Timeline for Character theory of finite groups

6 events

when toggle format	what		by	license	comment
Jan 29, 2017 at 7:18	comment	added	asad		@GeoffRobinson, Thank you very much!
Jan 22, 2017 at 13:53	comment	added	Geoff Robinson		(H) should of course have been $Z(H).$
Jan 22, 2017 at 13:48	comment	added	Geoff Robinson		I said $G \cong {\rm PSL}(2,11)$ so that it is not the case that $A \leq H.$ But yes, I was ambiguous because $G$ was defined in the question as the whole group .I was a little loose in that $A$ might not be complemented. Modulo this confusion, our answers are not incompatible.
Jan 22, 2017 at 13:07	comment	added	Ehud Meir		@GeoffRobinson: what do you mean by $AH$ and $(H)$? If I understand you correctly, $H$ contains $A$. Let me just add that another way to see it is the following: the extension $$1\to R(G)\to G\to G/R(G)\to 1$$ gives you a central extension of $G/R(G)$ by $\mathbb{C}^{\times}$. On this extension we can think of as a two cocycle $\alpha$. The degrees you are interested in are the degrees of the irreducible representations of the twisted group algebra $\mathbb{C}^{\alpha}G/R(G)$.
Jan 22, 2017 at 12:02	comment	added	Geoff Robinson		You may assume that $A = R(G)$ is cyclic ( after factoring out $ker \lambda$). You build a central extension $H$ of ${\rm PSL}(2,11)$ such that $\lambda$ extends to $AH.$ Then the irreducible constituents you seek are degrees of irreducible characters of $H$ which lie over a certain linear character of $(H)$.
Jan 22, 2017 at 10:13	history	asked	asad	CC BY-SA 3.0