Inverse Elements of Right Transversal is Left Transversal
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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 $4$