# Classes of finite non-abelian groups with non-trivial center .

I'm looking for an interesting class of finite non-abelian groups with non-trivial center whose representation theory is well-known. In particular, I'm interested in such groups with the additional condition that there is a linear character that is non-trivial on the center of the group.

For example: Consider $\mathcal{D}_{2n}=\left\langle a,b\mid a^n=b^2=1, bab=a^{n-1}\right\rangle$ where $n$ is divisible by $2$ once, then $\ker(\mathcal{D}_{2n})=\left\{1,a^{\frac{n}{2}}\right\}$. There are four linear characters determined by $a\mapsto \pm 1$ and $b\mapsto \pm 1$. The two characters that map $a$ to $-1$ have the value $-1$ on $a^{\frac{n}{2}}$ since $\frac{n}{2}$ is odd. Thus this class satisfies all conditions.

If some other class springs to mind, I would be very interested. I'm not interested in examples found by taking direct sums of easier classes.

• All the linear characters of a finite group $G$ are trivial exactly in the commutator subgroup $G':=[G,G]$, so you want groups such that $Z(G)$ is not contained in the derived group $G'$. If $A$ is an abelian group, then this holds for groups of the form $G=H\times A$ with $H$ arbitrary. – Jyrki Lahtonen Aug 3 '16 at 12:39
• This is an answer! And an excellent one! – P Vanchinathan Aug 3 '16 at 13:19
• @JyrkiLahtonen: Indeed, thank you. That's a nice way of generating examples. Although, not all examples occur in this fashion and I would also be interested in other examples that do not arise in a trivial fashion. But thumps up, very elegant reasoning and also helpful! – Mathematician 42 Aug 3 '16 at 13:45
• I think the problem with giving a general answer is that having a well-known representation theory is not really a well-defined notion. – Tobias Kildetoft Aug 3 '16 at 13:53

It is a fact that all the (complex valued) linear characters of a finite group $G$ are trivial exactly in the commutator subgroup $G':=[G,G]$. Because the target group $GL_1(\Bbb{C})=\Bbb{C}^*$ is abelian, the elements of $G'$ are in the kernel of all the 1-dimensional characters. The reverse inclusion follows from the fact that the quotient $G/G'$ is abelian, and a finite abelian group has enough 1-dimensional characters that only $1_G$ is in the intersection of their kernels.
So you want groups such that $Z(G)$ is not contained in the derived group $G'$. If $A$ is an abelian group, then this holds for groups of the form $G=H\times A$ with $H$ an arbitrary group. Here $G'=H'\times\{1_A\}$ but $\{1_H\}\times A\subseteq Z(G)$, so the criterion is satisfied. Admittedly this is not a very exciting family of examples.
Another family of examples are the groups $G=GL_n(K)$, where $K$ is a finite field. Here $Z(G)$ consists of scalar matrices, and $G'\le SL_n(K)$ (more often than not there is equality). Unless $|K|-1$ is a factor of $n$ there will be scalar matrices with determinants $\neq1$.
• @Mathematician42 The groups $GL_n(K), SL_n(K)$ themselves are finite, and they have interesting complex representations. Admittedly your topic may not be interesting with those groups because there are very few 1-dimensional characters (the derived group is large). – Jyrki Lahtonen Aug 3 '16 at 14:11