How to show $ G = \langle a,b|a^3 = b^2 = 1,a^2b = ba\rangle$ is isomorphic to $S_3$

Question

How to show G = $ \langle a,b|a^3 = b^2 = 1,a^2b = ba\rangle$ is isomorphic to $S_3$.

I am able to show G has six elements and how to send them to $S_3$. Then I can list the multiplication to prove this actually. But I just wonder how to prove this without multiplication table. Of cause, $S_3$ satisfies all the relation, but how can I know there does not exist other relation of elements in $S_3$?

Answer 1

Hint: $S_3$ is generated by $(1\ 2\ 3)$ and $(1\ 2)$. Map generator to generator, and show this is the desired isomorphism.