Is $\mathbb{Z}\times\mathbb{Z}\to S_4:(1\,0)\to(1\,2)(3\,4),(0\,1)\to(1\,3)(2\,4)$ a homomorphism? I know the answer is "yes" because the cycles $(1\,2)(3\,4)$ and $(1\,3)(2\,4$ are disjoint and thus commute, but I don't understand how commutativity alone satisfies the homomorphism property
 A: Define the function by $$f:\Bbb{Z}\times \Bbb{Z} \to S^4:(n,m)\mapsto (12)^n(34)^n(13)^m(24)^m.$$ Then $f(1,0)f(0,1)=f(0,1)f(1,0)$ because, as you said, the permutations commute, and $f(n,m)=f(1,0)^nf(0,1)^m$ by definition.
So $$f((a,b)+(c,d))=f(a+c,b+d)=f(1,0)^{a+c}f(0,1)^{b+d}=f(1,0)^af(1,0)^cf(0,1)^bf(0,1)^d=f(1,0)^af(0,1)^bf(1,0)^cf(0,1)^d=f(a,b)f(c,d)$$ and $f$ is indeed a homomorphism.
A: A categorical answer:
$\mathbb{Z}\times\mathbb{Z}$ is a coproduct "$\mathbb{Z}\oplus\mathbb{Z}$" in the category of abelian groups (and it isn't one in the category of groups). A homomorphism $\mathbb{Z}\oplus\mathbb{Z}\rightarrow A$ is then determined, and determined uniquely, by its values on the summands of the coproduct; that is, a homomorphism $\mathbb{Z}\oplus\mathbb{Z}\rightarrow A$ is equivalent to a pair of homomorphisms $\mathbb{Z}\rightarrow A$... for an abelian group $A$. Each homomorphism $\mathbb{Z}\rightarrow A$ is of course determined, and determined uniquely, by its value on $1$ ($\mathbb{Z}$ is a free abelian group on a single generator).
Therefore if the subgroup of $S_4$ generated by those permutations is an abelian subgroup, then the homomorphism you'd like to define exists.
