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One sentence from Amstrong's Group and Symmetry wrote the following to prove a group of order 6 is isomorphic to $\mathbf{Z}_6$:

The right cosets $\langle x\rangle$, $\langle x\rangle y$ give 6 elements $e$, $x$, $x^2$, $y$, $xy$, $x^2y$ which fill out $G$.

Where $x$ is of order $3$ and $y$ is of order 2. $x$, $y$ are both elements of $G$.

My confusion:

How can one guarantee that the right coset $\langle x\rangle$ and $\langle x\rangle y$ can fill out $G$? Because $\langle x\rangle$ and $\langle x\rangle y$ have no intersections? I am not sure about the properties of cosets. I also don't know if this is related to the fact that $|\langle x\rangle|=5$ and is precisely half the elements of $G$ and thus its left coset $y\langle x\rangle$ and right coset $\langle x\rangle y$ is exactly the same?

I don't know if you can understand my question. The question here actually arises from my vague understanding of "cosets". So everytime this term occurs I feel a steady uncertainty.

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    You may want to reread the question yourself and see where one or more letters can be added.2012-03-28
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    @KannappanSampath: An artifact of using `<` and `>` in text; parser thought they were mark-up.2012-03-28
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    This is specifically page 70 (and end of 69).2012-03-28
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    Exactly. It is from page 70. Sorry for the disorganized order, I will check that next time before I submit. And thanks again to @ArturoMagidin for the edition.2012-03-28
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    For more on cosets, see [this answer](http://math.stackexchange.com/a/16842/742).2012-03-28

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