0
$\begingroup$

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 $|G|-[|G:H|(|H|-1)+1]=|G|/|H|-1$

  • 0
    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.2012-06-19

1 Answers 1