Let \(G\) be a group of order \(2p\), where \(p\) is an odd prime. Assume that \(G\) has a normal subgroup of order \(2\). Prove that \(G\) is cyclic.
Solution
Let \[ N=\{e,z\} \] be the given normal subgroup of order \(2\).
1. The element \(z\) is central in \(G\).
For every \(g\in G\), normality gives \[ gzg^{-1}\in N. \] Conjugation preserves order, and \(z\) has order \(2\), so \(gzg^{-1}\) cannot be \(e\). Hence \[ gzg^{-1}=z, \] which is equivalent to \(gz=zg\).
2. There exists an element \(x\in G\) of order \(p\).
By Cauchy’s theorem, since \(p\) divides \(\abs G=2p\), the group \(G\) contains an element of order \(p\).
3. The element \(xz\) has order \(2p\).
By step 1, \(x\) and \(z\) commute. Hence \[ (xz)^{2p}=x^{2p}z^{2p}=e. \] Conversely, suppose \((xz)^m=e\). Then \[ x^m=z^{-m}. \] The left-hand side lies in \(\langle x\rangle\), while the right-hand side lies in \(N\). Since \[ \abs{\langle x\rangle}=p, \qquad \abs N=2, \] their intersection is trivial. Thus \(x^m=e\) and \(z^m=e\), so \[ p\mid m \qquad\text{and}\qquad 2\mid m. \] Because \(p\) is odd, \(2p\mid m\). Therefore the order of \(xz\) is exactly \(2p\).
4. The group \(G\) is cyclic.
By step 3, the cyclic subgroup \(\langle xz\rangle\) has order \(2p=\abs G\). Hence \[ \langle xz\rangle=G. \]
5. Q.E.D.
Step 4 proves the assertion.