Subgroup of a finite group with a certain property is a Hall subgroup? 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?
 A: 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 $G$.
A: 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.
