Two versions of modulo arithmetic: On the clock and on the complex unit circle.

 

We are used to the clock face, though I wonder for how much longer with so many people using their phones as time pieces. We have the numbers 1 through 12, with 12 at the "high noon" position, and also 60 hash marks for counting both minutes and seconds. The idea of clockwise movement comes from Northern Hemisphere sundials, because a Southern Hemisphere sundial travels counter-clockwise, the 12 being pointed due south instead of due north.

 

60 minutes is a remnant of the base 60 number systems used first in ancient Sumeria, then Babylon and passed down to us in how we cut up circles into 360 degrees.

 

  

Counting on the complex unit circle starts at 1, rightmost point on the real axis. Multiplying complex numbers has the physical representation of multiply the lengths and adding the angles.

 

On the complex unit circle, all lengths are 1, so it is just about adding the angles, where we count the positive x-axis as 0. The unit circle means the radius is 1, the diameter is 2 and the circumference is 2𝝅. 

 

I have labeled four of the points as 1, -1, i and -i, as they lie exactly on the real axis or the imaginary axis. The other eight points are given in the form e^(ni𝝅/6), where n is a number between 1 and 11, and if n = 2k, the form is e^(ki𝝅/3). When n = 12, we get e^(2i𝝅), which means we have gone full circle, and can identify the point with e^0 = 1.

 

Most people learn degrees long before they learn radians, so degrees feel more "natural" to us, but 360 degrees was an arbitrary decision based on how many divisors 360 has.

 

Divisors of 360: 1 & 360, 2 & 180, 3 & 120, 4 & 90, 5 & 72, 6 & 60, 8 & 45, 

9 & 40, 10 & 36, 12 & 30, 15 & 24, 18 & 20 

 

Let's use the definition of natural that means "based on nature". While it is a human concept, 𝝅 = circumference/diameter, and those two measurements are based on the nature of circles, so I would argue that radians are more natural than degrees. While perfect circles are hard to find in nature, if you pour a little oil into a still pan of water, the oil droplets will do their best to form circles, because a circle is the two-dimensional shape maximizes surface area and minimizes the perimeter. The surface area will be 𝝅r² and the perimeter 2𝝅r. No other shape will give us a better ratio.

 

Whether we start at the top and move clockwise around the circle starting at 12 or start at the right and move counterclockwise starting at 0, these are both representations of ℤ12, the most commonly used cyclic group around the world.

The complex unit circle, a group under mutiplication (corrected on 22 May)

 

Most of my posts so far have been dealing with finite groups, and in the last month, I have been concentrating on non-abelian groups. When I first learned the topic, I was fascinated that a math structure could be so easily defined, but that we could have a system where ab ≠ ba in every case. Today, we will learn about an infinite group, all numbers of the form a+bi, where sqrt(a² + b²) = 1 and  i² = -1. When we add real and imaginary numbers, we call the result complex numbers, and we represent them on a plane where the x-axis is purely real and the y-axis is purely imaginary. They are still complex, since 0+bi and a+0i still fit the form a+bi. 

 

The set we have defined is known as the complex unit circle. In the illustration above, let r = 1, which means the diameter d = 2 and the circumference = 2π. Because this is the unit circle, radians make more sense than angles, so we will write these points in the form cos(𝜃) + isin(𝜃). Our group operation will be multiplication, and complex multiplication has the lovely property that the product will be multiplying the lengths together and adding the angles. Since all lengths are 1, 

[cos(𝜃) + isin(𝜃)]*[cos(𝜶) + isin(𝜶)] = cos(𝜃+𝜶) + isin(𝜃+𝜶) and

[cos(𝜃) + isin(𝜃)]/[cos(𝜶) + isin(𝜶)] = cos(𝜃-𝜶) + isin(𝜃-𝜶).

 

When we add or subtract angles, the results could be greater than 2π or less than 0, but this is not a problem. If we divide by 2π and take the remainder, we will get a number 𝛽 such that 0 ≤ 𝛽 < 2π. 

 

While we have defined the group as a multiplicative group, we see that this multiplication is mapped onto the addition of angles, and additive groups are always abelian. More than this, any finite cyclic group is a subgroup of the complex unit circle. We will discuss this in greater detail tomorrow.

I came back to this post to correct it. The complex unit circle does wrap around on itself but it is NOT cyclic because it does not have a single generator. If some number a+bi generates an infinite subgroup of the complex unit circle, it can only be dense on the circle, and an infinite number of points will be missing. This is analogous to the rationals being dense among the reals.

 

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.

 

The Simplified Hasse diagram of A_5

 A5 is the alternating group on 5 elements. Since the order of S5 is 120, the order of A5 is half the size at 60. Here is the Simplified Hasse Diagram of the structure of the subgroups.

The red circles indicate the normal subgroups, which in this case are just the whole group and the identity. having just two normal subgroups means this is a simple group.

 

Red arrows can only originate in normal subgroups, and in this case, the identity is a normal subgroup of every subgroup of prime order.

 

There are 15 subgroups of order 2, all of the generators of the form (ab)(cd). 

 

There are 10 subgroups of order 3, all of generators of the form (abc).

The 6 subgroups of order 5 have generators of the form (abcde).

For the Klein-4 subgroups, we have the identity and three double transpositions that all keep the same element fixed. For example, (12)(34), (13)(24) and (14)(23) all leave 5 untouched. Likewise, the five copies of A4 all leave one element fixed.

The subgroups of order 5, 6 and 10 do not leave any element fixed, so they are not subgroups of A4. Of those three orders, only 6 divides 12, but the groups isomorphic to S3 are generated by a 3-cycle (abc) and a double transposition (ab)(de), so no element is fixed.

 

Every purple arrow goes from a subgroup of order n to a subgroup of order 2n. These subgroups are not normal, which is indicated by the blue rectangles, nor are they subnormal, since subnormality only occurs if a subgroup is part of a chain of subgroups, each normal in the next subgroup up in the chain and terminating in the entire group. A simple group that has any proper subgroup that is not the identity cannot contain a subnormal subgroup, so these subgroups that have purple arrows coming out of them are normalish, the phrase I am using until I learn the proper term.

 

The sum of all the numbers in red circles and blue rectangles is 59, the total number of subgroups.

 

The next Simplified Hasse diagram will be of S5, which will be a much larger undertaking with 156 total subgroups.

 

 

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 Simplified Enhanced Hasse diagram for the symmetries of the dodecahedron, isomorphic to the symmetries of the icosahedron and to A_5

 A week ago, I published a Hasse diagram for the subgroups of the symmetries of the cube, which is also the symmetries of the octahedron and is isomorphic to S4. There were two problems with my diagram.

1) It was hard to read.

 

2) There are actually 30 subgroups of this group instead of 27.

Is my face red? Yes, it is. Gotta fix that diagram.

In the group we deal with today, the order is 60 and there are 59 subgroups. Any regular Hasse diagram will be a mess. I hit upon a new way to simplify such a diagram, and I present it here.

How to read this diagram

The numbers on the left are the orders of the subgroups. 

 

The numbers in the squares or circles represent how many copies of that order exist that are isomorphic to each other. 

 

Examples: At order 60, there is only 1 subgroup, and that is the group itself.

 

At order 12, there are 5 isomorphic copies, at order 10, there are 6 isomorphic copies, etc.

A red circle indicates the subgroup is normal in the group. 

 

A blue square means the group isn't normal.

Red arrows can only start from red circles, indicating a subgroup that is normal in the entire group must be normal in any supergroup to which it belongs.

A blue arrow shows that a subgroup is not normal in the next supergroup above it.

A purple arrow indicates a subgroup is not normal in the entire group, but is normal in some supergroup above it, but not necessarily in every supergroup above it. This is not the definition of subnormal, so there should be a different word for it. I do not yet know that word, so I have coined normalish to describe this condition.

Take a look at level 3. Any subgroup of order 3 will be normal in a supergroup of order 6, hence the purple arrow, but in this case, the subgroups of order 3 are not normal in the subgroup of order 12, hence the blue arrow. Likewise, the subgroups of order 2 must be normal if they are contained in a supergroup of order 4, but in this group, they are not normal when contained supergroups of order 10 or 12.

Notice that there are divisors of 60 that are not orders of subgroups of this particular group. We have seen that the order of a subgroup H of a finite group G must divide the order of G. This is known as LaGrange's Theorem. The converse would be that if a divides b and b is the order of G, there must be a subgroup of order a in G. This example shows us the converse is not true.

 

Usually, when something is simplified, we have lost information, and that is true here. If each of the 59 subgroups got its own node in this graph, we could see exactly which subgroups of order 2 are contained in each of the subgroups of order 10. There are six subgroups of order 10, and each would have five blue arrows coming up from five of the fifteen order 2 subgroups. The diagram would turn into a tangled mass of multi-colored pasta, and while more information would be shown, it would be a challenge to decipher it.

 

This next week is going to be cleaning up previous diagrams, and seeing if someone more knowledgeable than myself knows the real word that is used where I am using normalish. It is a common occurrence in well-known groups of small order, someone must have noticed it before I did.