# For a pro-p, profinite group, abelianization being finitely generated is the same as being topologically finitely generated

I remember reading (without proof) that for $$\Gamma$$ a profinite, pro-$$p$$ group, the following are equivalent:

1) Every open subgroup $$\Gamma_0$$ is topologically finitely generated.

2) The abelianization of every open subgroup $$\Gamma_0$$ is finitely generated (as a $$\mathbb Z_p$$ module .

3) Every open subgroup $$\Gamma_0$$ only admits finitely many maps $$\Gamma_0 \to \mathbb F_p$$.

Clearly, $$1) \implies 2) \implies 3)$$ but how do you show equivalence?

• The cardinality of a minimal set of topological generators of a pro-$p$-group $G$ is $\mathrm{dim}\,\mathrm{H}^1(G,\mathbf{F}_p)$, see Cohomology of Number Fields, (3.9.1). – TKe Dec 2 '18 at 15:17

Following TKe's reference, here is the solution:

Suppose $$\mathscr I$$ is a set of elements in $$G$$, a pro-$$p$$ group.

Denote by $$G^*$$ the group generated by $$G^p$$ and $$[G,G]$$ together. Note that $$H^1(G,\mathbb F_p) = \operatorname{Hom}(G,\mathbb F_p) = \operatorname{Hom}(G/G^*,\mathbb Q_p/\mathbb Z_p)$$. Then, the claim is that $$\mathscr I$$ generates $$\Gamma$$ (topologically) if and only if $$\mathscr I$$ generates $$G/G^*$$ as a $$\mathbb F_p$$ vector space.

This is clearly sufficient to show that $$3) \implies 1).$$ The crucial idea is the following:

Lemma: Any maximal closed subgroup $$H \subset G$$ is normal of index $$p$$.

Proof: There exists an open subgroup $$U$$ such that $$HU > U$$ (otherwise $$H=G)$$. But then, by Sylow theory, the image of $$H$$ in $$G/U$$ is a maximal proper subgroup and is therefore of index $$p$$ and normal. Moreover, the pullback of this subgroup is in fact exactly $$H$$ since $$H$$ is maximal and we are done.

Finally, we have:

Lemma: If $$G,G'$$ are two $$pro-p$$ groups, then a map $$f: G \to G'$$ is surjective if and only if $$f^*: H^1(G',\mathbb F_p) \to H^1(G,\mathbb F_p)$$ is injective.

Proof: Clearly, if $$f$$ is surjective, $$f^*$$ is injective. Conversely, suppose $$f$$ is not surjective and let $$H = f(G)$$. Then, $$H$$ is closed (by compactness+hausdorff) and hence is contained in a maximal normal subgroup $$H'$$. $$G/H'$$ is then isomorphic to $$\mathbb F_p$$ by the previous lemma and hence gives an element in $$H^1(G',\mathbb F_p)$$ whose restriction along $$f^*$$ is $$0$$ and hence $$f^*$$ is not injective.

To complete the proof, consider a pro- $$p$$ group $$G$$ such that $$H^1(G,\mathbb F_p)$$ is finite dimensional and let $$H$$ be a finitely generated subgroup whose image in $$G/G^*$$ generates $$G/G^*$$. Then:

$$H \to G$$ is surjective $$\iff$$ $$H^1(G,\mathbb F_p) \to H^1(H,\mathbb F_p)$$ is injective $$\iff$$ $$H/H^* \to G/G^*$$ is surjective where the final equivalence is by Poincare duality.

• – TKe Dec 2 '18 at 16:21