Let $G = \langle a,b; a^2b^{-3} = 1 \rangle$ and consider $(1 2), (123) \in S_3$. I need to show there is a unique homomorphism from $G$ to $S_3$ which sends $a$ to $(12)$ and $b$ to $(123)$.
I'm not really good with free groups and group presentations, so hopefully this is on the right track. I used the universal property of free groups that given any function from a set $S$ to a group $G$, there is a homomorphism from the free group generated by that set to $G$. So I was thinking maybe if I had $f: \{a,b\} \to S_3$ with $f(a) = (12)$ and $f(b) = (123)$, I would obtain $\varphi: F(\{a,b\}) \to S_3$. Then $G = F(\{a,b\}) / K$ where $K$ is the subgroup normally generated by the relation $a^2b^{-3}=1$.
Is the homomorphism that problem asking for somewhat related to this, or am not heading in the right direction at all? This is from an algebraic topology class, and I know that $G$ is the presentation of a fundamental group of a certain space. Would that be helpful in answering this question?