Maps from a finitely generated pro-p group to $\mathbb F_p$ factors through Frattini quotient

Question

Let $G$ be a finitely generated pro-p group, these notes (p.99, Corollary 5.4.21) claims that all maps from $G$ to $\mathbb F_p$ factor through the Frattini quotient ($G/\Phi(G)$), where $\Phi(G)$ is the Frattini subgroup $\overline{[G,G]G^p}$. But I am very confused why it's true. I can't find any resources in these notes for an explanation. I would be very grateful if someone could help.

Answer 1

By map they actually mean a continuous homomorphism (otherwise the claim is wrong, just define $f(g) = 1$ for some $g\in \Phi(G)$ and zero otherwise).

Then it's not difficult, in fact it's true for any continuous homomorphism $\chi:G\rightarrow \mathbb{F}_p$ even when $G$ is any group.

Indeed,

Lemma 1: $[G,G]\subseteq \ker \chi$.

Proof: $\ker \chi$ is a group so it suffices to show that $\chi([g,h])=0$. Indeed, $$\chi([g,h]) = \chi(h^{-1}g^{-1}hg) = \chi(h^{-1})+\chi(g^{-1})+\chi(h)+\chi(g) = -\chi(h) - \chi(g) +\chi(h) + \chi(g)=0$$ where the last equality follows beacuase $\mathbb{F}_p$ is abelian.

Lemma 2: $G^p\subseteq \ker \chi$

Proof: $\chi(g^p) = p\cdot \chi(g)=0$.

Lemma 3: If $A,B\leq \ker \chi$ then $A\cdot B \leq \ker \chi$.

Proof: $\chi(a\cdot b) = \chi(a) + \chi(b) =0+0 =0$.

Finally,

Lemma 4: If $A\leq \ker \chi$, then $\overline{A} \leq \ker \chi$.

Proof: It suffices to show that $\ker \chi$ is closed, but this is immediate as $\ker\chi = \chi^{-1}(\{0\})$ is the pre-image of a closed set.