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