If finite group G has a center how does it influence the representations of this group ? And vice versa - can we see somehow the center (or some of its properties) from representations (from character table, from ring structure, ... whatever) ?

One has a natural map Z(G)-> G-> G/Z(G), so we can pull-back representations of G/Z(G) to G, but so what ? How far R(G) is from R(G/Z(G)) ? Can we claim that at least the dimensions of irreps of G are the same or just not bigger, than that of G/Z(G) ? (NO as Xogn Ambandl answer implies). (F. Ladisch comment below is some weaker indication that something like this might happen).

In any irrep of G center Z should act by scalars, so it defines some homomorhpism of Z to C^, is any such homomorphism is realized by some irrep V of G ? Probably not... Is it possible to characterize those Z->C^ which occur, depending on the group G ?


I just learnt from comments by F. Ladisch:

"It is a general fact that χ(1)^2≤|G:Z(G)| for any irred. character χ of a group G (see Isaacs' book on character theory, Corollary 2.30)."


Another relevant MO-discussion Which finite groups have faithful complex irreducible representations?. Let me quote: "Obvious necessary condition is that the center must be a cyclic group."

"For finite p-groups, it's a standard fact that having a faithful irreducible representation is equivalent to having a cyclic center. I'm not sure about the general case, but it's been discussed in many books and papers. My impression is that there is no known definitive structural condition for sufficiency. – Jim Humphreys Mar 2 2011 at 16:52"

And further - see answers by Andreas Thom and Rob Harron.

  • 1
    $\begingroup$ A comprehensive consideration of faithful representation one can fiund in Chapter 9. of the book Berkovich-Zhmud;, Character of Finite groups. Part.1. In particular, the following results there proved: (1) (Gasch\"utz): A group $G$ possesses a faithful irreducible character iff its solle is generated by $G$-class. (2) (Zhmud;) The number os kernels of irreducible characters of $G$ is equal to the number of normal closures of elements of $G$. $\endgroup$
    – yakov
    Commented Jul 8, 2016 at 13:52

4 Answers 4


The center of a group and its isomorphism type can be seen from the character table: The center is the set $\newcommand{\Irr}{\operatorname{Irr}}$ $$ Z(G) = \{ g\in G \mid \: |\chi(g)| = \chi(1) \text{ for all } \chi \in \Irr(G)\} .$$ Since $\chi_{Z(G)}=\chi(1)\lambda$, you see the linear characters of $Z(G)$ in the character table, and thus you see the isomorphism type of $Z(G)$. The irreps lying over two different characters of $Z(G)$ need not be related. To see this, consider a direct product $G=H\times K$. Then $Z(G)=Z(H)\times Z(K)$. The irreps of $G$ lying over $\lambda \times \mu$ are tensors of irreps of $H$ over $\lambda$ with irreps of $K$ over $\mu$. For characters, this reads $$\Irr(G\mid \lambda\times \mu) = \Irr(H\mid\lambda)\times \Irr(K\mid\mu).$$ Now taking $\lambda\neq 1 = \mu$ or vice versa and suitable examples for $H$ and $K$, we see that these sets can look quite different.
An interesting fact is that the linear characters of $Z(G)$ yield a grading of $\Irr(G)$: the linear characters define a partition of $\Irr(G)$ and if $\chi\in \Irr(G)$ lies over $\lambda$ and $\psi$ over $\mu$, then all irred. constituents of $\chi\phi$ lie over $\lambda \mu$. Moreover, it is not too difficult to show that every grading of $\Irr(G)$ by some group comes in fact from a subgroup $Z$ of the center, that is, $\chi$ and $\psi$ are in the same subset of the partition if the restrictions $\chi_Z$ and $\psi_Z$ contain the same character of $Z$. (This idea has been used by Gelaki and Nikshych to define nilpotency of arbitrary fusion categories: arXiv:math/0610726)


Let $\chi:Z(G)\rightarrow\mathbb{C}^\times$ be an homomorphism, viewed as a 1-dimensional representation, and let $\pi=Ind_{Z(G)}^G \chi$ be the induced representation from $Z(G)$ to $G$. Then $Z(G)$ acts in $\pi$ as scalar multiplication by $\chi$. Of course $\pi$ is not necessarily irreducible, but the same will hold for every irreducible component of $\pi$. So $\chi$ is realized by some irrep of $G$.

  • $\begingroup$ Aaah, so simple... Thank you ! Can this line of thought be continued ? Is it true that we have the same number of irreps of G, for each character of Z(G) ? In the other words how far irreps of G are different from irreps of (G/ZxZ) ? May be we can use, that Induction from regular rep = regular rep of G, so contains everything. $\endgroup$ Commented Sep 9, 2012 at 18:58
  • $\begingroup$ Oops, it cannot go this way... Consider quaternions G=Q_8, Z=[G,G]=Z/2Z={+1,-1}, so in all 4 1d irreps of G, center acts trivially, and only in 2d irrep of G it acts non-trivially. So induction from trivial character is sum of 4 1d irreps, while from non-trivial char. it is sum of 2 copies of 2d irrep... So quite far from Zx(G/Z)... Is there any control or it behaves randomly ? $\endgroup$ Commented Sep 9, 2012 at 19:42

There is an easy way to count the number of irreducible characters of $G$ that restrict to $Z = {\bf Z}(G)$ to yield a multiple of a given linear character $\lambda$ of $Z$. By a result of Gallagher, this number is the number of conjugacy classes of "$\lambda$-good" elements of $G/Z$, where an element $Zg$ of $G/Z$ is $\lambda$-good if for every element $x$ of $G$ such that $[g,x] \in Z$, it is true that $\lambda([g,x]) = 1$.

In particular, it follows that the number of irreducible characters of $G$ lying over a given linear character $\lambda$ of $Z$ is at most $|{\rm Irr}(G/Z)|$. It is also true that the sum of the squares of the degrees of the irreducible characters of $G$ lying over $\lambda$ is equal to $|G:Z|$. We conclude from this that the average of the squares of the degrees of these characters is at least the average of the squares of the degrees of all members of ${\rm Irr}(G/Z)$, so one might say that "on average" the irreducible character degrees of $G/Z$ are no less than the degrees of the irreducible characters lying over each linear character $\lambda$.


Take any group $G$ such that $G/Z(G)$ is abelian. Then any irrep og $G/Z(G)$ is one-dimensional, but $G$ has irreps of dimension >1.

  • $\begingroup$ Good catch :) So it implies that my "hope" that dims of irreps of G/Z are bigger than same for G, is not true. I edited quest. $\endgroup$ Commented Sep 9, 2012 at 14:09
  • $\begingroup$ And in fact, this dimensions can be as large as one wishes: An extraspecial $p$-group of order $p^{2n+1}$ has irreps of dimensions $1$ and $p^n$. $\endgroup$ Commented Sep 9, 2012 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.