I'm studying a course in abstract algebra. Recently I solved an exercise which said that if you had a group $G$, its commutator $[G:G]$, an abelian group $K$ and a group morphism $f:G \rightarrow K$ then you can factorize this group morphism uniquely by the quotient $G/[G:G]$. For this I used the universal property of quotient group.
My doubt is whether you can guarantee the existence of this morphism given any (say) finite group. I know that every finite group is isomorphic to a subgroup of $S_n$ where $n$ is the order of the group. What I'm asking is a little bit more specific. I checked Hungerford's and Artin's books and couldn't find an answer. Maybe I should check a book on group theory? Thanks!
You can/should also check that $G/[G,G]$ is abelian, itself.
Then, what do you mean "guarantee the existence of this morphism..."? If $G/[G,G]$ is not isomorphic to any non-trivial subgroup of abelian group $K$, then the only group hom $G\to K$ is the map to $\{e_K\}$. Yes, this does factor through $G/[G,G]$.
But, yes, any group hom $G\to K$ for abelian $K$ does factor uniquely through the quotient $G/[G,G]$. Existence of the $G\to K$ assures the existence of the associated map $G/[G,G]\to K$, if that's what you're asking.
EDIT: to incorporate @DanielSchepler's comment (since comments can and will get deleted at some point): yes, in particular, since $G/[G,G]$ is always abelian, if it is non-trivial then the quotient map $G\to G/[G,G]$ is a non-trivial hom to an abelian group... Is this the issue?
And, indeed, for simple $G$, $[G,G]=G$... since any hom $G\to H$ has a kernel, which is normal, which can only be $G$ or $\{e\}$.