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Olexandr Konovalov's user avatar
Olexandr Konovalov's user avatar
Olexandr Konovalov's user avatar
Olexandr Konovalov
  • Member for 11 years, 5 months
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Factorization of a finite group by two subsets
If you think that such subsets exist, perhaps you could find them with some random search. But if you think that the group is a candidate for the counterexample, it won't help.
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Factorization of a finite group by two subsets
NrCombinations([1..Size(PSL(2,8))],21); will tell you that the straightforward enumeration of all subsets of orders 21 and 24 is not going to work.
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Factorization of a finite group by two subsets
Here is a one-liner in GAP that shows no subgroups of index 21 in $PSL_2(8)$: G:=PSL(2,8);ForAny(ConjugacyClassesSubgroups(G),c->Index(G,Representative(c))=21);. Same for index 24.
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