symmetric 2-cocycle / many projective representations

Question

Let $G$ be a finite group, $k$ the field of complex numbers.

Are there (cohomologically nontrivial) group 2-cocycles $\sigma\in Z^2(G,k^\times)$ such that for all $g,h\in G$: $$\sigma(g,h)=\sigma(ghg^{-1},g)$$

I would like to rule out nontrivial $\sigma$ with that property for any nonabelian group or get hands on a counterexample. They are directly related to possible additional automorphisms in nonabelian Dijkgraaf-Witten theory resp. additional braided autoequivalences of the Drinfel'd double representations.

A conjugacy class $[g]$ is called $\sigma$-regular iff $\sigma(g,h)=\sigma(h,g)$ for all $h\in Cent(g)$ (this does not depend on the representing $g$ or $\sigma$). Hence a symmetric cocycle in the sense above would especially imply that all conjugacy classes $[g]$ are $\sigma$-regular. We can hence reformulate the question:

Can a nontrivial $k_\sigma[G]$ have the same number of irreducible representations as $k[G]$?

Thank you!

Added: The converse is also true! It seems standard (e.g. Ofir's reference [Higgs89] Lm. 1.2i) that in any class $[\sigma]$ there is a representative such that for any $\sigma$-regular $x$ and any $g$ $$\frac{\sigma(g,x)\sigma(gx,g^{-1})}{\sigma(g,g^{-1})}=1$$ An easy cohomology calculation shows indeed $$\frac{\sigma(g,x)}{\sigma(gxg^{-1},g)} =\frac{\sigma(g,x)}{\sigma(gxg^{-1},g)} \cdot \frac{\sigma(gx,g^{-1})\sigma(gxg^{-1},g)}{\sigma(gx,1)\sigma(g,g^{-1})}=1$$ hance if all $x$ are $\sigma$-regular (second question) then this equation holds for all $g,x$ (first question).

Answer 1

I will answer your second question. By generalized Maschke theorem the twisted group algebras $\mathbb{C}^fG$ are all semi-simple and the number of simple components is the number of the irreducible projective $f$-representations of $G$, ($|Irr_f(G)|$). Hence, $$|Irr_f(G)|=dim (Z(\mathbb{C}^fG)).$$ Now, an element $g\in G$ is called $f$-regular if for any $x\in C_G(g)$ $$f(g,x)=f(x,g).$$ It is also known that $Z(\mathbb{C}^fG)$ has a basis consisting of $f$-regular conjugacy classes. Clearly for trivial cocycle any conjugacy class is regular, hence I formulte your second question as follows:

Is there exist a group $G$ and a $f\in Z^2(G,\mathbb{C}^*)$ which is not cohomogaclly trivial such that any $g\in G$ is $f$-regular.

This elements are sometimes called "distinguish" and they may exist as in the example gave by R.J.Higgs in his paper from 1987: "projective characters of degree one and the inflation-restriction sequence".

Answer 2

Not a complete answer: I prefer to think of this question as a question about central extensions: it is equivalent to the following: Let $H$ be a finite group with $Z(H) \leq H^{\prime}$, and let $\lambda$ be a linear character of $Z(H)$. Is it possible that the number of complex irreducible characters of $H$ which lie over $\lambda$ (ie, have restrictions to $Z(H)$ which are multiples of $\lambda$) is the same as the number of complex irreducible characters of $H/Z(H)$? The former number is always less that or equal to the latter. When $Z(H)$ has prime order $p,$ we have the desired equality exactly when some non-identity element of $Z(H)$ is not a commutator (and, in fact, in that case, no non-identity element of $Z(H)$ is a commutator). Here, by a commutator, I mean an element of the form $[x,y] = x^{-1}y^{-1}xy.$ There are finite quasi-simple groups $H$ which contain central elements which are not commutators, and all such occurences were listed by H. Blau ( with later tables by M. Liebeck giving all non-commutators, central or not, in quasisimple groups). If I am reading these lists correctly, there are no quasisimple groups $H$ with $Z(H)$ of prime order with a central element which is not a commutator. However, I do not not know whether there are any perfect groups $H$ with $Z(H)$ of prime order in which non-identity elements of $Z(H)$ are non-commutators. In general, if there were a finite perfect group $H$ in which no non-identity element of $Z(H)$ was a commutator, then $G$ would have $|Z(H)|k(H/Z(H))$ complex irreducible characters, where $k(X)$ denotes the number of conjugacy classes of a finite group $X$, and furthermore, there would be $k(H/Z(H))$ irreducible characters of $H$ lying over each linear character of $Z(H)$. However, I do not know if there is such a perfect group $H$.