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
\[AB=\{ab \mid a\in A, b\in B\}.\]

Proof.

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

See the post↴
Each Element in a Finite Field is the Sum of Two Squares
for a proof of this problem.

The post If Two Subsets $A, B$ of a Finite Group $G$ are Large Enough, then $G=AB$ first appeared on Problems in Mathematics.

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 integers, there exists integers $s, t$ such that
\[sm+tn=1.\]

Then we have
\begin{align*}
a&=a^1=a^{sm+tn}=a^{sm}a^{tn} \tag{*}
\end{align*}
Note that since the order of the group $G$ is $n$, any element of $G$ raised by the power of $n$ is the identity element $e$ of $G$.
Thus we have
\[a^{tn}=(a^n)^t=e^t=e.\] Putting $b:=a^s$, we have from (*) that
\[a=b^me=b^m.\]

Now we show the uniqueness of such $b$. Suppose there is another $g’\in G$ such that
\[a=b’^m.\] Then we have
\begin{align*}
&\quad b^m=a=b’^m\\
&\Rightarrow b^{sm}=b’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow b^{1-tn}=b’^{1-tn}\\
&\Rightarrow b(b^n)^t=b'(b’^n)^t\\
&\Rightarrow b=b’ \quad \text{ since } b^n=e=b’^n.
\end{align*}
Therefore, we have $b=b’$ and the element $b$ satisfying $a=b^m$ is unique.

Proof 2.

Consider a map $f$ from $G$ to $G$ itself defined by sending $g$ to $f(g)=g^m$.
We show that this map is injective.
Suppose that $f(g)=f(g’)$.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
\[sm+tn=1.\]

We have
\begin{align*}
f(g)&=f(g’)\\
&\Rightarrow g^m=g’^m \\
&\Rightarrow g^{sm}=g’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow g^{1-tn}=g’^{1-tn}\\
&\Rightarrow g(g^n)^t=g'(g’^n)^t\\
&\Rightarrow g=g’ \quad \text{ since } g^n=e=g’^n, n=|G|.
\end{align*}

Therefore the map $f$ is injective. Since $G$ is a finite set, it also follows that the map is bijective.
Thus for any $a \in G$, there is a unique $b \in G$ such that $f(b)=a$, namely, $b^m=a$.

The post Finite Group and a Unique Solution of an Equation first appeared on Problems in Mathematics.