Page 98 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
P. 98
3.3 Finite simple groups
• Representative (∗)(∗ ∗)(∗ ∗): this class has size 15 and does not split into two conju-
gacy classes of A5;
• Representative (∗)(∗)(∗ ∗ ∗): this class has size 20 and does not split into two conju-
gacy classes of A5;
• Representative (∗ ∗ ∗ ∗ ∗): this class has size 24 and splits into two conjugacy classes
of A5, each of size 12.
A normal subgroup N of A5 would have to be a union of conjugacy classes and contain
the identity, plus its order would have to divide 60. Checking all the possibilities, we see
that either N is trivial or N = A5.
It turns out that A5 is the only simple group of order 60. A formal proof can be found
in [4]. Here is a proof using GAP:
gap> Filtered(AllSmallGroups(60), IsSimple);
[ Alt( [ 1 .. 5 ] ) ]
Theorem 3.3.13 If n ≥ 5, then An is simple.
PROOF. The proof goes by induction on n . The case n = 5 is covered by Proposition 3.3.12.
Suppose N is a non-trivial normal subgroup of An . Since An clearly acts doubly transi-
tively on X = {1, 2, . . . , n }, this action is primitive by 3.3.2. Therefore N acts transitively on
X by 3.3.5. It follows by Frattini’s argument that N An−1 = An . The intersection N ∩ An−1
is a normal subgroup of An−1. By assumption, either N ∩ An−1 = {1} or An−1 ≤ N . In
the latter case, An /N = N An−1/N ∼= An−1/(An−1 ∩ N ) = {1}, hence N = An . So assume
that N ∩ An−1 = {1}. In this case N acts regularly and so |N | = n by a discussion above.
By Lemma 3.2.30, N can be generated by at most log2 n elements. An automorphism
of N is determined by the images of generators, hence | Aut(N )| ≤ n log2 n . On the other
hand, An−1 acts faithfully on N by conjugation, so (n − 1)! ≤ n log2 n which is impossible
for n ≥ 6.
Corollary 3.3.14 Let n ≥ 5. Then the only normal subgroups of Sn are {1}, An and Sn .
PROOF. Let N be a normal subgroup of Sn . Then N ∩ An is a normal subgroup of An ,
hence either An ∩ N = {1} or An ≤ N . Suppose the first possibility holds. Then N =
N /(N ∩ An ) =∼ N An /An . If N is non-trivial then N An = Sn and hence N ∼= C2. This is
impossible as there would have to be a non-identity element of An in a conjugacy class
of size 1. The remaining possibility is An ≤ N , but in this case we either have N = An or
N = Sn , as An is a maximal subgroup of Sn .
• Representative (∗)(∗ ∗)(∗ ∗): this class has size 15 and does not split into two conju-
gacy classes of A5;
• Representative (∗)(∗)(∗ ∗ ∗): this class has size 20 and does not split into two conju-
gacy classes of A5;
• Representative (∗ ∗ ∗ ∗ ∗): this class has size 24 and splits into two conjugacy classes
of A5, each of size 12.
A normal subgroup N of A5 would have to be a union of conjugacy classes and contain
the identity, plus its order would have to divide 60. Checking all the possibilities, we see
that either N is trivial or N = A5.
It turns out that A5 is the only simple group of order 60. A formal proof can be found
in [4]. Here is a proof using GAP:
gap> Filtered(AllSmallGroups(60), IsSimple);
[ Alt( [ 1 .. 5 ] ) ]
Theorem 3.3.13 If n ≥ 5, then An is simple.
PROOF. The proof goes by induction on n . The case n = 5 is covered by Proposition 3.3.12.
Suppose N is a non-trivial normal subgroup of An . Since An clearly acts doubly transi-
tively on X = {1, 2, . . . , n }, this action is primitive by 3.3.2. Therefore N acts transitively on
X by 3.3.5. It follows by Frattini’s argument that N An−1 = An . The intersection N ∩ An−1
is a normal subgroup of An−1. By assumption, either N ∩ An−1 = {1} or An−1 ≤ N . In
the latter case, An /N = N An−1/N ∼= An−1/(An−1 ∩ N ) = {1}, hence N = An . So assume
that N ∩ An−1 = {1}. In this case N acts regularly and so |N | = n by a discussion above.
By Lemma 3.2.30, N can be generated by at most log2 n elements. An automorphism
of N is determined by the images of generators, hence | Aut(N )| ≤ n log2 n . On the other
hand, An−1 acts faithfully on N by conjugation, so (n − 1)! ≤ n log2 n which is impossible
for n ≥ 6.
Corollary 3.3.14 Let n ≥ 5. Then the only normal subgroups of Sn are {1}, An and Sn .
PROOF. Let N be a normal subgroup of Sn . Then N ∩ An is a normal subgroup of An ,
hence either An ∩ N = {1} or An ≤ N . Suppose the first possibility holds. Then N =
N /(N ∩ An ) =∼ N An /An . If N is non-trivial then N An = Sn and hence N ∼= C2. This is
impossible as there would have to be a non-identity element of An in a conjugacy class
of size 1. The remaining possibility is An ≤ N , but in this case we either have N = An or
N = Sn , as An is a maximal subgroup of Sn .
