Does it factor through? Let $f:F\to G$ and $g:F\to H$ be group homomorphism between groups. If $\ker f \subset \ker g$ then does there exists $h:G\to H$ such that $hf = g$?
I know the the above is true for vector spaces by extending linearly independent set into a basis. I think the above is not true in the category of groups. What about finitely generated abelian groups? And what about for abelian groups? Is there a name for this property?
 A: If we find groups $F$, $G$, and $H$ such that $G$ and $H$ both have subgroups isomorphic to $F$, but $G$ does not have an image isomorphic to $F$, then we can use the embedding homomorphisms $$i_G:F\rightarrow G \text{ and }i_H:F\rightarrow H.$$  In particular, given any group $G$ which has a subgroup $F$ and no quotient isomorphic to $F$, we can take $H=F$.
So, the simplest example would be $F=C_3$, $G=S_3$, and $H=C_3$.
It does work for finite abelian groups, on the other hand, since we can find normal complements for any subgroup. (In fact this argument applied to finitely generated abelian groups.)  I am not so sure about the non-finitely generated case, or even for divisible abelian groups.
A: Even for cyclic groups it's not true.
For example, take $F=H=C_2$, with $g:F\to H$ the identity map, and $G=C_4$ with $f:F\to G$ an inclusion.
A: This will not be true already for injective $f$ and $g$ (when the kernel condition is trivially satisfied). For example, the cyclic group of order $3$ admits monomorphisms to both groups of order $6$ but there is no such $h$. Namely, $S_3$ does not have a quotient of order $3$.  Therefore there is no such $h: S_3\to C_6$.
