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.