Partitioning Group by Cosets

Given a subgroup $H$ of a group $G$ , $G$ can be expressed as a disjoint union of cosets of $H$ in $G$ , either all left or all right cosets. By this theorem, these cosets are all of the same size.

This is clear to see from proving two related results (either of which would be sufficient).

Theorem

Given two left cosets of $H$ in $G$ , $g H$ and $g^{'} H$ , exactly one of the following is true:

$g H = g^{'} H$
$g H \cap g^{'} H = \emptyset$

That is, any two cosets are either equal or disjoint.

We are going to prove this by showing that:

g H \cap g^{'} H \neq \emptyset ⟺ g H = g^{'} H

That is, the first case is true if and only if the negation of the second is false (i.e. the second is true).

For the first implication, we will assume that $g H \cap g^{'} H \neq \emptyset$ , that is, there exists an element which is common to both cosets, or an $h, h^{'} \in H$ such that:

g h = g^{'} h^{'} .

It then follows from this that:

\begin{aligned} g h = g^{'} h^{'} \\ ⟹ & g = g^{'} (h^{'} h^{- 1}) \\ ⟹ & g \in g^{'} H \end{aligned}

and likewise that $g^{'} \in g H$ .

Now we will show $g H = g^{'} H$ by showing inclusions both ways.

Let $a \in g H$ be arbitrary, that is

a = g h_{1}

for some $h_{1} \in H$ . However since $g \in g^{'} H$ ,

g = g^{'} h_{1}^{'}

for some $h_{1}^{'} \in H$ .

Substituting the second of these expressions into the first we have that $a = g^{'} (h_{1}^{'} h_{1})$ , and hence $a \in g^{'} H$ . Since $a$ is chosen arbitrarily, $g H \subseteq g^{'} H$ . The reverse inclusion follows in the same way.

Corollary

$a \in b H$ for some cosets of $H$ in $G$ if and only if $a H = b H$ .

This follows simply from the fact that $a \in a H$ trivially, and hence $a H \cap b H$ at least contains $a$ . This means they are not disjoint, and hence $a H = b H$ .

Theorem

The operation

g \sim g^{'} ⟺ g \in g^{'} H

is an equivalence relation on $G$ .

Being an equivalence relation on $G$ this partitions $G$ into equivalence classes, which then correspond with the sets in the disjoint subset we use to express $G$ .

Symmetry

Let $g \sim g^{'}$ for $g, g^{'} \in G$ . Then, $g \in g^{'} H$ , and hence there exists an $h \in H$ such that $g = g^{'} h$ .

Therefore $g h^{- 1} = g^{'}$ , and $g^{'} \in g H$ since $h^{- 1} \in H$ .

Transitivity

Let $g \sim g^{'}$ and $g^{'} \sum g^{″}$ . Hence there exists $h^{'}, h^{″} \in H$ such that:

g = g^{'} h^{'} and g^{'} = g^{″} h^{″} .

Now, $g = g^{″} h^{″} h^{'}$ , and since $H$ is closed under multiplication $g \in g^{″} H$ , hence $g \sim g^{″}$ .

Reflexivity

Since $H$ is a subgroup, $id \in H$ , and hence for any $g \in G$ , $g = g id \in g H$ , thus $g \sim g$ .