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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.

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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Aug 3 '16 at 12:39
  • $\begingroup$ This is an answer! And an excellent one! $\endgroup$ – P Vanchinathan Aug 3 '16 at 13:19
  • $\begingroup$ @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! $\endgroup$ – Mathematician 42 Aug 3 '16 at 13:45
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    $\begingroup$ I think the problem with giving a general answer is that having a well-known representation theory is not really a well-defined notion. $\endgroup$ – Tobias Kildetoft Aug 3 '16 at 13:53
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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 $\neq1$.

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  • $\begingroup$ Upvote since it is interesting to see that I can use abelian groups to spice things up. I already completed the abelian case of whatever it is I'm interested in, but in the abelian case, the theory kind of splits over cyclic components. Mixing abelian and non-abelian groups via direct products leads to interesting phenomena. $\endgroup$ – Mathematician 42 Aug 3 '16 at 14:04
  • $\begingroup$ I like the latter class, but I'm working over an algebraically closed field of characteristic zero. I cannot get rid of this condition immediately. (Although this might be interesting to look at if at all possible). $\endgroup$ – Mathematician 42 Aug 3 '16 at 14:08
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    $\begingroup$ @Mathematician42 Just because the groups use finite fields to define does not mean you should only study their representations over that field. $\endgroup$ – Tobias Kildetoft Aug 3 '16 at 14:11
  • $\begingroup$ @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). $\endgroup$ – Jyrki Lahtonen Aug 3 '16 at 14:11
  • $\begingroup$ @TobiasKildetoft: Woops, you're right. And @ JyrkiLahtonen: The theory could still be interesting, in fact it seems that the theory becomes more interesting if the order of some fixed linear character evaluated at some group element in the center is large. I will look at these examples and figure out what's going on. Thank you, this is very useful. $\endgroup$ – Mathematician 42 Aug 3 '16 at 14:17

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