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Let $G$ be a finite group. Let $H \leq G$ such that $H^x \cap H = 1$ for all $x \in G \setminus H$. I wish to show that $H$ is a Hall subgroup of $G$.

Here's what I have so far. Let $|H|=n$ and $|G:H|=m$. So we want to show $\gcd(m,n)=1$. I try to look at $N_G(H)$. $H$ is not normal because otherwise $H^g=H$ for all $g \in G$. So $N_G(H)<G$, a strict inclusion. And of course we also have $H \lhd N_G(H)$, and $|G:H|=|G:N_G(H)||N_G(H):H|$. But I'm unable to deduce much about the indexes on the right hand side of this last equation.

How should I proceed to show that $H$ is a Hall subgroup?

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Let $p$ be a prime dividing $|H|$, $Q$ a Sylow p-subgroup of $H$, and $P$ the Sylow p-subgroup of $G$ containing $Q$. Think about the elements of $P-Q$. –  user641 May 10 '12 at 7:58
    
So, if we can show that $P=Q$ then the result follows. So suppose there exists $x \in P \setminus Q$. Then $|x|$ is a power of $p$. Any element in $H$ that has order a power of $p$ must lie in $Q$, so $x \not\in H$. Then $H^x \cap H =1$. But now I'm not sure about the next step. –  under May 10 '12 at 10:11
    
So $x \not\in N_G(H)$, otherwise $H^x=H$ and $H^x \cap H=H$. Not sure how to get a contradiction from here. –  under May 11 '12 at 10:06
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If $Q \neq P$ then $N_P(Q)$ is strictly bigger than $Q$, so pick an element $x\in N_P(Q)-Q$. Then $Q^x=Q$, so $Q\subset H^x\cap H$, and then... –  user641 May 11 '12 at 17:52
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What I mean is, if $Q \neq P$ we get that $Q \subset H^x \cap H = 1$ since $x \not\in H$, which would then imply $Q=1$, which is not possible. Therefore $Q=P$. –  under May 12 '12 at 10:36
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2 Answers

up vote 1 down vote accepted

This answer is based on the hints by Steve D.

Let $Q$ be a Sylow $p$-subgroup of $H$, and let $P$ be the Sylow $p$-subgroup of $G$ containing $Q$. The claim is that $P=Q$. Assume the contrary, i.e. $Q<P$. $P$ is nilpotent, so $N_P(Q)>Q$. Let $x \in N_P(Q)\setminus Q$. As $x \in P$, $x$ has order a power of $p$, so $x \not\in H$ for otherwise $x \in Q$. Since $Q \leq H$ and $Q=Q^x \leq H^x$, we have $Q \leq H^x \cap H=1$. This implies $Q=1$, which is impossible. Therefore $P=Q$. Then $p \nmid |G:Q|$, so $p \nmid |G:H|$. This holds for every prime $p$ dividing $|H|$. It follows that $\gcd(|H|,|G:H|)=1$, hence $H$ is a Hall subgroup of $G$.

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Are you taking MATHS720 at Auck uni this semester? I am taking it :D

There are several methods to prove this problem, the first two are based on the same fact.

The first way: As the hint Steve.D already made, one can show every Sylow subgroup of H must be a Sylow subgroup of G, here the idea is to prove by contradiction. Suppose $p|gcd(|H|,|G:H|)$, then by Cauchy's theorem showed in the lecture (as a corollary of SylowE theorem) H contains an element $h$ of order p, and G has a Sylow p-subgroup P which contains $h$. Next, one needs to show $Z(P) \leq C_H(x)$ for all $x \in P\setminus H$ , notice that by $C_G(x) \leq H$ for all $x \in H$\ {$1_G$} one can show $C_H(x)$ is trivial for all$x \in P\setminus H$, so it implies Z(P) is trivial; however every non-trival p-group has non-trivial centre, contradiction.

The second way: Observe that each non-trivial class of H has order divisible by the index $|G:H|$ , so it implies $|G:H|$ divides (|H|-1), thus H is a Hall subgroup of G.

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For a third approach you may consider the set $X:=(G\\( \cup H^x))\cup {1_G}$ where x runs over G, observe it has trivial intersection with H and $|H||X|=|G|$, then show |x|,|H| are coprime. –  Harry May 10 '12 at 12:31
    
Sorry, but I'm a bit confused... maybe I'm not clear with some basic concepts. For the first way: $Z(P)$ are those elements in $P$ that commute with those in $P$, while $C_H(x)$ are those elements in $H$ that commutes with $x$. So how do we know that an element of $P$ (in $Z(P)$) is an element of $H$ (in $C_H(x)$), because we don't know that $P \leq H$? (I'm even trying to prove $Z(P) \leq C_H(x)$ at this point, just trying to make sense of it). –  under May 11 '12 at 9:53
    
For the second way: can you explain what mean by "non-trivial class", and why is it divisible by $|G:H|$? I know that the number of conjugates of $H$ is equal to $|G:N_G(H)|$, but you seem to be referring to something else. I'm a little hazy on group actions... –  under May 11 '12 at 9:56
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