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I was wondering if it's really the case that, if $G$ is a group with subgroups $H$ and $N$ such that $H\unlhd N\unlhd G$ such that $G/N$ and $N/H$ is a $p$-group, then the intersection of all the conjugates of $H$ in $G$ is also of $p$th power index.

This problem actually comes from a Galois Theory problem, which asks to prove that if $F\hookrightarrow K\hookrightarrow L$ are field extensions such that $K$ is Galois over $F$, and $L$ is Galois over $K$, both with $p$ groups as Galois groups, then the Galois closure of $L$ over $F$ also has a $p$ group as a Galois group.

The Galois theory solution is to note that $L$'s Galois closure is the composite field of the conjugates of $L$ in an algebraic closure of $F$, and use the fact that the degree of a composite extension divides the product of the degrees of each composite factor. But I was wondering if there were a purely group-theoretic proof of (what I think is) the corresponding statement stated in terms of the Galois groups.

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What was that Galois theoretic statement again? The splitting field of $x^3-2$ over $\Bbb{Q}$ is the compositum of two cubic extensions, yet it is a sextic extension and $6\nmid 9$. –  Jyrki Lahtonen Jun 28 at 15:16
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The statement is that of $L_1,\ldots,L_n$ are Galois over $K$, then $[L_1\cdots L_n:K]$ divides $[L_1:K]\cdots[L_n:K]$. This does not apply to your example since the composite factors are not Galois. –  Nishant Jun 28 at 15:53
    
Ok. I think I got it :-) –  Jyrki Lahtonen Jun 28 at 16:04

1 Answer 1

up vote 3 down vote accepted

This follows from the following observation.

Lemma. Let $G$ be a finite group. Assume that $H_1\unlhd G$ and $H_2\unlhd G$ both have index that is a power of a prime $p$. Then $H_1\cap H_2\unlhd G$, and $[G:H_1\cap H_2]$ is also a power of $p$.

Proof. The normality of $H_1\cap H_2$ in $G$ is obvious. We also know that $H:=H_1H_2$ is a normal subgroup of $G$. Because $H_1\subseteq H\subseteq G$ the index $[H:H_1]$ is a power of $p$. By one of the fundamental isomorphisms we have $$ H/H_1\cong H_2/H_1\cap H_2. $$ Therefore $[H_2:H_1\cap H_2]$ is a power of $p$ as well. From the formula $$ [G:H_1\cap H_2]=[G:H_2]\cdot[H_2:H_1\cap H_2] $$ we then infer that $[G:H_1\cap H_2]$ is a power of $p$. Q.E.D.

An induction then proves the Corollary that any finite intersection of normal subgroups of $p$-power indices also has $p$-power index. The Lemma is both the base case and the inductive step.

Turning to the claim at hand. Conjugation by an element $g\in G$ is an automorphism of $N$, so all the groups $H^g$ are normal subgroups of $N$ of a $p$-power index. Applying the Corollary to $N$ and all the conjugates $H^g$ then shows that $[N:\operatorname{core}_G(H)]$ is a power of $p$. The main claim follows from this.


My earlier solution (that is conceptually more complicated) striked out below. Leaving it here, because some of my algebra students lurk (or even participate!) here, and I want them to 1) know that I have blind spots, 2) develop an urge to find solutions to HW assignments that are better than what I had in mind.


Let $q$ be a prime, $q\neq p$. Let $Q$ be a Sylow $q$-subgroup of $G$. Because $[G:N]$ is a power of $p$, the Sylow $q$-subgroups of $N$ have the same size as Sylow $q$-subgroups of $G$. Thus $N$ contains a conjugate of $Q$. But $N\unlhd G$, so actually $Q\le N$.

So all the Sylow $q$-subgroups of $G$ are contained in $N$. Hence they are Sylow $q$-subgroups of $N$.

Let $H ^g$ be an arbitrary conjugate of $H$. Because conjugation by $g$ is an automorphism of $N$, we can deduce that $H^g\unlhd N$ and $[N:H^g]$ is a power of $p$. Repeating the earlier argument tells us that $H^g$ contains all the Sylow $q$-subgroups of $N$, i.e. all the Sylow $q$-subgroups of $G$. As this holds for all $g$ the intersection $$ \operatorname{core}_G(H)=\bigcap_{g\in H}H^g $$ then contains all the Sylow $q$-subgroups of $G$. Thus $[G:\operatorname{core}_G(H)]$ is coprime to $q$.

The argument works for all primes $q\neq p$, so we can conclude that $[G:\operatorname{core}_G(H)]$ must be a power of $p$.

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Great, thanks! And of course, to finish the Galois correspondence, I'd assume that the core of $H$ is trivial, right? –  Nishant Jun 28 at 15:56
    
@Nishant: At the very least the core is a normal subgroup of $G$. I think that suffices for the Galois theoretic argument, right? –  Jyrki Lahtonen Jun 28 at 16:02
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Ah, so the core is a Galois extension of $L$ containing $F$, and this it must be the Galois closure! –  Nishant Jun 28 at 16:12
    
Duh. I knew there had to be something simpler. Just "having a blind spot". –  Jyrki Lahtonen Jun 30 at 9:03

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