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.