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

To put it another way:
 * 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:
 * $$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$$

Every element of $$G$$ is contained in exactly one of the left cosets, as cosets form a partition.

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

$$ $$ $$ $$ $$

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.

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 $$a H = b H$$.

$$ $$

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.