How "exotic" are non-local directly irreducible modular group algebras?

Question

Let $G$ be a finite group and $R = \Bbb F_{p^k} G$ a modular group algebra ($p$ divides $|G|$). I would like to know when $R$ is a directly indecomposable ring. It is quite well-known that $R$ is local iff $G$ is a $p$-group, but a finite noncommutative ring can, of course, have nontrivial idempotents without any of them being central.

I have a suspicion that Brauer's three theorems or related results might give a description of such $G$'s, or some strong properties of them (e.g. related to factorization of $|G|$), but I am not familiar with their modus operandi. For instance, if the $p$-Sylow is normal, then the first theorem (from what I'm reading on Wikipedia) is a tautology.

H. Meyer computed the central idempotents for $\Bbb F_2 S_n$, $n \le 54$, and apparently $\Bbb F_2 S_4$ has a single block (the expressions on pp. 3–5 have no $|_4$).

Answer 1

A necessary condition (as pointed out in comments by @rschwieb) is that $G$ has no nontrivial normal subgroup of order coprime to $p$.

Theorem 2.1 of

Fong, P.; Gaschütz, W., A note on the modular representations of solvable groups, J. Reine Angew. Math. 208, 73-78 (1961). ZBL0100.25801.

states that any solvable finite group satisfying this condition has only one $p$-block. This covers the case of $S_4$ in characteristic $2$.

The condition of solvability was weakened to being $p$-constrained by Cossey and Gaschütz in Theorem 1 of

Cossey, John; Gaschütz, Wolfgang, A note on blocks, Proc. 2nd internat. Conf. Theory of Groups, Canberra 1973, Lect. Notes Math. 372, 238-240 (1974). ZBL0295.20015.

(@rschwieb discovered this result in Blackburn and Huppert's Finite Groups II.)

Given that people were proving results like this in the 1970s I'm guessing that there's no simple necessary and sufficient condition for general finite groups. Being $p$-constrained is not a necessary condition, as I used GAP to check sporadic simple groups (which are never $p$-constrained if their order is divisible by $p$), and in characteristic $2$ the Mathieu groups $M_{22}$ and $M_{24}$ each have only one block.

Answer 2

There exists a full classification in terms of group properties of $G$ (over $\overline{\Bbb F_p}$). Can the principal block split when passing to the algebraic closure..?

Harris, M. E., On the $p$-deficiency class of a finite group, J. Algebra 94:2, 411–424 (1985) (bracketed comments mine):

Theorem 1. Let G be an arbitrary finite group. Then:
(a) if $p$ is an odd prime integer, then $B_0(G)$ [the principal block] is the only $p$-block of $G$ if and only if $F^*(G)$ [the generalized Fitting subgroup] $= O_p(G)$ [the $p$-core of $G$]; and
(b) $B_0(G)$ is the only $2$-block of $G$ if and only if (i) $O_{2'}(G) = 1$ and (ii) all components of $G$ are of type $M_{22}$ or $M_{24}$.

(I found the article in references of Liu, Y.; Willems, W., Lie-type-like groups, J. Algebra 447, 432–444 (2016).)

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In his latter comment about simple groups, @JeremyRickard likely alludes to this theorem in Harris's paper (by the comment after Definition 1, $p$-deficiency class $0$ is equivalent to having a single $p$-block):

Theorem 4. (a) $M_{22}$ and $M_{24}$ are the only finite simple groups of $2$-deficiency class $0$.

and the paper Brockhaus, P.; Michler, G. O., Finite simple groups of Lie type have non-principal $p$-blocks, $p \not = 2$, J. Algebra 94:1, 113–125 (1985) which is used there.

If I understood the definitions of $F^*$ and $O_p$ correctly, Theorem 1 (a) implies that there cannot be simple groups with one $p$-block in characteristic $p > 2$.