More Simplified Hasse Diagrams, S_3, A_4 and S_4.

 

Today's post is all about Simplified Hasse Diagrams. I didn't label this one to specify the group, but the only group of order 6 that is non-abelian is S3. In this group, subgroups are either normal, indicated by red ovals, on not normal, indicated by the blue rectangle.

The next diagram is for A4, order 12 and slightly more complex than S3. Notice the purple arrow that connects the order 2 subgroups to the Klein-4 group at order 4. The order 2 groups are subnormal, because the are normal in the next group up, not normal in the entire group, and there is a chain starting at any order 2 subgroup, up to the Klein-4 and ending at A4.

 

 

 

 

 

 

 

 

 

 

 

 

And then we have the Simplified Hasse diagram for S4, which is much more complicated than the preceding diagrams.

 

First complication: We have order 4 subgroups that are not isomorphic to one another, 1 Klein-4 and 3 copies of Z4.

 

Second complication: while all groups of order 2 are isomorphic, three are generated by even permutations, specifically (12)(34), (13)(23) and (14)(23), while six are generated by odd permutations, (12), (13), (14), (23), (24) and (34).

 

We now have four purple arrows. The even permutation subgroups of order 2 are still subnormal, starting a chain that goes through the Klein-4 subgroup, up to the A4 and on to the S4. But there are other purple arrows that lead to non-normal subgroups and no chain leads up to the entire group.

 

Specifically, the order 3 subgroups are normal in the order 6 subgroups, but the order 6 subgroups have both even and odd permutations, so they are not subgroups of A4. They are non-normal subgroups of S4.

 

Likewise, the Z4 subgroups are normal in their D4 supergroups, but the D4 subgroups are not normal is S4.

 

So we have the groups with a single generator, Z3 and Z4, normal in the groups just above them, S3 and D4 respectively, but not part of a subnormal chain. Until I can find out what the correct word is, I am calling these subgroups normalish. I assume there is a term because they can be found in very well known groups of small order.

 

The "culprits" that keep these subgroups from being subnormal are the subgroups that have both even and odd permutations in them. In S4, only the subgroups of all even permutations are normal.

 

Next up: A5 and S5, which are order 60 and 120 respectively. These will clearly be more convoluted. 

The symmetries of the dodecahedron, which are the same as the symmetries of the icosahedron.

 Reviewing what we have learned about group theory and the platonic solids so far.

The tetrahedral symmetries are isomorphic to A4, the alternating group on four elements. The tetrahedron has 4 faces, 4 vertices and 6 edges, and it is self-dual. We create the dual by switching the faces and vertices. The group's order is 12.

The cube (6 faces, 8 vertices, 12 edges) and the octahedron (8 faces, 6 vertices, 12 edges) are duals, and their groups of symmetries have order 24 and are isomorphic to S4. A4 is a subgroup of S4 and a regular tetrahedron can be embedded in a cube. The cube can be embedded in a regular octahedron and vice versa, which is always the case with dual polyhedra.

 

So now we move on to the two largest Platonic solids, the 12-sided regular dodecahedron and the 20-sided regular icosahedron.

 

As you might expect, the dodecahedron (12 faces, 20 vertices, 30 edges) and the icosahedron (20 faces, 12 vertices, 30 edges) are duals, so I will represent them as a subgroup of S12, the smaller of our two choices of symmetric groups. The order of the group is 60.

 

Being a nerd, of course I had a 12-sided die just lying around, and I used the numbering convention from one of these. (This picture was nicked from Dice Game Depot, a fine web establishment for all your dice needs.)

 

Opposite sides always sum to 13, so the opposite pairs are 12 & 1, 11 & 2, 10 & 3, 9 & 4, 8 & 5 and 7 & 6. I changed 10, 11 and 12 to a, b and c, so that every face was represented by a single character.

 

Here are our conjugacy classes.

 

As always, the identity stands alone.

(1)

 

==

There are 15 elements of order 2, and they are six transpositions. To visualize this, think about holding the dodecahedron with you forefinger and thumb on two opposite edges, and spin the die 180°. Since there are 30 edges, there are 15 pairs we can use.

(1c)(3a)(47)(28)(96)(b5)

(1c)(2b)(9a)(57)(43)(86)

(1c)(49)(27)(8a)(b6)(53)

(1c)(67)(23)(48)(ba)(95)

(1c)(58)(7a)(29)(63)(b4)

(58)(67)(12)(4a)(cb)(93)

(2b)(58)(13)(46)(ac)(97)

(3a)(58)(14)(26)(c9)(b7)

(2b)(49)(15)(6a)(c8)(73)

(3a)(2b)(16)(54)(c7)(89)

(49)(58)(17)(2a)(c6)(b3)

(3a)(67)(18)(24)(c5)(b9)

(2b)(67)(19)(a5)(c4)(38)

(49)(67)(1a)(25)(c3)(b8)

(3a)(49)(1b)(65)(c2)(78)

 

Every such permutation will send two different pairs of faces to their opposite numbers, and those two transpositions are always the first two in the list.

 

These are even permutations.

 

==

 

Next are the 20 elements of order 3. To visualize these, spin the die around two opposite vertices. The inverses are written on the same lines with the "&" between them.

 

(124)(cb9)(357)(a86) & (142)(c9b)(375)(a68)

(146)(c97)(a8b)(352) & (164)(c79)(ab8)(325)

(165)(c78)(239)(ba4) & (156)(c87)(293)(b4a)

(15a)(c83)(4b7)(926) & (1a5)(c38)(47b)(962)

(1a2)(c3b)(689)(754) & (12a)(cb3)(698)(745)

(a95)(348)(2c6)(b17) & (a59)(384)(26c)(b71)

(95b)(482)(713)(6ca) & (9b5)(428)(731)(6ac)

(5b6)(827)(ac4)(319) & (56b)(872)(a4c)(391)

(6b3)(72a)(198)(c45) & (63b)(72a)(189)(c54)

(346)(a97)(c25)(1b8) & (364)(a79)(c52)(18b)

 

Yet again, these are all even permutations.

 

==

 

Lastly, there are 24 elements of order 5, which can be visualized as spinning the die while holding two opposite faces, and these permutations are also all even. Again, inverses are listed together on the same line, separated by the ampersand. 

 

Two faces are fixed by these permutations and the fixed faces are listed above the four permutations created by the rotations.

 

1,c fixed:

(2465a)(b9783) & (26a45)(b7398)

(254a6)(b8937) & (2a564)(b3879)

 

2,b fixed:

(1487a)(c9563) & (18a47)(c5396)

(174a8)(c6935) & (1a784)(c3659)

 

3,a fixed:

(46bc8)(97215) & (48cb6)(95127)

(4b86c)(92571) & (4c68b)(91752)

 

4,9 fixed:

(12836)(cb5a7) & (16382)(c7a5b)

(18623)(c57ba) & (18623)(c57ba)

 

5,8 fixed:

(1a9b6)(c3427) & (16b9a)(c7243)

(196ab)(c4732) & (1ba69)(c2374)

 

6,7 fixed:

(15b34)(c82a9) & (143b5)(c9a28)

(1b453)(c298a) & (1354b)(ca892)

 

A group of order 60 with all even permutations sounds suspiciously like A5, the alternating group of five elements, and sure enough, these groups are isomorphic. I leave it as an exercise to the reader how A5 can get mapped onto the symmetries of our 12-sided die. 

In the next post, I will have a Hasse diagram for this group that is both simplified and enhanced.

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.