Permutation of Cosets

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

Let $\mathbb S$ be the set of all distinct left cosets of $H$ in $G$.

Then:
 * 1) For any $g \in G$, the mapping $\theta_g: \mathbb S \to \mathbb S$ defined by $\theta_g \left({x H}\right) = g x H$ is a permutation of $\mathbb S$.
 * 2) The mapping $\theta$ defined by $\theta \left({g}\right) = \theta_g$ is a homomorphism from $G$ into the Symmetric Group on $\mathbb S$.
 * 3) The kernel of $\theta$ is the subgroup $\displaystyle \bigcap_{x \in G} x H x^{-1}$.

Proof
First we need to show that $\theta_g$ is well-defined, and injective.

Thus $\theta_g$ is well-defined, and injective.

Then we see that $\forall x H \in \mathbb S: \theta_g \left({g^{-1} x H}\right) = x H$, so $\theta_g$ is surjective.

Thus $\theta_g$ is a well-defined bijection on $\mathbb S$, and therefore a permutation on $\mathbb S$.

Next we see:

This shows that $\theta_{u v} = \theta_u \theta_v$, and thus $\theta \left({u v}\right) = \theta \left({u}\right) \theta \left({v}\right)$.

Thus $\theta$ is a homomorphism.

Now to calculate $\ker \left({\theta}\right)$.

as required.