Equivalence of definitions of left coset

This article gives a proof/explanation of the equivalence of multiple definitions for the term left coset
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

We are given a group $G$ and a subgroup $H$. We'd like to know whether a subset $S$ of $G$ is a left coset of $H$. We want to show that the following three descriptions are equivalent:

1. $x^{-1}y$ is in $H$ for any $x,y\in S$, and for any fixed $x$, the map $y\mapsto x^{-1}y$ is a surjection from $S$ to $H$
2. There exists $g\in G$ such that $S=gH$
3. For any $x\in S$, $S=xH$

Proof

We clearly have (3) implies (2) (because $S$ is nonempty). Let's show that (2) implies (1), and (1) implies (3) (the order of proof isn't really important, and once you see the proof, you'll see that it works all ways).

(2) implies (1)

Suppose $S=gH$. Then, pick elements $x,y\in S$. By definition $x=ga$ and $y=gb$ with $a,b\in H$. So the element $x^{-1}y$ is $(ga)^{-1}(gb)=a^{-1}g^{-1}gb=a^{-1}b$. This element is in $H$.

Also, given any $h\in H$, we have $y=xh\in S$ and $h=x^{-1}y$, so every $h\in H$ occurs as $x^{-1}y$ for some choice of $x$ and $y$. So the map from $S$ to $H$ is a surjection.

(1) implies (3)

Suppose it is true that $x^{-1}y\in H$ for any $x,y\in S$. Then, pick any $x\in S$. We want to show that $S=xH$.

First, observe that $S\subset xH$. That's because given $y\in S$, $x^{-1}y=h\in H$, so $y=xh$.

We now want to show that $xH\subset S$. In other words, we want to show that any element of the form $xh$ lives inside $S$. But this follows from the fact that for any $h\in H$, there exists $y\in S$ such that $x^{-1}y=h$, so we get that $xh\in S$.