Let $G$ be a finite group and let $A, B$ be subsets of $G$ satisfying
\[|A|+|B| > |G|.\]
Here $|X|$ denotes the cardinality (the number of elements) of the set $X$.
Then prove that $G=AB$, where
\[AB=\{ab \mid a\in A, b\in B\}.\]

Since $A, B$ are subsets of the group $G$, we have $AB\subset G$.
Thus, it remains to show that $G\subset AB$, that is any element $g\in G$ is of the form $ab$ for some $a\in A$ and $b\in B$.
This is equivalent to finding $a\in A$ and $b\in B$ such that $gb^{-1}=a$.

Consider the subset
\[B^{-1}:=\{b^{-1} \mid b \in B\}.\]
Since taking the inverse gives the bijective map $B \to B^{-1}$, $b \mapsto b^{-1}$, we have $|B|=|B^{-1}|$.

Also consider the subset
\[gB^{-1}=\{gb^{-1} \mid b\in B\}.\]
Note that multiplying by $g$ and by its inverse $g^{-1}$ give the bijective maps
\[B^{-1} \to gB^{-1}, b^{-1} \mapsto gb^{-1} \text{ and } gB^{-1} \to B^{-1}, gb^{-1} \mapsto b^{-1}.\]
Hence we have
\[ |B|=|B^{-1}|=|gB^{-1}|.\]

Since $A$ and $gB^{-1}$ are both subsets in $G$ and we have by assumption that
\[|A|+|gB^{-1}|=|A|+|B| > |G|,\]
the intersection $A\cap gB^{-1}$ cannot be empty.

Therefore, there exists $a \in A\cap gB^{-1}$, and thus $a\in A$ and $a=gb^{-1}$ for some $b\in B$.
As a result we obtain $g=ab$.
It yields that $G\subset AB$, and we have $G=AB$ as a consequence.

Related Question.

As an application, or use the similar technique, try the following

Problem.
Every element in a finite field $F$ is the sum of two squares in $F$.

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