proof: A5 is generated by (123) and (12345) both are commutators:
gap> a := (1,5,2);; b := (4,2,3);; a^(-1)*b^(-1)*a*b; (1,2,3) gap> a := (1,2,3)*(3,4,5);; b := (1,4,2)*(3,5,2);; a^(-1)*b^(-1)*a*b; (1,2,3,4,5)
Theorem A5 is simple.
proof: The most basic proof using cycles directly can be found in Goodman.
proof: The conjugacy classes have sizes: 1, 15, 20, 12, 12. No sum of these that includes 1 is a divisor of 60 so there are no normal subgroups (which would necessarily be a union of conjugacy classes).
proof: A perfect group is not solvable, and every smaller group whose order divides |A5|=60 is solvable so A5 has no normal subgroups (else it would be solvable too!)
gap> List(AllSmallGroups(2), StructureDescription); [ "C2" ] gap> List(AllSmallGroups(2^2), StructureDescription); [ "C4", "C2 x C2" ] gap> List(AllSmallGroups(2*3), StructureDescription); [ "S3", "C6" ] gap> List(AllSmallGroups(2*5), StructureDescription); [ "D10", "C10" ] gap> List(AllSmallGroups(2*3*5), StructureDescription); [ "C5 x S3", "C3 x D10", "D30", "C30" ] gap> List(AllSmallGroups(2^2*3), StructureDescription); [ "C3 : C4", "C12", "A4", "D12", "C6 x C2" ] gap> List(AllSmallGroups(2^2*5), StructureDescription); [ "C5 : C4", "C20", "C5 : C4", "D20", "C10 x C2" ]
Theorem An is simple.
proof: Induction on n with base case 5. An is n−2 transitive in the natural action (by the multiple-transitivity section) so for n>5 this action is (at least) 2-transitive so primitive (by primitivity section) which by the powerful corollary about transitivity with regular normal subgroups tells us that a regular normal subgroup would have to be C22 in the case of A6 and there isn't one otherwise but C22 doesn't have enough elements to be transitive on 6 points so it can't be regular - so An has no regular normal subgroups: therefore by the proposition in that section it's simple.
No comments:
Post a Comment