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 $F$.

Finite Group and a Unique Solution of an Equation
Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]
We give two proofs.
Proof 1.
Since $m$ and $n$ are relatively prime […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

Group Homomorphism, Preimage, and Product of Groups
Let $G, G'$ be groups and let $f:G \to G'$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
\[f^{-1}(f(H))=HN.\]
Proof.
$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.
It follows that there exists $h\in H$ […]

The Product of a Subgroup and a Normal Subgroup is a Subgroup
Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.
The product of $H$ and $N$ is defined to be the subset
\[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\]
Prove that the product $H\cdot N$ is a subgroup of […]

The Product of Distinct Sylow $p$-Subgroups Can Never be a Subgroup
Let $G$ a finite group and let $H$ and $K$ be two distinct Sylow $p$-group, where $p$ is a prime number dividing the order $|G|$ of $G$.
Prove that the product $HK$ can never be a subgroup of the group $G$.
Hint.
Use the following fact.
If $H$ and $K$ […]

Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57
Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.
Then determine the number of elements in $G$ of order $3$.
Proof.
Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.
By […]

If the Order of a Group is Even, then the Number of Elements of Order 2 is Odd
Prove that if $G$ is a finite group of even order, then the number of elements of $G$ of order $2$ is odd.
Proof.
First observe that for $g\in G$,
\[g^2=e \iff g=g^{-1},\]
where $e$ is the identity element of $G$.
Thus, the identity element $e$ and the […]

Nontrivial Action of a Simple Group on a Finite Set
Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.
Proof.
Since $G$ acts on $X$, it […]