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It is well known that for $n\geq2$ the group $PSL(n,q)$ is simple except for $PSL(2,2)=S_3$ and $PSL(2,3)=A_4$.

Let $G$ be one of the simple groups $PSL(n,q)$. From the ATLAS of finite group, we guess that for every prime $p\in\pi(G)$ there exists some $\mathbb{C}$-representation $\mathfrak{X}$ such that its degree $\chi(1)$ satisfies $\chi(1)_p=|G|_p$.

Now I want know some information about this conjecture.

Thanks.

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    $\begingroup$ You're looking for the Steinberg character. $\endgroup$
    – Jay Taylor
    Commented Oct 31, 2017 at 13:58
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    $\begingroup$ In principle, the extensive work of Lusztig (following the basic Deligne-Lusztig paper in 1976) allows one to compute all degrees of irreducible complex characters of finite groups of Lie type. (I assume your question concerns irreducible characters over $\mathbb{C}$.) But it may be nontrivial to answer your question, since you allow $p$ to range over all primes dividing the group order, not just the defining characteristic which is usually called $p$ (with $q$ being a power of $p$). In the literature other primes are usually denoted $\ell$ or such. $\endgroup$
    – Jim Humphreys
    Commented Oct 31, 2017 at 15:02
  • $\begingroup$ @Jay: Note that the question here allows for any prime dividing the group order. For the defining prime, certainly it's just the Steinberg character. $\endgroup$
    – Jim Humphreys
    Commented Oct 31, 2017 at 15:03
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    $\begingroup$ @JimHumphreys Ah, thanks! I missed that. I struggle to not read $p$ as the defining prime. Maybe, one could also say here that it's sufficient to find a $p$-block with trivial defect. $\endgroup$
    – Jay Taylor
    Commented Oct 31, 2017 at 17:04
  • $\begingroup$ At least for $n = 2$ this is true. In this case for $q$ even, $G$ has order $(q-1)q(q+1)$ and the character degrees of $G$ are $1$, $q$ and $q \pm 1$. For $q$ odd, $G$ has order $\frac{1}{2}(q-1)q(q+1)$ and the character degrees are $1$ , $q$, $(q \pm 1)$, and $\frac{1}{2}(q+1)$ or $\frac{1}{2}(q-1)$, depending whether $q \equiv 1 \mod{4}$ or $q \equiv 3 \mod{4}$. $\endgroup$
    – spin
    Commented Oct 31, 2017 at 20:14

1 Answer 1

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I only happen to know the answer to this question because I had occasion to look up similar things recently. As Jay mentions, asking for a character of this type is the same as asking for a block of $p$-defect zero (which then contains exactly one such character). For finite simple groups of Lie type, the existence of such a block was shown by Michler for all odd primes $p$, and by Willems for $p=2$.

Michler, Gerhard O. A finite simple group of Lie type has p-blocks with different defects, p≠2. J. Algebra 104 (1986), no. 2, 220–230.

Willems, Wolfgang. Blocks of defect zero in finite simple groups of Lie type. J. Algebra 113 (1988), no. 2, 511–522.

A complete list of nonabelian finite simple groups $G$ for which there exists a prime $p$ such that $G$ does not have a $p$-block of defect zero can be found in Corollary 2 here:

Granville, Andrew; Ono, Ken. Defect zero p-blocks for finite simple groups. Trans. Amer. Math. Soc. 348 (1996), no. 1, 331–347.

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  • $\begingroup$ These references look quite relevant. I haven't kept track of such developments in finite group theory but was vaguely aware of progress on existence of blocks of defect 0. It's useful to note that the articles you mention are now freely available online: see for example sciencedirect.com/science/journal/00218693?sdc=1 $\endgroup$
    – Jim Humphreys
    Commented Nov 1, 2017 at 15:43

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