# Coset

## Definition

Let [ilmath](G,\times)[/ilmath] be a group and [ilmath](H,\times)[/ilmath] a subgroup we denote cosets as follows:

Given any $g\in G$ the

• Left coset (the left coset of [ilmath]H[/ilmath] in [ilmath]G[/ilmath] with respect to [ilmath]g[/ilmath])
is denoted $gH=\{gh|h\in H\}$
• Right coset (the right coset of [ilmath]H[/ilmath] in [ilmath]G[/ilmath] with respect to [ilmath]g[/ilmath])
is denoted $Hg=\{hg|h\in H\}$

It is quite simply: the set of everything in [ilmath]H[/ilmath] (pre/post) multiplied by [ilmath]g[/ilmath]

## Properties

These will be stated for the left coset definition, but the right version is basically the same

### Membership

To say $x\in gH$ is to say $\exists y\in H:x=gy$ that is:

• $[x\in gH]\iff[\exists y\in H:x=gy]$

### Cosets are either disjoint or equal

Given two cosets, $g_1H$ and $g_2H$ we have either $g_1H=g_2H$ or $g_1H\cap g_2H=\emptyset$

Suppose they are not disjoint$\implies$ they are equal

If they are not disjoint, then take any $k\in g_1H\cap g_2H$, this means [ilmath]k[/ilmath] is in both [ilmath]g_1H[/ilmath] and [ilmath]g_2H[/ilmath] so:
• $k\in g_1H\implies \exists z_1\in H:k=g_1z_1$
• $k\in g_2H\implies \exists z_1\in H:k=g_2z_2$
This means $g_1z_1=g_2z_2$ allowing us to say:
• $g_1=g_2z_2z_1^{-1}$ where $z_2z_1^{-1}\in H$
• $g_2=g_1z_1z_2^{-1}$ where $z_1z_2^{-1}\in H$
We now need to show equality of sets
1. $g_1H\subseteq g_2H$, that is $x\in g_1H\implies x\in g_2H$
Take $x\in g_1H$, this means $x=g_1y_1$ for some $y_1\in H$
Using $g_1=g_2z_2z_1^{-1}$ we see $x=g_2z_2z_1^{-1}y_1$
but $z_2z_1^{-1}y_1\in H$ so let $h_1=z_2z_1^{-1}y_1\in H$ then
$x=g_2h_1\in g_2H$
we have shown $x\in g_1H\implies x\in g_2H$ if they are not disjoint, thus $g_1H\subseteq g_2H$
2. $g_2H\subseteq g_1H$, that is $x\in g_2H\implies x\in g_1H$
Take $x\in g_2H$, this means $x=g_2y_2$ for some $y_2\in H$
Using $g_2=g_1z_1z_2^{-1}$ we see $x=g_1z_1z_2^{-1}y_2$
but $z_1z_2^{-1}y_2\in H$ so let $h_2=z_1z_2^{-1}y_2\in H$ then
$x=g_1h_2\in g_1H$
we have shown $x\in g_2H\implies x\in g_1H$ if they are not disjoint, thus $g_2H\subseteq g_1H$
Combining these we see $g_1H\cap g_2H\ne\emptyset\implies g_1H=g_2H$

Suppose they are disjoint$\implies$ they are not equal

If they are disjoint then trivially the sets $g_1H$ and $g_2H$ are not equal.

So we may conclude $[g_1H=g_2H]\iff[g_1H\cap g_2H\ne\emptyset]$