# 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 $xh \in S$.