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

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 $\forall x H \in \mathbb S: \map {\theta_g} {g^{-1} x H} = x H$, 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:
 * $\map \theta {u v} = \map \theta u \, \map \theta v$

Thus $\theta$ is a homomorphism.

Now to calculate $\map \ker \theta$:

as required.