Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 3 / 1D (Groups)

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 $C_4$.

True. The group has order six. A Sylow $3$-subgroup $P$ has order three, while the given normal subgroup $N$ has order two. Their intersection is trivial and $NP=G$. Conjugation by $P$ on $N$ is trivial because $\mathrm{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

False. Let $G=C_4$ and $H=C_2\times C_2$, and in each choose a subgroup of order two. Then the chosen normal subgroups are isomorphic and both quotients are $C_2$, while $G$ and $H$ are not isomorphic.