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.
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.
