Left and Right Coset Spaces are Equivalent

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

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

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

is a bijection.

That is, the left and right coset spaces are equivalent.

Thus, the number of right cosets is the same as the number of left cosets of $$G$$ with respect to $$H$$.

Proof

 * 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, $$h^{-1} \in H$$ and so $$x H = y H$$ by Congruence Class Modulo Subgroup is 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.