- Let \(G\) be a cyclic group and let \(a,b\in G\) be elements that are not squares. Prove that \(ab\) is a square in \(G\).
- Give an example showing that the conclusion can fail when \(G\) is not cyclic.
Solution
1. If \(G=\langle g\rangle\) is cyclic, then the set of squares in\(G\) is the subgroup\[ G^2 \coloneqq \{x^2:x\in G\} = \langle g^2\rangle. \]
Every element of \(G\) has the form \(g^m\), and \[ (g^m)^2=g^{2m}. \] Thus every square lies in \(\langle g^2\rangle\). Conversely, every element \(g^{2m}\) of \(\langle g^2\rangle\) is the square of \(g^m\).
2. The quotient \(G/G^2\) has at most two elements.
The quotient is generated by the coset \(gG^2\), and \[ (gG^2)^2=g^2G^2=G^2. \] Hence the generator has order at most \(2\), so the quotient has order at most \(2\).
3. Part 1: if \(a\) and \(b\) are nonsquares, then \(ab\) is a square.
By step 1, being a nonsquare means lying outside \(G^2\). Since nonsquares exist, step 2 implies that \(G/G^2\) has exactly two elements. Therefore \(aG^2\) and \(bG^2\) are both its unique nonidentity element. Their product is the identity coset: \[ abG^2 = (aG^2)(bG^2) = G^2. \] Thus \(ab\in G^2\), so step 1 shows that \(ab\) is a square.
4. Part 2: the conclusion fails for\[ G=C_2\times C_2. \]
Write \[ G=\{1,a,b,ab\}, \] where \[ a^2=b^2=1 \] and \(a\neq b\). Every element of \(G\) has square equal to \(1\), so the only square in \(G\) is \(1\). Hence \(a\) and \(b\) are nonsquares, while their product \(ab\) is also a nonidentity element and therefore also a nonsquare. Thus the conclusion of part 1 fails in this noncyclic group.
5. Q.E.D.
Step 3 proves the cyclic case, and step 4 supplies the requested noncyclic counterexample.