# About the degree of character of $PSL(n,q)$

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 $\chi(1)_p=|G|_p$.

Thanks.

• You're looking for the Steinberg character. – Jay Taylor Oct 31 '17 at 13:58
• 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. – Jim Humphreys Oct 31 '17 at 15:02
• @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. – Jim Humphreys Oct 31 '17 at 15:03
• @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. – Jay Taylor Oct 31 '17 at 17:04
• 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 $q \equiv 3 \mod{4}$. – spin Oct 31 '17 at 20:14

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 $p=2$.
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: