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 $\displaystyle \bigcap_{x \in G} x H x^{-1}$.

Corollary

Let $H \le G$ such that $\left[{G : H}\right] = n$ where $n \in \Z$.

Then $\exists N \triangleleft G: N \triangleleft H: n \backslash \left[{G : N}\right] \backslash n!$.

Proof

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x H\)	\(=\)	\(\displaystyle y H\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle y^{-1} x\)	\(\in\)	\(\displaystyle H\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left({g y}\right)^{-1} g x\)	\(=\)	\(\displaystyle y^{-1} x \in H\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \theta_g \left({y H}\right)\)	\(=\)	\(\displaystyle \theta_g \left({x H}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \theta_{u v} \left({x H}\right)\)	\(=\)	\(\displaystyle u v x H\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \theta_u \left({v x H}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \theta_u \left({\theta_v \left({x H}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $\ker \left({\theta}\right)$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \ker \left({\theta}\right)\)	\(=\)	\(\displaystyle \left\{ {g \in G: \theta_g = I_\mathbb S}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g \in G: \forall x \in G: \theta_g \left({x H}\right) = x H}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g \in G: \forall x \in G: g x h = x H}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g \in G: \forall x \in G: x^{-1} g x h = H}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g \in G: \forall x \in G: x^{-1} g x \in H}\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ {g \in G: \forall x \in G: g \in x H x^{-1} }\right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \bigcap_{x \in G} x H x^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

as required.

$\blacksquare$

Proof of Corollary

Apply the main result to $H$ and let $N = \ker \left({\theta}\right)$.

Then $N \triangleleft G$ and $N \triangleleft H$ so $H / N \le G / N$ such that $\left[{G / N : H / N}\right] = n$ from the Correspondence Theorem.

Thus $n \backslash \left[{G : N}\right]$.

Also by the main result, $\exists K \in S_n: G / N \cong K$.

Thus $\left[{G : N}\right] \backslash n!$ as required.

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\S 9$: Proposition $9.22$, Corollary $9.23$