For any finite groups L and G, let $h(L,G)$ denote the number of homomorphisms from L to G and $i(L,G)$ denote the number of monomorphisms from L to G. Proof that
$$h(L,G)=\sum_{N \triangleleft \ L} i(L/N,G) $$
N is normal subgroup of L (denoted by $N \triangleleft \ L$).
I think that is a answer for this question.
If we have a monomorphism from $L/N$ to $G$ for any $N$, just make a composition with the $\pi: L \rightarrow L/N$, so we obtain a morphisms $L \rightarrow G$.
From the otherside, if we have $f:L \rightarrow G$ a morphism, we know that image of $f$ is a monomorphism $m: Im(f) \rightarrow G$, with property to decomposite $f$. For the first isomorphism theorem, $L/Ker(f) \simeq Im(f)$. For each $N$ normal, we make $\pi: L \rightarrow L/N$, and $\psi: L/N \rightarrow L/Ker(f)$. Thats compositions make the relation above presented.
If anyone have other ideia, show us please.