Abelianization of the direct product is the direct product of the abelianizations? I'm trying to prove that $(H \times K)^{ab} \cong H^{ab} \times K^{ab}$, by invoking the universal property of abelianization. 
As $(H^{ab} \times K^{ab})$ is an abelian group, we know that, for the abelianization map $ \phi : (H \times K) \rightarrow (H \times K) ^{ab}$, we can define a group homomorphism $f : (H \times K) \rightarrow (H^{ab} \times K^{ab})$, with the existence of a unique $F: (H \times K) ^{ab} \rightarrow (H^{ab} \times K^{ab})$ such that $f = F \circ \phi$ ensured by the universal property. 
So F ensures that $(H \times K)^{ab}$ is homomorphic to $H^{ab} \times K^{ab}$ pretty straight-forwardly. I'm not exactly sure how to go further, though, and show that $f$ is actually an isomorphism. Theoretically, I could show that $\phi$ and $f$ are both isomorphic, and then the condition about F would follow (as $F = f \circ \phi^{-1}$, when $\phi^{-1}$ is defined). But I don't have much intuition about doing that. Some guidance would be appreciated. 
EDIT: $\phi$ has a trivial kernel iff the commutator subgroup is trivial ... which would seem to mean that $\phi$ can only be an isomorphism, logically, when $(H \times K)$ is abelian. We're working with arbitrary $H, K$ here, so I'm guessing my strategy isn't on point. 
 A: Let's state the universal property for abelianization.

For the abelianization of a group $G$, denoted $G^{\text{ab}}$, we have that if a homomorphism $f: G\to H$ exists where $H$ is abelian, then there exist a unique homomorphism $\varphi$ such that $\varphi\circ\pi = f$ where $\pi:G\to G^{\text{ab}}$

For the group $G\times K$ we have that $G^{\text{ab}}\times K^{\text{ab}}$ is abelian and we know there exist an epimorphism between them, $f:G\times K \to G^{\text{ab}}\times K^{\text{ab}}$. This means we have a homomorphism $\varphi$ from $(G\times K)^{\text{ab}}$ to $G^{\text{ab}}\times K^{\text{ab}}$. However as $f$ is surjective that means that $\varphi$ must also be surjective. Now we argue similarly for $G$, $G^{\text{ab}}$ and $(G\times K)^{\text{ab}}$, then $K$, $K^{\text{ab}}$ and $(G\times K)^{\text{ab}}$, except we don't argue the case of surjectivity but rather existence. The step after that, as we established homomorphisms from $G^{\text{ab}}$ to $(G\times K)^{\text{ab}}$ and from $K^{\text{ab}}$ to $(G\times K)^{\text{ab}}$ we have that there exist a homomorphism from $G^{\text{ab}}\times K^{\text{ab}}$ to $(G\times K)^{\text{ab}}$.
This one must too be surjective as we have $G\times K \to (G\times K)^{\text{ab}}$ being surjective and we then have a homomorphism chain
$$G\times K \to G^{\text{ab}}\times K^{\text{ab}} \to (G\times K)^{\text{ab}}$$
which must be equivalent. We have then established 2 epimorphisms going in both directions, which means they must be isomorphisms.
