Given a field $k$ with characteristic $p$, let $G$ be a transitive permutation group on $4p$ points. Let $P$ be a Sylow $p$-subgroup of $G$ and $Q\leq P$ is a $p$-subgroup of $P$ of index $p$. Now denote $H:=N_G(Q)/Q$. Could anyone provide me with an counterexample suth that the dimension of the projective cover $P_k$ of the trivial $kH$-module $k$ does not divide $p(p-1)$?
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4$\begingroup$ Why would you believe that should be true? $\endgroup$– Geoff RobinsonCommented Aug 11, 2021 at 15:17
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3$\begingroup$ You really have to provide more context with questions like this. What is the significance of the permutation representation of degree at most $4p$? Why $4p$ in particular? Why $p(p-1)$ in particular. Anyway, with a routine computer search, I found a counterexample of order $660$ with $p=3$, where the projective cover has dimension $12$. $\endgroup$– Derek HoltCommented Aug 11, 2021 at 16:17
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$\begingroup$ @ Geoff Robinson @ Derek Holt Thanks. I asked it since there are very little examples in my mind. I just know that it is transitive permutation group of degree less than $4p$ and has cyclic Sylow $p$-subgroups of order $p$. $\endgroup$– Master GangCommented Aug 12, 2021 at 0:48
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$\begingroup$ Maybe it is appropriate for me to just ask for an counterexample. And I have editted the question to make it more clearly. $\endgroup$– Master GangCommented Aug 12, 2021 at 1:25
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$\begingroup$ @DerekHolt May I ask which software did you use? $\endgroup$– მამუკა ჯიბლაძეCommented Aug 12, 2021 at 5:47
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1 Answer
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Here is a Magma calculation that shows that the group ${\rm PSL}(2,11)$ is a counterexample to your question
> G := PSL(2,11);
> I := AbsolutelyIrreducibleModules(G,GF(3));
> I;
[
GModule of dimension 1 over GF(3),
GModule of dimension 5 over GF(3),
GModule of dimension 5 over GF(3),
GModule of dimension 10 over GF(3),
GModule of dimension 12 over GF(3^2),
GModule of dimension 12 over GF(3^2)
]
//So GF(9) is a splitting field
> P := ProjectiveCover(TrivialModule(G,GF(9)));
> Dimension(P);
12
> CompositionFactors(P);
[
GModule of dimension 1 over GF(3^2),
GModule of dimension 10 over GF(3^2),
GModule of dimension 1 over GF(3^2)
]
answered Aug 12, 2021 at 8:09
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3$\begingroup$ Another way to see this is is that $G = {\rm PSL}(2,11)$ has a subgroup $H$ of order $55$ . When char $k =3$, the module ${\rm Ind}_{H}^{G}(k)$ is projective, and has the projective cover of the trivial module as a summand . The associated (complex) permutation character is the sum of the trivial and an irreducible character of degree $11$, as the permutation action is doubly transitive. It is not possible to decompose this character as the sum of two characters which each vanish on $3$-singular elements. Hence ${\rm Ind}_{H}^{G}(k)$ is the projective cover of the trivial module. $\endgroup$ Commented Aug 12, 2021 at 9:01
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$\begingroup$ Thanks for your nice counterexample and useful comments. $\endgroup$ Commented Aug 12, 2021 at 9:50
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1$\begingroup$ @GeoffRobinson In fact that is exactly how the algorithm computes the projective indecomposable, using induction from the subgroup of order $55$. $\endgroup$ Commented Aug 12, 2021 at 10:34