Condition for Subset of Group to be Right Transversal

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 $G$ :
 * $\forall x, y \in S: x \ne y \implies x y^{-1} \notin H$

Proof
From, $S$ is a right transversal for $H$ in $G$ 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 $S$ :


 * $\exists x, y \in S: x \ne y \land x y^{-1} \in H$

This is a consequence of Elements in Same Right Coset iff Product with Inverse in Subgroup.