Let $H$ be a subgroup of $G$ such that for all $x \in G-H$ and all $y \in G$ there exist $u \in H$ such that $y^{-1}xy=u^{-1}xu$ ... Let $G$ be a group and $H$ be a subgroup of $G$ such that for each $x \in G-H $
and each $y \in G$, there is a $u \in H$ that $y^{-1}xy = u^{-1}xu$. Prove that $H \lhd G$,
and $G/H$ is abelian.
I saw this problem in a mathematical competition for university students in Iran. I'm completely confused. I don't know how to start. Could you please give me some hint. I'm not asking for the whole proof, a piece of hint or direction to start would be enough.
 A: HINT: First prove that $H$ is normal in $G$. Try to prove using the assumption that conjugation in $G$ sends things outside $H$ to things outside $H$, and hence must send things inside $H$ to things inside $H$.
To show that $G/H$ is abelian, try to show that in $G/H$, conjugation is trivial, and note that a group is abelian iff conjugation acts trivially in that group. 
****FULL PROOF****
Note that for $h\in H$, and $g\in G$, $hgh^{-1}\in H$ iff $g\in H$. Your assumption then says that for any $x\in G - H$, $g\in G$, $gxg^{-1} = hxh^{-1}$ for some $h\in H$, but $x\notin H$, so $gxg^{-1} = hxh^{-1} \in G - H$. This shows that conjugation in $G$ sends things outside $H$ to things outside $H$, but of course this also means that it must send things inside $H$ to things inside $H$ (if it sent something in $H$ to something outside $H$, then conjugation by the inverse sends something outside $H$ to something inside $H$, which we just proved can't happen).
This shows that $H$ is normal in $G$.
Now the assumption that for any $g\in G,x\in G-H$, we have $gxg^{-1} = hxh^{-1}$ for some $h\in H$, can be viewed mod $H$, where it becomes the statement: Conjugation in $G/H$ of nontrivial elements the same as conjugating by the identity, ie, conjugation in $G/H$ of nontrivial elements is trivial. Of course, conjugating the identity by anything is trivial, so this shows that conjugation in $G/H$ is trivial. The only groups where this is true are abelian groups.
