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Let $A$ be a (possibly infinite) group.

Consider subgroups $C\lhd B\lhd A$, and assume that $A/B$ and $B/C$ are both finite $p$-groups.

Is there necessarily a subgroup $D$ normal in $A$ and contained in $C$ such that $A/D$ is a finite $p$-group?

This is related to another question:

If the subgroup $H$ of $G$ is open in pro-$p$ topology, does it inherit the pro-$p$ topology?

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Milford
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1 Answers1

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Since the core of $C$ in $A$ has finite index in $A$, we can assume without loss of generality that $A$ is finite.

But then we can just choose $D=O^p(A)$, which is the smallest normal subgroup of $A$ such that the quotient is a $p$-group or, alternatively, the subgroup of $A$ generated by the $p'$-elements of $A$.

So the answer is yes.

Derek Holt
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  • I understand the passage to finite groups. What are $p'$-elements? – Milford Jun 20 '15 at 19:09
  • I'm confused. I'm not sure why this $D$ is contained in $C$. – Milford Jun 20 '15 at 19:20
  • Yes, I've replaced $G$ by $A$. A $p'$-element is an element whose order is not divisible by $p$. It is not difficult to see that if you define $D$ to be the subgroup of $A$ generated by its $p'$-elements, then $A/D$ is a $p$-group and, if $E$ is any normal subgroup of $A$ such that $A/E$ is a $p$-group, then $D \le E$, so $D$ is the unique smallest normal sugbroup whose quotient is a $p$-group. – Derek Holt Jun 20 '15 at 19:21
  • Ok. What remains for me is to understand why all $p'$-elements are contained in $A$. Probably raising them to appropriate powers until they're in $B$ and then in $A$ will work. I'll think about it. – Milford Jun 20 '15 at 19:26
  • Yes, it seems to work. Let $x$ be a $p'$-element of $A$. Then after raising $x$ to a certain power of $p$ we get an element of $B$, and after raising again to some power of $p$ we get an element of $C$. So $x^{p^m}$ is an element of $C$. Now we find the inverse of $p^m$ modulo the order of $x$, denote it by $t$, and see that $x=x^{tp^m}$ is an element of $C$. – Milford Jun 20 '15 at 19:32
  • Is that the argument you had in mind? – Milford Jun 20 '15 at 19:33
  • In the answer itself, you say "largest normal subgroup". This should be "smallest". – Milford Jun 20 '15 at 19:38
  • Yes, I meant smallest not largest. – Derek Holt Jun 20 '15 at 22:31