# Left and Right Coset Spaces are Equivalent

## Theorem

Let $\struct {G, \circ}$ 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:

- $\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$

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 Left Coset Space forms Partition, every element of $G$ is contained in exactly one of the left cosets.

Let $x \in G$.

Then, for $1 \le i \le r$:

\(\ds x\) | \(\in\) | \(\ds H a_i^{-1}\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x \paren {a_i^{-1} }^{-1}\) | \(\in\) | \(\ds H\) | Element in Right Coset iff Product with Inverse in Subgroup | ||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x a_i\) | \(\in\) | \(\ds H\) | Inverse of Group Inverse | ||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x^{-1} }^{-1} a_i\) | \(\in\) | \(\ds H\) | Inverse of Group Inverse | ||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x^{-1}\) | \(\in\) | \(\ds a_i H\) | Element in Left Coset iff Product with Inverse in Subgroup |

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:

- $\forall g \in G: \map \phi {g H} = H g^{-1}$

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

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

That is, that $a H = b H \implies \map \phi {a H} = \map \phi {b H}$.

Suppose $a H = b H$.

\(\ds a H = b H\) | \(\iff\) | \(\ds a^{-1} b \in H\) | Left Cosets are Equal iff Product with Inverse in Subgroupâ€Ž | |||||||||||

\(\ds H a^{-1} = H b^{-1}\) | \(\iff\) | \(\ds a^{-1} \paren {b^{-1} }^{-1} \in H\) | Right Cosets are Equal iff Product with Inverse in Subgroup |

But $a^{-1} \paren {b^{-1} }^{-1} = a^{-1} b \in H$ 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 $\exists x, y \in G: \map \phi {x H} = \map \phi {y H}$.

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:

- $y^{-1} x \in H$

So by Left Cosets are Equal iff Product with Inverse in Subgroup:

- $x H = y H$

Thus $\phi$ is injective.

Next we show that $\phi$ is surjective:

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

Since $x = \paren {x^{-1} }^{-1}$, $H x = \map \phi {x^{-1} H}$ and so $\phi$ is surjective.

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

$\blacksquare$

## Also see

- Index of Subgroup, which is the number of left (or right) cosets of a subgroup.

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 37 \beta$