So the question I ran into while working through some group theory problems was, is the map $Z\times Z \to S_3$ defined by $ (i,j) \to (12)^i (123)^j $, a homomorphism?
By looking at the problem, I assume that it is not, by the sole fact that $Z \times Z$ is abelian and $S_3$ is not, but I applied the homomorphism property just to be sure. I am unsure if I am applying it right, and if I am, how do I finish. Any ways to proceed are helpful.
\begin{equation} \Phi ((i_1,j_1)+(i_2,j_2)) =(1,2)^{i_1 + i_2}(123)^{j_1 + j_2} \\ \Phi(i_1,j_1) \circ \Phi(i_2,j_2)= (1,2)^{i_1}(123)^{j_1} \circ (1,2)^{i_2}(123)^{j_2} = ??? \end{equation}
If you guess that $\Phi$ is not a homomorphism, you need to find points $(i_1,j_1)$ and $(i_2,j_2)$ in $\mathbb{Z}\times\mathbb{Z}$ such that $$ \Phi((i_1,j_1) + (i_2,j_2)) \neq \Phi(i_1,j_1)\Phi(i_2,j_2) $$ Keep it super-simple when looking for contradictions. We know that in $\mathbb{Z}\times\mathbb{Z}$ that $(1,0)$ and $(0,1)$ commute. Do their images $\Phi(1,0) = (12)$ and $\Phi(0,1) = (123)$ commute?