Subgroup of finite group has a left transversal that is also a right transversal

Statement
Suppose $$G$$ is a finite group and $$H$$ is a subgroup of $$G$$. Then, $$H$$ has a left transversal that is also a right transversal. In other words, there is a subset $$S$$ of $$G$$ such that the intersection of $$S$$ with any left coset of $$H$$ has exactly one element and the intersection of $$S$$ with any right coset of $$H$$ has exactly one element.

Applications

 * Subgroup of finite index has a left transversal that is also a right transversal

For more, see subgroup having a left transversal that is also a right transversal.

Facts used

 * 1) Konig's theorem from combinatorics, which states that if a finite set has two partitions, both into $$m$$ subsets of equal size, then we can pick a subset that acts as a transversal for both partitions.
 * 2) uses::left cosets partition a group, uses::right cosets partition a group
 * 3) uses::left cosets are in bijection via left multiplication, uses::right cosets are in bijection via right multiplication

Abstract proof
The proof follows directly from fact (1), applied to the left coset partition and the right coset partition for $$H$$ in $$G$$. Facts (2) and (3) guarantee the necessary conditions to apply fact (1).

More explicit proof
A more explicit and concrete proof is as follows:


 * First, consider the partition of the group into double cosets of the subgroup.
 * Now note that each double coset is a union of left cosets as well as a union of right cosets, and further, that within each double coset, every left coset intersects every right coset and the number of left cosets equals the number of right cosets.
 * Within each double coset, choose an arbitrary bijection between the left cosets and the right cosets, and then pick a representative for each intersection of a left coset and the corresponding right coset.
 * Take the union of all these sets of representatives over all the double cosets.