Suppose \(K\) has characteristic different from \(2\) and contains a primitive \(n\)th root of unity for odd \(n\). Prove that \(K\) contains a primitive \(2n\)th root of unity.
Solution
Let \(\zeta\in K\) have order \(n\), with \(n\) odd. We claim that \(-\zeta\) has order \(2n\).
Certainly \[ (-\zeta)^{2n}=1. \] Suppose \((-\zeta)^m=1\). Then \[ \zeta^m=(-1)^m. \] If \(m\) is odd, this says that the odd-order group \(\langle\zeta\rangle\) contains \(-1\), an element of order \(2\), impossible because \(\operatorname{char}K\ne2\). Hence \(m\) is even. Then \(\zeta^m=1\), so \(n\mid m\). Since \(n\) is odd and \(m\) is even, \(2n\mid m\).
Thus the least positive exponent annihilating \(-\zeta\) is \(2n\), so \(-\zeta\) is a primitive \(2n\)th root of unity in \(K\).