# G minus the union of conjugates of a self-normalizing subgroup

Let $G$ be a finite group with $H \leq G$ such that $H \cap H^g=1$ for all $g \in G-H$ .

Prove there are exactly $|G|/|H|-1$ elements of $G$ lying outside of all conjugates of $H$.

My query: Group action shows the number of conjugates of $H$ is $|G|/|H|$, so it remains to show $H^x \cap H^y$ is trivial for each pair of distinct $x ,y \in G-H$: Assume there is $1 \neq z \in H^x \cap H^y$, then there are $h_1, h_2 \in H-{1}$ such that $z=h_1^{x}=h_2^{y}$, this gives $h_1^{xy^-1}=h_2 \in H$ thus $xy^{-1} \in H$<--- I intended to get a contradiction from here but it seems not obvious.

Note: Once one can show every pair of conjugates of $H$ has trivial intersection then the answer is $|G|-[|G:H|(|H|-1)+1]=|G|/|H|-1$

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It is not true that any distinct pair of elements $x$ and $y$ in $G-H$ will give distinct conjugates: if $Hx = Hy$ (and hence $x^{-1}H = y^{-1}H$), then $x^{-1}Hx = x^{-1}Hy = y^{-1}Hy$, so $H^x = H^y$. But that is the only situation in which you can have any elements in common. – Arturo Magidin Jun 19 '12 at 3:41

## 1 Answer

You don't actually want to prove that if $x\neq y$ are both in $G-H$ then $H^x \cap H^y = \{1\}$. (It's not true, anyway: if $h\neq 1$ then setting $y=hx$ gives $H^y = H^x$, even though $y\neq x$ and $x,y\in G-H$). What you actually want to prove is that if $H^x\cap H^y\neq\{1\}$, then $H^x = H^y$. Then you'll take the sum over the distinct conjugates of $H$.

And this follows from what you have: if $z\in H^x\cap H^y$, then $xzx^{-1}\in H\cap H^{yx^{-1}}$, and since $z\neq 1$, it follows that $yx^{-1}\in H$, hence $Hy = Hx$, hence $H^x = H^y$, as above.

So the pairwise distinct conjugates of $H$ have trivial intersection, and we have exactly $|G|/|H|$ conjugates. Now you can conclude as you want: the number of elements in the union of distinct conjugates of $H$ is exactly $[G:H](|H|-1) + 1 = |G| - |G|/|H| + 1$.

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Just had a mental block when I got $xy^{-1} \in H$.. which implies $H^x=H^y$ so contradiction arized by my choice that $x \neq y$. Thanks for clarifying! – user31899 Jun 19 '12 at 3:57