TITLE: If $A,B \triangleleft G$ where $G/A$ is perfect and $G/B$ is solvable, then $AB=G$
QUESTION [5 upvotes]: I'm having trouble proving the following and would appreciate a hint to nudge me in the right direction.
Let $A \triangleleft G$ and $B \triangleleft G$, where $G/A$ is perfect and $G/B$ is solvable.  Show that $AB=G$.
As a hint, I am told to show that $G/AB$ is both perfect and solvable.  I do see that this implies that $G/AB$ is trivial and therefore $AB=G$.  But, I have not been able to prove it.
I figure there are several ways we could conceivably show this.  One would be to somehow construct $G/AB$ from $G/A$ and $G/B$ and show that the desired properties transfer over.  However, I do not see a good way to do this as $G/A$ and $G/B$ are larger than $G/AB$, and I couldn't figure out how to anything with the intersection either.
My other idea was to use the properties of $G/A$ and $G/B$ to work backwards and find properties of $A$ and $B$, then try to combine those properties into $AB$, then try to transfer them to the quotient.  I also can't find a way to do this, and I'm not so sure this is a good route to take.  For example: since abelian groups are solvable, $B$ could be as general as any normal subgroup of prime or prime-squared index.
I know that if $\phi: G \to H$ is a surjective homomorphism, then $\phi(G^{(n)})=H^{(n)}$.  Take the canonical homomorphism $\pi: G \to G/A$.  Then since $(G/A)'=G/A$ we have $\pi(G')=G/A$ but since $\pi(G')=G'/A$, this seems to imply (to me) $|G'/A|=|G/A|$ and $G=G'$.  Is this correct?
Again, I'd appreciate some help to get me going in the correct direction.  Thanks.

REPLY [5 votes]: The hint you were given is a good one.  Note that $G/AB$ is a homomorphic image of both $G/A$ and $G/B$, by the third isomorphism theorem, because $AB$ contains both $A$ and $B$.  Homomorphic images of solvable groups are solvable, and homomorphic images of perfect groups are perfect.  (Both results follow, essentially, from the fact that homomorphisms take commutators to commutators.)  This implies that $G/AB$ is both soluble and perfect, hence, trivial.