Center and representations of finite group - how are related ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:07:11Z http://mathoverflow.net/feeds/question/106729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106729/center-and-representations-of-finite-group-how-are-related Center and representations of finite group - how are related ? Alexander Chervov 2012-09-09T13:47:21Z 2012-09-09T20:28:44Z <p>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) ?</p> <p>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).</p> <p>In any irrep of G center Z should act by scalars, so it defines some homomorhpism of Z to C^<em>, is any such homomorphism is realized by some irrep V of G ? Probably not... Is it possible to characterize those Z->C^</em> which occur, depending on the group G ? </p> <p>PS</p> <p>I just learnt from <a href="http://mathoverflow.net/questions/106521/representation-theory-of-p-groups-in-particular-upper-tringular-matrices-over-f-p/106605#106605" rel="nofollow">comments</a> by F. Ladisch: </p> <p>"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)." </p> <p>PSPS</p> <p>Another relevant MO-discussion <a href="http://mathoverflow.net/questions/57129/which-finite-groups-have-faithful-complex-irreducible-representations" rel="nofollow">Which finite groups have faithful complex irreducible representations?</a>. Let me quote: "Obvious necessary condition is that the center must be a cyclic group." </p> <p>"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"</p> <p>And further - see answers by Andreas Thom and Rob Harron.</p> http://mathoverflow.net/questions/106729/center-and-representations-of-finite-group-how-are-related/106731#106731 Answer by anton for Center and representations of finite group - how are related ? anton 2012-09-09T13:57:23Z 2012-09-09T13:57:23Z <p>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.</p> http://mathoverflow.net/questions/106729/center-and-representations-of-finite-group-how-are-related/106742#106742 Answer by Alain Valette for Center and representations of finite group - how are related ? Alain Valette 2012-09-09T17:26:33Z 2012-09-09T17:26:33Z <p>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 $G$.</p> http://mathoverflow.net/questions/106729/center-and-representations-of-finite-group-how-are-related/106750#106750 Answer by F. Ladisch for Center and representations of finite group - how are related ? F. Ladisch 2012-09-09T20:28:44Z 2012-09-09T20:28:44Z <p>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.<br> 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: <a href="http://arxiv.org/abs/math/0610726" rel="nofollow">arXiv:math/0610726</a>)</p>