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 ?
PS
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)."
PSPS
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.
Best Answer
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)