A problem from abstract algebra of finite group?

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A and B are to nonempty subset of a group G with $|A|+|B| > |G|$ then G = AB.

I am a bit confused of the order of a subset of a group. Generally a subset of a group is not a subgroup but when it occurs $|A|+|B| > |G|$, I want to figure out the relation between A and B and if A will have the inverse of some elements of A if $|A| > \frac{|G|}{2}$?

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Let $g\in G$.

Claim $g\in AB$.

Let $S=\{a^{-1}g\mid a\in A\}$.

Then $|S|=|A|$, so $|S|+|B| > |G|$.

By the pigeonhole principle, the sets $S,B$ cannot be disjoint.

Thus, we have $b\in S$, for some $b\in B$.

But then $b=a^{-1}g$, for some $a\in A$, so $ab=g$, hence $g\in AB$, as claimed.

It follows that $AB=G$.