Let be a subgroup of $G$ . Then for any two left cosets of $H$ in $G$ , there is a bijective function between the two cosets.

Proof

Let $a H, b H$ be two cosets. Define the function $f : a H \to b H$ by $x \mapsto b a^{- 1} x$ .

This has the correct codomain: if $x \in a H$ (so $x = a h$ , say), then $b a^{- 1} a x = b x$ so $f (x) \in b H$ .

The function is injective: if $b a^{- 1} x = b a^{- 1} y$ then (pre-multiplying both sides by $a b^{- 1}$ ) we obtain $x = y$ .

The function is surjective: given $b h \in b H$ , we want to find $x \in a H$ such that $f (x) = b h$ . Let $x = a h$ to obtain $f (x) = b a^{- 1} a h = b h$ , as required.

Parents:

Group coset

Let be a subgroup of $G$ . Then for any two left cosets of $H$ in $G$ , there is a bijective function between the two cosets.

Proof

Let $a H, b H$ be two cosets. Define the function $f : a H \to b H$ by $x \mapsto b a^{- 1} x$ .

This has the correct codomain: if $x \in a H$ (so $x = a h$ , say), then $b a^{- 1} a x = b x$ so $f (x) \in b H$ .

The function is injective: if $b a^{- 1} x = b a^{- 1} y$ then (pre-multiplying both sides by $a b^{- 1}$ ) we obtain $x = y$ .

The function is surjective: given $b h \in b H$ , we want to find $x \in a H$ such that $f (x) = b h$ . Let $x = a h$ to obtain $f (x) = b a^{- 1} a h = b h$ , as required.

Parents:

Group coset

AI ALIGNMENT FORUM
AF

Left cosets are all in bijection

Proof

AI ALIGNMENT FORUM
AF

Left cosets are all in bijection

Proof