The Enhanced Hasse Diagrams for D_4 and the Quaterion group Q_8

The two non-abelian groups of order 8, D4 and Q8, do not need Simplified Enhanced Hasse diagrams. There are so few subgroups, simplification seems unnecessary. So instead, we get Enhanced Hasse Diagrams, which show the differences in their structures.

 

D4 has a total of ten subgroups, six of them are normal and the other four subnormal. There are two different copies of the Klein-4 group, {(1), (12)(34), (13)(24), (14)(23)} and {(1), (13), (24), (13)(24)}. They are isomorphic, but the second version has two odd permutations and two even, while the first version is all even permutations.

 

All subgroups, other than the Klein-4 subgroups and D4 itself,  are defined by a single generator.

The Q8 diagram much less cluttered, with only six subgroups total and all of them normal. In abelian groups, all subgroups are normal, but it is rare when this is true in a non-abelian group. Only Q8 itself is not define by a single generator.

 

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 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.

  

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.