Product of subsets whose total size exceeds size of group equals whole group
Note that this statement as given here applies only to the product of two subsets.
- Every element of a finite field is expressible as a sum of two squares
- Subgroup of size more than half is whole group: This is usually proved in some other way, but it can be easily seen from this fact by taking the product of the subgroup with itself.
If , it is not necessary that . An example is when are both equal to each other and a subgroup of index two in .
Given: A finite group , (possibly equal, possibly distinct) subsets of such that . An element .
To prove: There exists such that .
|Step no.||Assertion/construction||Given data used||Previous steps used||Explanation|
|1||The set is a subset of of the same size as .||inversion is one-one, and left multiplication by is one-one, so the composite is one-one.|
|2||is non-empty. Let be an element of the intersection.||Step (1)||By Step (1) and the given information, . By the principle of inclusion-exclusion, the subsets must intersect.|
|3||for the obtained in Step (2).||Step (2)||Direct from Step (2), by multiplying both sides on the right by .|