Let $G$ be a group with $A, B$ subgroups. Define $AB$ as {$ab: a\in A, b\in B$}. Let $f$ be a homomorphism from $G$. My question is, when does $f(AB)=f(A)f(B)$?
That is, when does a homomorphism preserve multiplication of subgroup products as it does for the products of individual elements of a group.
I think that as long as $H\cap K$ is trivial, it will preserve it. This is because, in this case, any element of $AB$ can be written uniquely in the form $ab$ for some $a\in A$ and $b\in B$. But I'm not sure if this condition is necessary.
(Also, I don't think it it, but I'm also not sure if it's relevant whether or not $AB$ is a subgroup.)
No conditions are required - this is one of those "follow your nose" proofs. Make sure you know your definitions well.
Suppose $ x \in f(AB)$. Then for some $y \in AB$, we have $f(y) = x$. We can by definition write $y = ab$, where $a \in A$ and $b \in B$. Thus we have $f(ab) = x$. Since $f$ is a homomorphism, then $x = f(ab) = f(a) f(b) \in f(A) f(B)$.
If $x \in f(A) f(B)$, then $x = a' b'$, where $a' = f(a)$ and $b' = f(b)$ for some $a \in A, b\in B$. Thus $x = f(a) f(b) = f(ab) \in f(AB)$.
This establishes $f(AB) \subseteq f(A) f(B)$ and $f(A) f(B) \subseteq f(AB)$.