A function $f': G/H \rightarrow G'$ such that $f = f' \circ p$ exists iff $H \subset \ker f$

Question

Suppose you have $f : G \rightarrow G'$ a group homomorphism and $H$ a subgroup of $G$. Let $p: G \rightarrow G/H$ be a canonical projection.

Show that there exists a function $f' : G/H \rightarrow G'$ such that $f = f' \circ p$ if and only if $H \subset\ker f$.

Now, I started of by showing that the existence of the function $f'$ implies $H \subset \ker f$. I decided to just pick $x \in H$ and apply it to $f$. That is, $f(x) = f' \circ p(x) = f' (\{xx' | x \in H \})$, and from here I don't know what to do. Anything would be helpful!

Answer 1

Let $G$ be a set and $P$ and $P'$ two partitions of $G$ such that $\exists\varphi:P'\to P$ whereby $P'=\{\varphi^{-1}(y)|\forall y\in P\}$, that is $P$ is a refinement of $P'$.

Then the canonical projections $p\colon G\to P$ and $p'\colon G\to P'$ of $G$ over $P$ and $P'$ are related by: $$p'=\varphi\circ p\tag{1}$$

Back to your problem, the proof follows from the:

Lemma. The partition $P=G/H$ is a refinement of $P'=G/\ker f$ iff $H\subset\ker f$

Indeed let $p'\colon G\to G/\ker f$ the natural homomorphic projection of $G$ over $G/\ker f$ and $F:G/\ker f\to G'$ the isomorphism such that $f=F\circ p'$, then applying $(1)$:

$$f=F\circ p' = F\circ\varphi\circ p=f'\circ p$$

P.S. Let me know whether you need a proof of the lemma.

Answer 2

If $f=f'\circ p$, then, for each $h\in H$, $f(h)=f'\bigl(p(h)\bigr)=f'(e)=e$. Therefore, $H\subset\ker f$.

On the other hand, if $H\subset\ker f$, the you can define $f'\colon G/H\longrightarrow G'$ by $f'(gH)=f(g)$. Does this make sense? Yes, because if $gH=g'H$, then $g^{-1}g'\in H\subset\ker f$ and therefore $f\bigl(g^{-1}g'\bigr)=e$, which means that $f(g)=f(g')$. But then$$g\in G\implies f'\bigl(p(g)\bigr)=f'(gH)=f(g),$$by the definition of $f'$.