Surjection of $\text{Hom}(-,I)$ where $I$ is injective

Question

Let $I$ be an injective abelian group and $G\to H$ an injective morphism of groups, not necessarly abelian. If $G,H$ were abelian, then we would have a surjection $$\text{Hom}(H,I)\to \text{Hom}(G,I).$$ Can we still say something about $\text{Hom}(H,I)\to \text{Hom}(G,I)$ if we do not assume that $G,H$ are abelian. Since $I$ is assumed to be abelian, this is the same as the morphism $$\text{Hom}(H^{ab},I)\to \text{Hom}(G^{ab},I)$$ but the abelianization functor $(-)^{ab}$ is only right-exact, but not left-exact. We have a left exact sequence $0\to G\cap[H,H]/[G,G]\to G^{ab}\to H^{ab}$ which yields $$\text{Hom}(H^{ab},I)\to \text{Hom}(G^{ab},I)\to \text{Hom}(G/G\cap[H,H],I)\to \text{Ext}^1(G^{ab},I)=0.$$ It is probably to much to hope for $\text{Hom}(G\cap[H,H]/[G,G],I)=0$, but are there "nice" things to assume make it true?