Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 3 / 1D / i / Solution

False. Take $G=C_2\times C_2$ and let $N$ be either factor. Then $N\cong C_2$ and $G/N\cong C_2$, but $G$ is not cyclic and hence is not $C_4$.

Solved by gpt-5.6-sol high.