Inverse Elements of Right Transversal is Left Transversal

Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$.

Let $S \subseteq G$ be a right transversal for $H$ in $G$.

Let $T$ be the set defined as:

$T := \set {x^{-1}: x \in S}$

where $x^{-1}$ is the inverse of $x$ in $G$.

Then $T$ is a left transversal for $H$ in $G$.

Proof

Let $g \in G$.

We show that $g H$ contains exactly $1$ element of $T$.

Since $S$ is a right transversal:

$\exists ! x \in S: x \in H g^{-1}$

By Right Cosets are Equal iff Element in Other Right Coset:

$H x = H g^{-1}$

By Right Cosets are Equal iff Left Cosets by Inverse are Equal:

$x^{-1} H = g H$

We have from definition $x^{-1} \in T$.

The result follows from uniqueness of $x$.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $6$: Cosets: Exercise $4$