proof: $A_5$ is generated by $(1\;2\;3)$ and $(1\;2\;3\;4\;5)$ 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 $A_5$ 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 $|A_5| = 60$ is solvable so $A_5$ 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 $A_n$ is simple.
proof: Induction on $n$ with base case $5$. $A_n$ 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 $C_2^2$ in the case of $A_6$ and there isn't one otherwise but $C_2^2$ doesn't have enough elements to be transitive on 6 points so it can't be regular - so $A_n$ has no regular normal subgroups: therefore by the proposition in that section it's simple.
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