# Showing $\mathbb{Z}_{2} * \mathbb{Z}_{3} \cong\ (a, b\ |\ a^2 = b^3 = e)$

Let $G = (a, b\ |\ a^2 = b^3 = e)$. I recognize there must be an epimorphism $\phi : G \rightarrow \mathbb{Z}_{2} * \mathbb{Z}_{3}$ (the free product) by the Van Dyck theorem, but I must show an isomorphism. Essentially, I believe we can do $\psi : G \rightarrow \mathbb{Z}_{2} * \mathbb{Z}_{3}$, $\psi(a^{p_{1}} b^{q_{1}} ... a^{p_{n}} b^{q_{n}}) = (1_{2})^{p_{1}} (1_{3})^{q_{1}} ... (1_{2})^{p_{n}} (1_{3})^{q_{n}}$ where $1_{2}$ is the generator of $\mathbb{Z}_{2}$ and $1_{3}$ the generator of $\mathbb{Z}_{3}$. But I'm having difficulty showing that this works, i.e. it is well-defined and an isomorphism. If we could say that $\psi = \phi$, i.e. it's an epimorphism, I believe seeing injectiveness is easy (by triviality of the kernel). But the proof of Van Dyck is not explicit, so I can't tell what $\phi$ is.

• I would use $g_n$ for the generator of $\mathbb Z_n$, just because it is potentially confusing when moving between mutliplication and additive groups to have $1$ ever not be the group identity... – Thomas Andrews Jul 10 '13 at 17:54
• If $\phi(a) = 1_2$ and $\phi(b) = 1_3$, then $\phi(aba) = \phi(a) \phi(b) \phi(a) = 1_2 1_3 1_2$. $\psi=\phi$. – Jack Schmidt Jul 10 '13 at 17:56
• But can we show that $\phi(a) = 1_{2}$ and $\phi(b) = 1_{3}$ must occur? – Pedro Jul 10 '13 at 18:00
• Ah, I believe I can recover that from the proof of Van Dyck. – Pedro Jul 10 '13 at 18:02