Romanian Mathematical Olympiad

a) If $AA = A$, then $|AA| = |A|$. For any $x \in A$ we have $|xA| = |A| < \infty$ and $xA \subseteq AA = A$, so that $xA = A$. But then $x \in xA$, hence $e = x^{-1} \cdot x \in x^{-1} \cdot xA = A$. Reciprocally, if $e \in A$ and $|AA| = |A|$, then $A = e \cdot A \subseteq AA$ and since $|A| = |AA| < \infty$, it follows that $AA = A$.

b) Let $G = U_4 = \{1, i, -1, -i\}$ be the group of all 4th roots of unity. Considering $A = \{i, -i\}$, we have $AA = \{i^2, i \cdot (-i), (-i)^2\} = \{1, -1\} = U_2 < U_4$, $|AA| = 2 = |A|$ and $AA \neq A$ (furthermore, $AA \cap A = \emptyset$).