Let $G$ be a finite group such that $G$ has a normal subgroup $N$ of order $p(p^2+1)/2$, where $p>13$ is an odd prime and $p\ne 239$. Also $G/N\cong \text{PSL}(2,p)$. Can we say that there exists a prime divisor $t$ of $|N|$ such that $2t\not\mid \chi(1)$ for every $\chi\in \text{Irr}(G)$?

Of course we know that $N$ is isomorphic to ${\Bbb Z_p}\times H$, where $H$ is a solvable group of order $(p^2+1)/2$.

Thanks for your helps.

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    $\begingroup$ Where does this question arise? -- You are making some very specific choices, like $|N| = p(p^2+1)/2$ and $G/N \cong {\rm PSL}(2,p)$. What is the motivation for these choices? $\endgroup$ – Stefan Kohl Mar 15 '14 at 14:05
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    $\begingroup$ BHZ: I think @StefanKohl means that you should explain honestly what led you to consider these particular parameters and examples, because then people will be able to follow your train of thought, or offer solutions to related questions. $\endgroup$ – Yemon Choi Mar 15 '14 at 16:48
  • $\begingroup$ The most obvious candidate would seem to be to try the case $t = p.$ $\endgroup$ – Geoff Robinson Mar 15 '14 at 20:11
  • $\begingroup$ I apologize but we must consider $ t\ne p $ in the problem. Since we need to discuss on the nonabelian part of $ N $. $\endgroup$ – BHZ Mar 15 '14 at 22:13

I suppose that $p >3$ in my answer. Note that $|H|$ and $|{\rm PSL}(2,p)|$ are coprime. It follows by transfer and a Theorem of Gaschutz that $N$ is complemented in $G,$ and that $O_{p}(N)$ is a direct factor of $G.$ Hence we might as well concentrate on $H,$ rather than $N.$ So we look at the group $HX$ where $X \cong {\rm PSL}(2,p).$ I claim that $X$ acts trivially on $H.$ If not, then there is a chief factor $V$ "within" $H$ such that $X$ acts faithfully on $V,$ and $V$ is a $q$-group for some prime $q.$ Since the action is coprime, and $X$ has no non-trivial complex irreducible character of degree less than $\frac{p-1}{2},$ we have $|V| \geq q^{\frac{p-1}{2}}.$ Also, we have $q \geq 5$ as ${\rm PSL}(2,p)$ has order divisible by $6.$ But $|V| \leq |H|,$ so we certainly obtain $2^{p-1} < \frac{p^{2}+1}{2},$ a contradiction as $2^{p} > p^{2}+1$ for $p \geq 5.$

Thus $G \cong X \times N.$ Since you do not allow $t =p,$ and since $|X|$ and $|H|$ are relatively prime, your question is now really entirely about the irreducible characters of $H.$ But since $X$ has irreducible characters of even degree, if I understand what you are asking, you want every irreducible character of $H$ to have degree prime to $t$ for some prime divisor $t$ of $H.$ That means that you need $H$ to have an Abelian normal Sylow $t$-subgroup for some such prime $t.$ So this question seems to reduce to one you have asked before, which is purely number-theoretic, and probably difficult.

  • $\begingroup$ Thank you very much for your very complete and detailed answer. $\endgroup$ – BHZ Mar 16 '14 at 2:35

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