I've been stuck on this for a long time ... I'm reading a textbook which simply states "this subgroup is proper" but it doesn't make sense to me.
Context: I have a pro-$p$ group $G$, which just means that for every open normal subgroup $N$, the quotient $G/N$ is a finite $p$-group. I want to prove the converse: if $N$ has finite index then $N$ is open. If it helps, I also know that $G$ is compact Hausdorff.
So let $m=[G:N]$ be finite. I have two lemmas:
- Fact #1: $G^p[G,G]$ is open (hard to prove, but let's take it for granted).
- Fact #2: $[G:N]$ must be a power of $p$.
By Fact #1, the subgroup $G':=G^p[G,G]N$ must be open in $G$. That makes sense to me. Now the author simply says "by Fact #2, we have $G'\subsetneq G$. If we have this we're done: then $[G':N]<m$ so $N$ is open in $G'$ by induction, which implies it's open in $G$.
But I don't understand that remark. All we know is that $N\subseteq G'\subseteq G$, and $G/N$ is a finite $p$-group. Why does that imply $G'\subsetneq G$? There's something like $$[G:N]=[G:G'][G':N]$$ but it's totally possible that $[G:G']=1$, so this wouldn't help. Would it?
For reference, this is the proof of Proposition 1.19 in Analytic Pro-$p$ Groups by Dixon et al.