# If cosets $xH \ne Hy$ then, their intersection is empty?

I am revising basic Group Theory and was hoping to check if my understanding about this is correct.

http://www.proofwiki.org/wiki/Left_and_Right_Coset_Spaces_are_Equivalent says that the left and right coset spaces are equivalent.

Does this mean if cosets $xH \ne Hy$ then, their intersection is empty?

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No, it means that there is a bijection between the set of left cosets and the set of right cosets. It is entirely possible that $xH\neq Hy$ yet their intersection is nonempty. For example, for a non-normal subgroup $H$ we have $xH\neq Hx$ for some $x$ yet $x\in xH\cap Hx$.

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I see I confused the two. Thanks for the answer! – Legendre Sep 28 '12 at 17:26

Not necessarily, unless $H$ is a normal subgroup.

For example, consider the group $G=S_3=\langle g,h|g^3=h^2=ghgh=e\rangle=\{e,g,g^2,h,gh,g^2h\}$, and let $H=\langle h\rangle=\{e,h\}$. It's a pretty easy multiplication table to construct (if you need to). List out the left cosets and right cosets of $H$, and you'll find several examples of $x,y\in G$ such that $xH\neq Hy$ and $xH\cap Hy\neq \emptyset$.

Note however that there are the same number of left cosets as right cosets. This holds in general for any group $G$ and any subgroup $H$. Try showing that $xH\mapsto Hx^{-1}$ is a (well-defined) bijection from the left cosets of $H$ in $G$ to the right cosets of $H$ in $G$. That is, show that $xH=yH$ if and only if $Hx^{-1}=Hy^{-1}$ (well-defined and injective), and observe that surjectivity is simple since $G$ is a group. (Hint for the first part: Don't forget that $H$ is a group!)

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Thanks for your helpful example. :) – Legendre Sep 28 '12 at 17:30

Edit: Oops, I made a mistake. Sorry. $xG = yG = G$ since $x^{-1}$ and $y^{-1}$ are elements of $G$. I'm keeping this answer here so maybe you could learn from my mistake :)

Original post:

I'd like to throw in another example that you may learn something from: Let $G$ be the free group generated by two elements, $x$ and $y$. Then $xG$ is the coset of all words that start with $x$ and $Gy$ is the coset of all words that end with $y$; these two are clearly not the same, but they have a non-empty intersection - for example, $xy$ lies in both cosets.

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Nice example. Thanks! +1 – Legendre Sep 28 '12 at 17:51
Interesting example! I wonder, though, if that's truly a coset in the same sense. After all, $G$ isn't partitioned into $G$, $xG$, and others. – Cameron Buie Sep 28 '12 at 17:53
@CameronBuie You're right; it's not a coset in the same sense - a coset would have to be $xH$ for a subgroup $H$ of $G$. It's not hard to find "proper" examples using the free group, though. – Yoni Rozenshein Sep 28 '12 at 17:57
Actually my whole example is wrong (edited to reflect this). Sorry! :( – Yoni Rozenshein Sep 28 '12 at 18:01

Just to add to what's already been said, you might be interested to hear that a lot of research has been done by people who asked questions similar to yours. A big deal is often made in advanced group theory about $xH \cap Hx$ (usually written in the form $H \cap x^{-1}Hx)$, which is called the "twist" of $H$ by $x$. Furthermore, "double cosets" are also a thing in the big leagues, which are written $HxK$ and work kind of like "shared" cosets between two groups $H$ and $K$.

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