Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.9 AIME, harder Prove it Romania

Let $(G, \cdot)$ be a group, with the unit element $e$ , and $A$ a non-empty subset of $G$ . We denote $AA = \{xy \mid x, y \in A\}$ .

a) Show that if $G$ is finite, then $AA = A$ if and only if $e \in A$ and $|AA| = |A|$ .

b) Give an example of a group $G$ and a subset $A \subseteq G$ , such that $AA \neq A$ , $|AA| = |A|$ and $AA < G$ .

(The notation $H < G$ means that $H$ is a proper subgroup of the group $G$ , i.e., a subgroup of $G$ different from $G$ itself.)

Solution

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, $AA \cap A = \emptyset$ ).

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