Let $G$ be a finite group of size $p^a\cdot r$. Does there exist a simple way to calculate radical of the group algebra $F_p[G]$?
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2$\begingroup$ Karpilovski wrote a whole book, The Jacobson Radical of Group Algebras, about the subject... What do you mean by «simple»? $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 27, 2011 at 17:55
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1$\begingroup$ Thanks for reference. For example when G have normal p-group P then I know simple answer. In this case as I understand radical is $Ind_P^G\Delta(P)$, where $Delta(P)$ are all elements of $F_p[P]$ with sum zero. Does there exists similar answer when P is not normal subgroup? – student 4 mins ago $\endgroup$– studentCommented Sep 27, 2011 at 18:18
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4$\begingroup$ When G has a normal p-Sylow subgroup P, then the group homomorphism $G\rightarrow G/P$ induces the semisimple quotient at the level of algebras (since $F_p[G/P]$ is semisimple by Maschke's theorem). So in this case the radical is generated by all elements of $F_p[G]$ such that the coefficients of the elements of $P$ sum to zero. In general, $G$ has a largest normal $P$-subgroup $N$, but $G/N$ is not semisimple if $N$ is not $p$-Sylow. Thus $F_p[G]\right F_p[G/N]$ is not the semisimple quotient. Clearly though it suffices to compute $rad(F_p[G/N])$ so you are reduced to the case $N$ is trivial $\endgroup$– Benjamin SteinbergCommented Sep 27, 2011 at 18:30
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$\begingroup$ Benjamin: the case in which G has a normal p-Sylow subgroup is known as Wallace Theorem. $\endgroup$– Salvatore SicilianoCommented Jan 1, 2012 at 12:23
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I think the answer is "no, in general", though the terms in the question are not really well-defined. For example, if you knew the radical of the group algebra, then you could easily compute the dimension of the absolutely irreducible modules in characteristic $p.$ When $G$ is the symmetric group, this is already a notoriously difficult (and as yet unsolved) problem.
answered Dec 31, 2011 at 23:26