# Is this proof that $\widehat{G_p}$ is pro-$p$ free correct?

Let $$G$$ be an abstract group with the following presentation:

$$G \simeq \langle x,y \mid x^2y^2 = 1 \rangle$$

Let $$p \neq 2$$ be an odd prime. I want to show that $$\widehat{G_p} \simeq \mathbb{Z}_p$$ (here, $$\widehat{G_p}$$ denotes the $$p$$-completion of $$G$$, and $$\mathbb{Z}_p$$ denotes the $$p$$-adic integers).

Proof. First, notice that $$x^2y^2 \equiv (xy)^2 \mod [G,G]$$, so the relation can be written as $$(xy)^2w(x,y) = 1$$ with $$w(x,y) \in [G,G]$$. By introducing the letter $$t = xy$$, we can rewrite the presentation as $$G \simeq \langle t,y \mid t^2w(ty^{-1},y) = 1\rangle\,.$$

This proves that the abelianization of $$G$$ is $$G^{\text{ab}} = G/[G,G] \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$$. The $$p$$-completion and abelianization functors commute, so we have $$\widehat{G^{\text{ab}}_p} \simeq \mathbb{Z}_p$$

Now, we have that the Frattini quotient $$\widehat{G_p}/\Phi(\widehat{G_p})$$ factors trough $$\widehat{G_p^{\text{ab}}}$$, so this means that this quotient is cyclic. It is well known that $$\widehat{G_p}$$ and $$\widehat{G_p}/\Phi(\widehat{G_p})$$ have the same rank (as pro-$$p$$ groups), which proves that $$\widehat{G_p} \simeq \mathbb{Z}_p$$.

Given that $$G$$ is the fundamental group of a surface, I was expecting $$\text{cd}(\widehat{G_p}) = 2$$ (i.e. not a pro-$$p$$ free group) but I can't find any error in the argument, which is making me suspicious.