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Let $G$ be a finite group, $R(G)$ be the solvable radical of $G$, and $G/R(G)$ be isomorphic to $PSL_2(11)$. Also let $\lambda$ be a non-trivial complex linear character of $R(G)$ such that $\lambda$ is invariant in $G$. If $\lambda$ is not extendible in $G$, what can be said about character degrees of the constituents of $\lambda^G$?

Any help or references or comment are appreciated.

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    $\begingroup$ 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 $(H)$. $\endgroup$
    – Geoff Robinson
    Commented Jan 22, 2017 at 12:02
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    $\begingroup$ @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 $\mathbb{C}^{\alpha}G/R(G)$. $\endgroup$
    – Ehud Meir
    Commented Jan 22, 2017 at 13:07
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    $\begingroup$ 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. $\endgroup$
    – Geoff Robinson
    Commented Jan 22, 2017 at 13:48
  • $\begingroup$ (H) should of course have been $Z(H).$ $\endgroup$
    – Geoff Robinson
    Commented Jan 22, 2017 at 13:53
  • $\begingroup$ @GeoffRobinson, Thank you very much! $\endgroup$
    – asad
    Commented Jan 29, 2017 at 7:18

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