Let $G$ be a finite group, $H$ be a subgroup of $G$ with $[G:H]=2$ and for all $h \in H-{1}$, $C_G(h) \leq H$. Prove $G-H$ forms a conjugacy class.

Question

I'm studying for an exam and I am very much stuck on this problem. I wasn't able to find anything similar enough to this problem to help me yet, so I thought I would make a post.

Let $G$ be a finite group and $H$ a subgroup of $G$ with $[G:H]=2$. In addition, suppose that for all $h \in H-\{1\}$ that $C_G(h) \leq H$. Prove that the elements of $G-H$ form a conjugacy class of $G$.

So, I am starting to wonder if I am approaching this the wrong way. I've tried to prove this using double containment, starting both ways to see if one is easier, but I get stuck early on in both ways. I'll show what I have below.

Proof: Let $x \notin H$. We want to show $Cl(x) = G-H$.

$(\subseteq)$ Let $a \in Cl(x)$. Then there exists $g \in G$ such that $gxg^{-1}=a$. For the sake of contradiction, suppose $a \in H$. Then, $gx = ag$. If $g \in H$ then $gx \notin H$ as $[G:H]=2$ implies $G/H = \{H, Hx\}$. Yet, $ag \in H$ which would create a contradiction. Hence, $g \notin H$. [This is where I get stuck in this direction.]

$(\supseteq)$ Let $a \in G-H.$ Then, $a = hx$ for some $h \in H$. [This is where I get stuck here. I know we want to find some $g \in G$ such that $gxg^{-1} = hx$ but I am having trouble finding any such $g$.]

Answer 1

Let $x\not\in H$. Following @jacob, if $h_1\ne h_2\in H$, then $h_1xh_1^{-1}=h_2xh_2^{-1}\implies xh_2^{-1}h_1x^{-1}=h_2^{-1}h_1\implies x\in C_G(h_2^{-1}h_1)\subset H$.

Thus we have an injection from $H$ to the set of conjugates of $x$ by elements of $h$.

So $|\rm{cl}(x)|=|H|$.

Answer 2

Another way to look at it: consider the action of $H$ on $G$ by conjugation. For $x \in G-H$, $stab(x)$ is trivial by the given condition on centralizers. By orbit-stabilizer, $orbit(x)$ has size $|H|$. But $orbit(x)$ is just $cl(x)$.