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.