Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Prove it Romania

Let $(G, \cdot)$ be a group with the unit element $e$ , and $H$ and $K$ two proper subgroups of $G$ , such that $H \cap K = \{e\}$ and the set $(G \setminus (H \cup K)) \cup \{e\}$ is closed with respect to the operation in $G$ . Show that $x^2 = e$ , for any $x \in G$ .

Solution

Consider $L = (G \setminus (H \cup K)) \cup \{e\}$ . Because $x \in H \cup K \iff x^{-1} \in H \cup K$ , it follows that $x \in L \iff x^{-1} \in L$ , so that $L$ is a proper subgroup of $G$ .
Also, $L \cap H = L \cap K = H \cap K = \{e\}$ , $G = H \cup K \cup L$ and it follows that for any permutation $\{A, B, C\} = \{H, K, L\}$ , if $a \in A \setminus \{e\}$ and $b \in B \setminus \{e\}$ , then $ab \in C \setminus \{e\}$ .

Then, in the same conditions as above, $a^2b = a(ab) \in B \setminus \{e\}$ , so that $a^2 \in A \cap B = \{e\}$ . Hence, $x^2 = e$ for any $x \in G \setminus \{e\}$ .
Since $e^2 = e$ , it follows that $x^2 = e$ for any $x \in G$ .

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