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 Enhanced Hasse diagram of the subgroup structure of the symmetries of the cube.

 This one took some work. Let me explain it.

 

The numbers along the left side give the order of the subgroups. There is one subgroup of order 1, nine subgroups of order 2, four subgroups of order 3, four subgroups of order 4, four subgroups of order 6, three subgroups of order 8, and only one subgroup of order 12 and one of order 24. 

 

Circles indicate normal subgroups. There are four, and their orders are 1, 4, 12 and 24.

 

Squares indicate subgroups that are not normal, and the ovals are subnormal.

 

Of the nine subgroups of order 2, three are generated by even permutations and six by odd permutations. The even permutation subgroups each connect to two subgroups of order 4. Each one is matched directly to one of the subnormal subgroups above it, while all are subgroups of the normal subgroup.

The six odd permutations generate subgroups of one of the order 8 subgroups and two of the order 6 subgroups. This creates a lot of traffic in the middle of the diagram, so I assigned a color of the rainbow to each of the six order 2 subgroups and all the arrows leading up to the higher levels from a single order 2 subgroup are the same color. For example, the first odd permutation order 2 subgroup is connected with red arrows to the second order 8 subgroup and the to the first and third order 6 subgroups.

 

None of these six subgroups are normal or even subnormal.

 

My next task is explaining the symmetries of the dodecahedron, a group that is isomorphic to the symmetries of the icosahedron. It is a group of order 60 and will be represented as a subgroup of S12. The subgroup structure is much more complex than the group we have been working on, and I'm not sure I'm up to making an enhanced Hasse diagram of it.

 

Time will tell.

  

The subgroups of the symmetries of the cube

I have made progress since the last post and here is what I have. I assume this is a complete list, but I cannot dismiss the possibility that I have missed something.

 

Order = 1: 

(1) even

There is always only one subgroup of order 1, and the identity is an even permutation, since it is made by zero transpositions. Easy peasy lemon squeezy.

 

Order = 2:

Every subgroup of order 2 has a single generator, In this group, there are nine subgroups of order 2.

(25)(34)      even

(16)(34)      even

(25)(16)      even

(16)(24)(35)  odd

(16)(23)(45) odd

(25)(13)(46) odd

(25)(14)(36) odd

(34)(12)(56) odd

(34)(15)(26) odd

 

 

Order = 3:

Every subgroup of order 3 has a single generator, In this group, there are four subgroups of order 3.

 

(123)(654) even

(135)(642) even

(154)(623) even

(142)(635) even

 

Order = 4:

There are four subgroups of order 4, three isomorphic to Z4 and one isomorphic to Z2xZ2. The last one is normal, the first three are not.

 

{(1), (2354), (25)(34), (2453)}

{(1), (1364), (16)(34), (1463)}

{(1), (1265), (16)(25), (1562)}

{(1), (25)(34), (16)(34), (25)(16)} The normal subgroup of order 4.

 

Order 6 groups:

There are four subgroups of order 6, all of them isomorphic to S3, none of them normal.

 

{(1), (123)(654), (132)(645), (34)(15)(26), (14)(25)(36), (16)(24)(35)}

{(1), (135)(642), (153)(624), (14)(25)(36), (12)(34)(56), (16)(23)(45)}

{(1), (154)(623), (145)(632), (12)(34)(56), (16)(24)(35), (13)(25)(46)}

{(1), (142)(635), (124)(653), (34)(15)(26), (13)(25)(46), (16)(23)(45)}

 

Order 8 groups:

There are three. Any subgroup of order 4 that has generators leaves two opposite faces fixed. For example, {(1), (1364), (16)(34), (1463)}, does not touch face #2 or face #5. Add in all four permutations that include (25). These groups are isomorphic to D4, and there are three of them, and none of them is normal.

_{{(1), (1364), (16)(34), (1463), (25)(34), (14)(25)(36), (25)(16), (25)(13)(46)}}

_{{(1), (2354), (25)(34), (2453), (16)(34), (25)(16),} (16)(24)(35), 

(16)(23)(45)_} 

{(1), (1265), (16)(25), (1562),(25)(34), (16)(34), (34)(12)(56), (34)(15)(26)}

 

Order 12 groups:

There is only one subgroup of order 12, which is all the even  permutations. It has to be normal because 24/12 = 2, and it is the only way to make a subgroup of order 12 out of the union of conjugacy classes.

Order 24 groups:

The entire group. Easy peasy yadda yadda. 

 

My next post will draw the enhanced Hasse diagram of the subgroup structure. I hope to have it finished by Sunday.

My slow progress on the subgroups of the symmetries of the cube

 My first inclination when given a math problem is to try to solve it myself, and it feels like defeat if I have to look it up. I have a lot of projects taking up time right now, and the group theory blog posts take considerable time.

Here is what I have so far when it comes to subgroups, listed by their order.

Order = 1: 

(1) even

There is always only one subgroup of order 1, and the identity is an even permutation, since it is made by zero transpositions. Easy peasy lemon squeezy.

 

Order = 2:

Every subgroup of order 2 has a single generator, In this group, there are nine subgroups of order 2.

(25)(34)      even

(16)(34)      even

(25)(16)      even

(16)(24)(35)  odd

(16)(23)(45) odd

(25)(13)(46) odd

(25)(14)(36) odd

(34)(12)(56) odd

(34)(15)(26) odd

 

 

Order = 3:

Every subgroup of order 3 has a single generator, In this group, there are four subgroups of order 3.

 

(123)(654) even

(135)(642) even

(154)(623) even

(142)(635) even

 

Order = 4:

As far as I have been able to determine, there are four subgroups of order 4, three isomorphic to Z4 and one isomorphic to Z2xZ2. The last one is normal, the first three are not.

 

{(1), (2354), (25)(34), (2453)} The generator is odd.

{(1), (1364), (16)(34), (1463)} The generator is odd.

{(1), (1265), (16)(25), (1562)} The generator is odd.

{(1), (25)(34), (16)(34), (25)(16)} The normal subgroup of order 4.

 

Order 6 groups:

Still hunting for these.

 

 

Order 8 groups:

So far I have found three. Consider the subgroups of order 4 that have generators. For example, {(1), (1364), (16)(34), (1463)}, does not touch face #2 or face #5. Find any permutation that includes (25) and do the group operation with every one of our four original elements. This group is isomorphic to D4, and there are three of them.

 

Order 12 groups:

So far, I have found one subgroup of order 12, which is all the even  permutations. There are two ways to see it is normal in the whole group, the "easy" way is that 12 = 24/2 and any subgroup that is half the order of the whole group must be normal, and the "not as easy" way is to see the subgroup is made up of three complete conjugacy classes.

Order 24 groups:

The entire group. Easy peasy yadda yadda. 

My next post will be filling in the blanks in the list.

More on the subgroups of the symmetries of the cube, and answers to the questions from the previous post.

 

Question 1.

Can you find five perfect squares such that 1²+a²+b²+c²+d²=24?

Answer: Yes. 1²+1²+2²+3²+3²=24

Question 2.

If yes to Question 1, does this mean there are exactly five conjugacy classes?

Answer: No, there might be some conjugacy classes that get split up.

 

Question 3.

Find the conjugacy classes of the symmetries of the cube, and find the k squares that sum to 24.

Answer: It takes some work, but in fact, all of our original conjugacy classes remain intact, and the sum is the answer to Question 1.

 

I am going to move slowly through this answer because there are a lot of things to take into consideration. I start with a Hasse diagram of the factors of 24. Note that the left column is 1, 1x2, 2x2 and 4x2, while the second column is 3, 3x2, 6x2 and 12x2. Any two circles connected by diagonal arrows can be stated as x in the left column and 3x in the right column.

When this gets expanded into a subgroup diagram, there will be normal subgroups of order 1, 4, 12 and 24. As for subnormal subgroups, if a subgroup of order n is contained in a subgroups of order 2n, the smaller subgroup must be subnormal in the larger group.

In the next post, I will list all the subgroups of every order.

Have a nice weekend.

The symmetries of the cube and the octahedron

 In the previous post, I stated that the cube and the octahedron are duals of one another. A cube has six faces and eight vertices, while an octahedron has eight faces and six vertices. Since six is less than eight, my natural laziness says it should be represented as a subset of S6 instead of S8, which means for a cube, we are permuting the faces, while for an octahedron, we are permuting the vertices. Same diff, the group will still have 24 elements.

 

Six-sided dice have a convention for the placement of the numbers represented, opposite sides have a sum of 7, meaning 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4. 

 

The 12 even permutations first.

 

The identity is always even.

(1)

 

Pick two opposite faces to be fixed, and turn the cube 180° on the axis that runs through the centers of the fixed faces. There are three pairs of opposite faces, so three elements here.

(25)(34), (16)(34), (25)(16)

 

 

Pick two opposite vertices to be the axis of rotation, and turn the cube 120° and 240°. There are four pairs of opposite vertices, so this accounts for eight elements.

 

Spin 1-2-3 and 4-5-6

(123)(654), (132)(645)

 

Spin 1-3-5 and 2-4-6

(135)(642), (153)(624)

 

Spin 1-5-4 and 3-2-6

(154)(623), (145)(632)

 

Spin 1-4-2 and 5-3-6

(142)(635), (142)(653) 

 

Composing two even permutations creates an even permutation since even+even = even. This means these 12 permutations are a subgroup because of closure under the group operation and the existence of the identity in the set. Because 12 = 24/2, This must be a normal subgroup of the the larger group.

The 12 odd permutations.

Pick two opposite faces to be fixed, and turn the cube 90° or 270° on the axis that runs through the centers of the two fixed faces. There are three pairs of opposite faces, so this accounts for six permutations.

(2354), (2453), (1364), (1463), (2156), (2651)

 

Swap two opposite faces, and turn the cube 90° or 270° on the axis that runs through the centers of the swapped faces. There are three pairs of opposite faces, so this accounts for six permutations.

(16)(24)(35)

(16)(23)(45)

(25)(13)(46)

(25)(14)(36)

(34)(12)(56)

(34)(15)(26)

 

If we add diagonals to all six faces with the proviso that every vertex is either the endpoint of three diagonals or the endpoint of none, we get a regular tetrahedron embedded in our cube. Even permutations will permute the pink tetrahedron, while odd permutations will send the pink tetrahedron's vertices to the vertices that have zero diagonals in the original position.

 

This means the symmetries of the tetrahedron is a normal subgroup of the symmetries of the cube, and by duality, a normal subgroup of the symmetries of the octahedron.

 

Symmetries within symmetries. The beauty of group theory.

 

We learned about the group representation trick that tells us if we have k conjugacy classes, we can find k perfect squares that add up to the order of the group, with the proviso that at least one of the squares is 1², which is matched with the identity, always standing alone as a singleton conjugacy class. We have at least five conjugacy classes, based on the five different permutation shapes listed above.

Question 1.

Can you find five perfect squares such that 1²+a²+b²+c²+d²=24?

Question 2.

If yes to Question 1, does this mean there are exactly five conjugacy classes?

 

Question 3.

Find the conjugacy classes of the symmetries of the cube, and find the k squares that sum to 24.

 

Answers on Friday, when I will post more information about the structure of this group.