\b[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 $\operatorname{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

Solution

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

 Solution (source code)