I'm given two groups, say $G$ and $H$, and two homomorphism, say $f$ and $g$ both from $G$ to $H$. Define $A=\{ x\in G | f(x)=g(x)\}$, I've managed to prove that $A$ is a subgroup of $G$ and that if $H$ is abelian $A$ is also normal then I'm stucked in proving that if $G=\ker (f) A$ then $f$ coincides with $g$ (it is not specified if this happens when $H$ is abelian so I don't know if I have to suppose that from the first point that I've mentioned).
Done that I have to work in the case in which $G$ has order $36$ while $H$ has order $3$ (so it is isomorphic to $\mathbb Z_3$) and $9$ divides the order of $A$ to conclude again that $f$ and $g$ must coincide. In this case $H$ is abelian so I was thinking to prove that $G/A$ is isomorphic to $\ker f$ but I don't know how to proceed any further (and if it is the right way in the first place).
The conditions given in the first paragraph are not enough, even if we assume that neither $\mathrm{ker}(f)$ nor $A$ are the the whole group (or nontrivial); let $G=H=C_2\times C_2$, with $C_2$ generated by $x$ and written multiplicatively. Let $f,g\colon G\to H$ be given by $f(a,b) = (1,b)$, and $g(a,b)=(a,b)$. Then $A=\{(1,1), (1,x)\}$, and $\mathrm{ker}(f) = \{(1,1),(x,1)\}$; therefore, $\mathrm{ker}(f)A=G$, but $f\neq g$.
In the specific case that $H\cong C_3$, $|G|=36$, and it is known that $9\bigm||A|$, the situation is pretty easy.
Note that any $2$-Sylow subgroup of $G$ must be in the kernel of any homomorphism from $G$ to $C_3$; in particular, both $f$ and $g$ must agree on all $2$-Sylow subgroups of $G$. Thus, $A$ contains a subgroup of order $4$. Since we also know that the order of $A$ is a multiple of $9$, it follows that the order of $A$ must be $36$; that is, $A=G$, so $f=g$.