The Alternating Groups A_4 and A_5

 The symmetric group Sn is the group of all permutations of n elements and the alternating group An is the group of all even permutations of n elements. The order of Sn is n! and for n > 1, the order of An is n!/2. If we look at An as a subgroup of Sn, it is normal, as are all subgroups whose order is exactly half the order of the group.

Recall that a normal subgroup N of a group G has several properties, but here are the two we will look at in this post.

1. For finite groups, the order of N must divide the order of G. (This is true of non-normal subgroups as well.)

2. A normal subgroup N must be the union of conjugacy classes, and one of those classes must be the identity element, which is a singleton class because the identity commutes with all elements.

Let us start with A4.

 

The identity: 

(1)

 

The permutations of the form (ab)(cd): 

(12)(34), (13)(24), (14)(23)

 

The 3-cycles:

(123), (132), (124), (142), (134), (143), (234), (243)

 

The order of A4 is 12, and the cardinalities of the conjugacy classes are 1, 3 and 8 respectively. To add these numbers up when we must include the 1, only 1+3 = 4 will divide 12. As it happens, the identity and the double transpositions do form a subgroup isomorphic to the Klein four-group, and it is a normal subgroup of A4.

Because A4 has a normal subgroup that is not either the whole group or the identity, A4 is not simple.

 

A5 is much larger, with 60 elements. Again we will list the conjugacy classes.

The identity: 

(1)

(1 total)

 

The permutations of the form (ab)(cd): 

(12)(34), (13)(24), (14)(23), 

(12)(35), (13)(25), (15)(23),

(12)(45), (14)(25), (15)(24),

(13)(45), (14)(35), (15)(34),

(23)(45), (24)(35), (25)(34)

(15 total)

 

The 3-cycles:

(123), (132), (124), (142), (125), (152),

(134), (143), (135), (153), (145), (154),

(234), (243), (235), (253), (245), (254),

(345), (354)

(20 total)

The 5-cycles:

(12345), (12354), (12435), (12453), (12534), (12543), 

(13245), (13254), (13425), (13452), (13524), (13542), 

(14235), (14253), (14325), (14352), (14523), (14532), 

(15234), (15243), (15324), (15342), (15423), (15432)

(24 total)

Now the numbers we must use to sum to a divisor of 60 are 1, which is mandatory, and 15, 20 and 24. Clearly, the sum of any three will be greater than 30 and less than 60, and there are no proper divisors of 60 greater than 30. Our two number sums are

1+15 = 16

1+20 = 21

1+24 = 25

None of these numbers divide 60 evenly, so the union of these classes cannot be a subgroup.

A5 is our first example of a simple non-abelian group. This is an important fact in Abel's proof of the insolubility of the quintic.