Subgroup of finite group has a left transversal that is also a right transversal
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup of finite group) must also satisfy the second subgroup property (i.e., subgroup having a left transversal that is also a right transversal)
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Suppose is a finite group and is a subgroup of . Then, has a left transversal that is also a right transversal. In other words, there is a subset of such that the intersection of with any left coset of has exactly one element and the intersection of with any right coset of has exactly one element.
- Konig's theorem from combinatorics, which states that if a finite set has two partitions, both into subsets of equal size, then we can pick a subset that acts as a transversal for both partitions.
- left cosets partition a group, right cosets partition a group
- left cosets are in bijection via left multiplication, right cosets are in bijection via right multiplication
The proof follows directly from fact (1), applied to the left coset partition and the right coset partition for in . 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.