Condition for Subset of Group to be Right Transversal
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Theorem
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index in $G$ is $n$:
- $\index G H = n$
Let $S \subseteq G$ be a subset of $G$ of cardinality $n$.
Then $S$ is a right transversal for $H$ in $G$ if and only if:
- $\forall x, y \in S: x \ne y \implies x y^{-1} \notin H$
Proof
From Definition of Right Transversal, $S$ is a right transversal for $H$ in $G$ if and only if every right coset of $H$ contains exactly one element of $S$.
Since there are $n$ right cosets of $H$ and $S$ has cardinality $n$, if $S$ is not a right transversal for $H$ in $G$, at least one right coset of $H$ contain more than one element of $S$.
Thus the contrapositive of the theorem is as follows:
- At least one right coset of $H$ contain more than one element of $S$ if and only if:
- $\exists x, y \in S: x \ne y \land x y^{-1} \in H$
This is a consequence of Elements in Same Right Coset iff Product with Inverse in Subgroup.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $6$: Cosets: Exercise $5$