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I am revising basic Group Theory and was hoping to check if my understanding about this is correct. 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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up vote 6 down vote accepted

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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