The Simplified Hasse diagram of A_5

 A5 is the alternating group on 5 elements. Since the order of S5 is 120, the order of A5 is half the size at 60. Here is the Simplified Hasse Diagram of the structure of the subgroups.

The red circles indicate the normal subgroups, which in this case are just the whole group and the identity. having just two normal subgroups means this is a simple group.

 

Red arrows can only originate in normal subgroups, and in this case, the identity is a normal subgroup of every subgroup of prime order.

 

There are 15 subgroups of order 2, all of the generators of the form (ab)(cd). 

 

There are 10 subgroups of order 3, all of generators of the form (abc).

The 6 subgroups of order 5 have generators of the form (abcde).

For the Klein-4 subgroups, we have the identity and three double transpositions that all keep the same element fixed. For example, (12)(34), (13)(24) and (14)(23) all leave 5 untouched. Likewise, the five copies of A4 all leave one element fixed.

The subgroups of order 5, 6 and 10 do not leave any element fixed, so they are not subgroups of A4. Of those three orders, only 6 divides 12, but the groups isomorphic to S3 are generated by a 3-cycle (abc) and a double transposition (ab)(de), so no element is fixed.

 

Every purple arrow goes from a subgroup of order n to a subgroup of order 2n. These subgroups are not normal, which is indicated by the blue rectangles, nor are they subnormal, since subnormality only occurs if a subgroup is part of a chain of subgroups, each normal in the next subgroup up in the chain and terminating in the entire group. A simple group that has any proper subgroup that is not the identity cannot contain a subnormal subgroup, so these subgroups that have purple arrows coming out of them are normalish, the phrase I am using until I learn the proper term.

 

The sum of all the numbers in red circles and blue rectangles is 59, the total number of subgroups.

 

The next Simplified Hasse diagram will be of S5, which will be a much larger undertaking with 156 total subgroups.

 

 

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.