A $Z$-group is a group whose Sylow subgroups are all cyclic groups. I know from here that if two $Z$-groups have the same character table, then they are in fact isomorphic groups.
To my understanding, all finite $Z$-groups are a semidirect product of two cyclic groups of the form $\mathscr{C}_n \rtimes \mathscr{C}_m$, since they have presentation $\langle a, b \,\vert\, a^n=b^m=1, b^{-1}ab = a^r\rangle$ where $gcd((r-1)n, m)=1$ and $r^n\equiv 1\pmod{m}$.
I know for $n$ odd the dihedral groups $D_{2n}$, and dicyclic groups $D_{4n}$ are $Z$-groups, and the character tables are known for all $n$. Is the character table for all $Z$-groups known, and where could I find it?
Best Answer
The character tables of split metacyclic groups $G=N\rtimes H\cong C_m\rtimes C_n$ can be constructed easily via Clifford theory: The irreducible characters of $N$ are well-known. For $\psi\in\mathrm{Irr}(N)$, let $H_\psi$ be the stabilizer of $\psi$ in $H$. Note that $[N/\ker(\psi),H_\psi]=1$. The extensions of $\psi$ to $NH_\psi$ are the inflations of $\psi\times\lambda\in\mathrm{Irr}((N/\ker\psi) \times H_\psi)$. Then the induced character $(\psi\times\lambda)^G$ is irreducible. If $\psi$ runs through a set of representatives of the $H$-orbits on $\mathrm{Irr}(N)$ and $\lambda$ runs though $\mathrm{Irr}(H_\psi)$, you get every irreducible character of $G$ exactly once.