. Since you defined $\bar f(gH)=f(g)$ where in the right hand side $g$ occurs alone (not in the context of its coset $gH$), you must check that whenever $gH=g'H$ it follows that $f(g)=f(g')$, or otherwise your definition would be meaningless. Very often the check of meaningfulness for functions defined like this is the hardest part.

Answer 2

Hint: You need to check your map $\overline{f}$ is well defined. Right now it seems to depend on the choice of a representative of a coset. $G/H$ consists of equivalence classes, and equivalence classes may have different representatives. It is here that you need $H \subset ker(f)$.