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Revising for Group Theory at the moment and I'm sort of stumped on this exercise:

Given a subset $C$ of $G$, prove that $C$ is a left coset of a certain subgroup of $G$ iff $$xy^{-1}z \in C $$ when $$ x,y,z \in C.$$

I'm not sure how to start solving this, so any hints are welcome.

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@J.D. I appreciate that you changed title in effort to make it more informative, but I think that your version of the title was different from the actual question. –  Martin Sleziak Aug 3 '12 at 18:34
    
@MartinSleziak Oh, now I see. I changed the question! Thanks for the feedback. –  user2468 Aug 3 '12 at 18:36
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2 Answers

up vote 4 down vote accepted

If $C=cH$, for some subgroup $H$, then you can determine $H$ by taking the set of all values $y^{-1}z$ where $y,z\in C$.

So let's try that. Given a $C$ with the property above, define: $$H=\{y^{-1}z: y,z\in C\}$$

Step: Prove $H$ is a subgroup - that it is non-empty, and it is closed under multiplication and inverses. (You need to add the condition that $C$ is non-empty to prove this.)

Step: Given any element $c\in C$, all elements can be written as $ch$ for some $h\in H$

So if $C$ has this property and is non-empty then $C$ is a left coset.

If $C$ is a left coset, it is pretty clear it is non-empty, and it's direct to prove that $C$ has this property. Specifically, if $C=cH$, then $x=ch_1$, $y=ch_2$ and $z=ch_3$, for $h_1,h_2,h_3\in H$, and we get $$xy^{-1}z = ch_1h_2^{-1}c^{-1}ch_3 = c(h_1h_2^{-1}h_3)\in C$$

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You probably need the assumption that $C$ is non-empty, too. The empty set fulfills this condition, but it is not a left coset.

$\boxed{\Rightarrow}$ If $C=aH$ for some subgroup $H$ then every element of $C$ can be written as $ah$ for some $h\in H$. If you write $x=ax_1$, $y=ay_1$, $z=az_1$, where $x_1,y_1,z_1\in H$, what can you say about $xy^{-1}z$?

$\boxed{\Leftarrow}$ Suppose that $C$ fulfills the condition $x,y,z\in C$ $\Rightarrow$ $xy^{-1}z\in C$. Fix arbitrary $a\in C$. Denote $H=\{x\in G; ax\in C \}$. Can you show that $H$ is a subgroup? Can you show that $C=aH$?

(Note that we have used $C\ne\emptyset$ at the step where we have picked an arbitrary element $a\in C$.)

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