Let $(G,+)$ and $(H,\star)$ be groups (with respectively noted relations $+$ and $\star$). Let $f$ and $g$ be group homomorphisms $f,g:G \to H$.
Claim: If $H$ is abelian, then the product map $f \cdot g,\,x \mapsto f(x) \star g(x)$ is also a homomorphism.
I was able to prove this, but now I am wondering if there exists a concrete but simple counterexample for when $H$ is not abelian... (e.g. with whole numbers under addition or similar...)
Well, you could consider the left multiplications $\ell_\pi:S_n\rightarrow S_n:\sigma\mapsto\pi\sigma$ with given $\pi\in S_n$, where $S_n$ is the symmetric group on $n$ places.
The set of left multiplications $L(S_n)$ forms a group under function composition, since $\ell_\pi\ell_\sigma=\ell_{\pi\sigma}$, $(\ell_\pi)^{-1} = \ell_{\pi^{-1}}$ and $\ell_e= id$ is the identity mapping, where $e$ is the identity permutation.
Consider the mapping $F:S_n \rightarrow L(S_n):\pi\mapsto \ell_\pi$, which is a homomorphism, since $F(\pi\sigma) =\ell_{\pi\sigma} = \ell_\pi\ell_\sigma = F(\pi)F(\sigma)$.