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. 

Permutation matrices

 The symmetric group Sn for n > 0 is the set of permutations of n distinct elements. We have seen the cycle notation representation, but we can also use permutation matrices, which are matrices whose entries are all ones and zeros, with the rule that every row and every column has a single one in it, all the other entries equal to zero.

Lemma: The identity matrix for In for n x n matrices is a permutation matrix.

Proof: By definition, the identity matrix has ones on the main diagonal and zeros in every other position.

Here is a table showing how S3 can be written as 3x3 permutation matrices. For example the permutation (123) is represented by the matrix which moves column 1 of the identity to column 2, column 2 of the identity to column 3, and column 3 of the identity to column 1.

In some books, rows are moved instead columns, but this way, multiplying the matrices left to right gives the same result as combining the permutations left to right.

 

Cycle notation is more compact than permutation matrix representation, but multiplying 0-1 matrices is a very easy operation, especially since we know if we get a one in a row or column, we can immediately fill in zeros in the rest of both the row and column.

Next time, we will use the Cayley table in a special form that will let us turn any group of order n into a permutation group representation of nxn matrices.