# Left and Right Coset Spaces are Equivalent

## Theorem

Let $\left({G, \circ}\right)$ be a group.

Let $H$ be a subgroup of $G$.

Let:

$x H$ denote the left coset of $H$ by $x$
$H y$ denote the right coset of $H$ by $y$.

Then:

$\left|{\left\{{x H: x \in G}\right\}}\right| = \left|{\left\{{H y: y \in G}\right\}}\right|$

That is:

The number of right cosets is the same as the number of left cosets of $G$ with respect to $H$.
The left and right coset spaces are equivalent.

## Proof 1

Let there be exactly $r$ different left cosets of $H$ in $G$.

Let a complete repetition-free list of these left cosets be:

$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$

From Coset Spaces form Partition, every element of $G$ is contained in exactly one of the left cosets.

Let $x \in G$. Then, for $1 \le i \le r$:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle x$$ $$\in$$ $$\displaystyle$$ $$\displaystyle H a_i^{-1}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle x \left({a_i^{-1} }\right)^{-1}$$ $$\in$$ $$\displaystyle$$ $$\displaystyle H$$ $$\displaystyle$$ $$\displaystyle$$ Elements in Coset iff Product with Inverse in Coset $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle x a_i$$ $$\in$$ $$\displaystyle$$ $$\displaystyle H$$ $$\displaystyle$$ $$\displaystyle$$ Inverse of Group Inverse $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle \left({x^{-1} }\right)^{-1} a_i$$ $$\in$$ $$\displaystyle$$ $$\displaystyle H$$ $$\displaystyle$$ $$\displaystyle$$ Inverse of Group Inverse $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle x$$ $$\in$$ $$\displaystyle$$ $$\displaystyle a_i H$$ $$\displaystyle$$ $$\displaystyle$$ Elements in Coset iff Product with Inverse in Coset

Since $x^{-1} \in a_i H$ is true for precisely one value of $i$, it follows that $x \in H a_i^{-1}$ is also true for precisely that value of $i$.

So there are exactly $r$ different right cosets of $H$ in $G$, and a complete repetition-free list of these is:

$H a_1^{-1}, H a_2^{-1}, H a_3^{-1}, \ldots, H a_r^{-1}$

The result follows.

$\blacksquare$

## Proof 2

Let $G$ be a group and let $H \le G$.

Consider the mapping $\phi$ from the left coset space to the right coset space defined as:

$\forall g \in G: \phi \left({g H}\right) = H g^{-1}$

We need to show that $\phi$ is a bijection.

First we need to show that $\phi$ is well-defined.

That is, that $a H = b H \implies \phi \left({a H}\right) = \phi \left({b H}\right)$.

Suppose $a H = b H$.

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle a H = b H$$ $$\iff$$ $$\displaystyle$$ $$\displaystyle a^{-1} b \in H$$ $$\displaystyle$$ $$\displaystyle$$ Equal Cosets iff Product with Inverse in Coset‎ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle H a^{-1} = H b^{-1}$$ $$\iff$$ $$\displaystyle$$ $$\displaystyle a^{-1} \left({b^{-1} }\right)^{-1} \in H$$ $$\displaystyle$$ $$\displaystyle$$ Equal Cosets iff Product with Inverse in Coset‎

But $a^{-1} \left({b^{-1}}\right)^{-1} = a^{-1} b \in H$ as $a H = b H$.

So $H a^{-1} = H b^{-1}$ and $\phi$ is well-defined.

Next we show that $\phi$ is injective:

Suppose $\exists x, y \in G: \phi \left({x H}\right) = \phi \left({y H}\right)$.

Then $H x^{-1} = H y^{-1}$, so $x^{-1} = e_G x^{-1} = h y^{-1}$ for some $h \in H$.

Thus $h = x^{-1} y \implies h^{-1} = y^{-1} x$.

As $H$ is a subgroup, $h^{-1} \in H$.

Thus $y^{-1} x \in H$ and so $x H = y H$ by Equal Cosets iff Product with Inverse in Coset.

Thus $\phi$ is injective.

Next we show that $\phi$ is surjective:

Let $H x$ be a right coset of $H$ in $G$.

Since $x = \left({x^{-1}}\right)^{-1}$, $H x = \phi \left({x^{-1} H}\right)$ and so $\phi$ is surjective.

Thus $\phi$ constitutes a bijection from the left coset space to the right coset space, and the result follows.

$\blacksquare$