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Let $G$ be a group, and let $H$ be a subgroup of $G$. Let $x,y\in G$ be such that $xH\subseteq Hy$. Can we conclude that $xH=Hy$? this certainly holds if $H$ is finite because both sets will have the same finite cardinality. But what about the case $|H|=\infty$? I am especially interested in the case $x=y$.

EDIT: As Thomas Andrews pointed out in the comments, it is suffices to consider the case $x=y$, because the hypothesis $xH\subseteq Hy$ implies $x\in Hy$, hence $Hy=Hx$. And so, the question becomes:

Can a subgroup (of a group, of course) strictly contain some of its conjugates?

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Yeah, I think @KarolisJuodelė is confused. – Thomas Andrews May 21 '14 at 5:41
If $x\in Hy$ then $Hy=Hx$. So you can reduce the question to $x=y$. – Thomas Andrews May 21 '14 at 5:44
The question then becomes: Can $xHx^{-1}\subsetneq H$. – Thomas Andrews May 21 '14 at 5:45
I think that you should remove the "EDIT", as with it your post is a duplicate (of, say, this one). Without it it is interesting :-) – user1729 May 21 '14 at 7:48
up vote 8 down vote accepted

Yes. $xH\subsetneq Hx$ is equivalent to $xHx^{-1}\subsetneq H$, so we want to find a subgroup strictly containing one of its conjugates. This is possible with infinite groups. For instance, let $\psi$ act on $\Bbb Q$ by $x\mapsto2x$, and form the semidirect product $G=\Bbb Q\rtimes\langle\psi\rangle$. Then $\psi \Bbb Z\psi^{-1}=2\Bbb Z\subsetneq\Bbb Z$. I also gave this example over at this question, and by googling "subgroup strictly contains conjugate" you can find more.

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There are also lots of examples here and in the questions linked therein. – user1729 May 21 '14 at 7:46

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