Product of subsets whose total size exceeds size of group equals whole group

Statement
Suppose $$G$$ is a finite group and $$A,B$$ are (possibly equal, possibly distinct) subsets of $$G$$ such that $$|A| + |B| > |G|$$. Then, every $$g \in G$$ can be written as $$ab$$, where $$a \in A$$, $$b \in B$$. In other words, the product of subsets $$AB$$ equals $$G$$.

Note that this statement as given here applies only to the product of two subsets.

Applications

 * 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.

Strictness
If $$|A| + |B| = |G|$$, it is not necessary that $$AB = G$$. An example is when $$A,B$$ are both equal to each other and a subgroup of index two in $$G$$.

Similar facts

 * Cauchy-Davenport theorem
 * Kemperman's theorem
 * Kneser's theorem

Proof
Given: A finite group $$G$$, (possibly equal, possibly distinct) subsets $$A,B$$ of $$G$$ such that $$|A| + |B| > |G|$$. An element $$g \in G$$.

To prove: There exists $$a \in A, b \in B$$ such that $$g = ab$$.

Proof: