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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.

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