## Group Theory – Building abelian groups

I propose to find all the abelian groups up to order 100. It’s pretty easy, and there’s one nice idea that will simplify things. An abelian group, I hope you recall, is one that is commutative.

(The smallest non-abelian group is D3, the dihedral group of order 6… which is isomorphic to S3, the symmetric group on 3 symbols. All other dihedral groups are non-abelian, and the quaternion group of order 8 is non-abelian. But we’re going to look for abelian groups.)

There’s also one nice tool that will help. I have said that Mathematica® version 8 has “finite group data”. Well, it’s actually in version 7! I still think that the Cayley tables in the free Absract Algebra package are nicer, but I will be exploring the tools available in Mathematica proper. (Not in this post. You can look at Finite Group Data in the Mathematica help system.)

In particular, however, Mathematica knows how many abelian groups there are of order n. We’ll be in the position of filling in the details, as it were. To be specific, the command

tells us that there are 5 abelian groups of order 16.

(Yes, there is a related command for the total number of groups of order n…

We can infer these two answers that there are 14 – 5 = 9 nonabelian groups of order 16. (And I no longer have to rely on a table I got from a book.)

Let me review what we’ve already seen.

First, I claim without proof that every finite abelian group is the direct product of finite cyclic groups; and conversely, every direct product of finite cyclic groups is abelian.

Actually, it’s not difficult to show the converse. The point is that the direct product has components – it’s made up of ordered n-tuples – and within each component the group operation is commutative.

Anyway, that means that we can find all abelian groups of order n by finding all possible direct products of cyclic groups that result in order n.

Second, there exists a finite cyclic group of order n, for every positive integer n.

Admittedly, C1 is the trivial group, consisting of the identity alone. But for n greater than 2, we can construct Cn as the group of rotations in the plane of a regular n-gon. For n = 2, take {-1,1} with the usual multiplication; or {0,1} with addition mod 2.)

Third, if n is prime, then there is exactly one distinct cyclic group of order n, namely Cn.

Fourth… If n is not prime, say n = rs, and if r and s are relatively prime, then Cr x Cs ~ Crs. (Two numbers are relatively prime if their greatest common divisor is 1… and ~ denotes isomorphism.)

In particular, C2 x C3 ~ C6 and C3 x C4 ~ C12. (Yes, 3 and 4 are relatively prime.)

Look at that a different way. When we’re looking for abelian groups of order 12, we will want to write down both C12 and C3 x C4, amog other possibilities. We need to recognize that these two are the same group, otherwise our count will be high.

That result generalizes to any number of relatively prime factors. For example, C2 x C3 x C5 ~ C30.

With that, let me show you the idea that will help organize our investigation.

We’ve seen that there are two distinct abelian groups of order 4:

C4 and C2 x C2.

We know they are distinct because C4 has an element of order 4, while C2 x C2 (the Klein 4-group) does not.

What we did amounts to asking for partitions of 2. The first line gets me the partitions… the second line writes powers for an arbitrary prime p… and the third line specialized to the case p = 2:

That is, 2 = 2 + 0 and 2 = 1 + 1 are all the ways of writing positive integer exponents that add up to 2. From p^2 we get C4, and from {p, p} we get C2 x C2.

This generalizes to any prime. For p = 3 and p = 5 we get

which give rise to C9 and C3 x C3 of order 9, and C25 and C5 x C5 of order 25.

You’ll see below why I only care about powers of primes.

Let’s review the abelian groups up to order 10. Here’s how many there are. The key is the call to FiniteAbelianGroupCount. The first line contains orders, the second line contains the number of abelian groups:

We see “1” for order 1 and for each of the primes. The non-primes are 6, 8, 9, 10… of them, 6 and 10 are 2×3 and 2×5, so C6 ~ C2 x C3 is the unique abelian group of order 6; and C10 ~ C2 x C5 is the unique abelian group of order 10. We saw above that 9 = 3^2 has 2, namely C3 x C3 and C9.

Although we’ve seen in a previous post why order 8 = 2^3 has three distinct abelian groups, let’s use the partitions of 3 explicitly:

As before, the first line is the partitions of 3… the second line shows what the partitions give us for subscripts… the third line specializes to p = 2.

And, as before, the second line applies to any prime – so we should expect that for order 27 = 3^3 there will also be three abelian groups.

To finish this up, the last line tells us that the three abelian groups of order 8 are
C8
C4 x C2
C2 x C2 x C2.

OK?

Let’s get the next ten. To make it easy to see similarities, I will add to the list:

The primes are 11, 13, 17, 19 – each with 1. Then 14 = 2×7 and 15 = 3×5… each have 1. But 12 = 2^2 3, 18 = 2 3^2, and 20 = 2^2 5 are each of the form p^2 q… each with 2, namely Cp x Cp x Cq and Cp^2 x Cq.

You got that? If the order of a group is of the form p^2 q, with p and q prime, then the p^2 gives us 2 possibilites – but multiplying both of those possibilities by Cq still only leaves us with 2 possibilities.

So, the abelian groups of order 12 are
C2 x C2 x C3
C4 x C3

Then we want to recognize that C4 x C3 ~ C12.

Incidentally, we could also recognize that
C2 x C2 x C3 ~ C2 x C6…

We’ll come back to that… for now, I just want to be sure that we find Cn among the products of prime-order cyclic groups.

Similarly, the abelian groups of order 18 are

C2 x C3 x C3
C9 x C2 ~ C18.

Finally, the abelian groups of order 20 are
C2 x C2 x C5
C4 x C5 ~ C20.

That leaves 16 = 2^4. What are the partitions of 4?

That is, there are 5 partitions, and they give rise to 5 distinct abelian groups:
C16
C8 x C2
C4 x C4
C4 x C2 x C2
C2 x C2 x C2 X C2.

(The same thing will happen when p = 3… there will be 5 distinct abelian groups of order 27 = 3^3.)

We begin to see why I’m interested in powers of a prime. If we write 16 = 4^2 and ask for the partitions of 2, we only get C16 and C4 x C4. Instead, we need to write 16 = 2^4 and get the partitions of 4.

The next 10?

Everything with just 1 is either prime, or of the form pq, or of the form pqr (30 = 2*3*5). 25 is of the form p^2 and 28 = 2^2 7 is of the form p^2 q, each with 2 possibilities. We get

C5 x C5
C25

and

C2 x C2 x C7
C4 x C7 ~ C28.

27 = 3^3 is of the form p^3, hence 3 possibilities, just like 8 = 2^3. Here are the partitions of 3… the corresponding prime powers… specialized to p = 3:

That is, we get

C27
C9 x C3
C3 x C3 x C3.

Let’s look at 24 = 8*3 = 2^3 3.

It will have the same number of abelian groups as anything of the form p^3, namely 3. To be specific,

C8 x C3
C4 x C2 x C3
C2 x C2 x C2 X C3

and we recognize that

C8 x C3 ~ C24.

Through 40?

OK. By now you probably understand why the 1s occur. For the other entries… 32 = 2^5 is of the form p^5 and has 7 partitions:

Specialized to p = 2, we get

That is,
C32
C16 x C2
C8 x C 4
C8 x C2 x C2
C4 x C4 x C2
C4 x C2 x C2 x C2
C2 x C2 x C2 x C2 x C2.

It’s beginning to be very nice to let Mathematica compute the partitions, instead of doing it myself.

Next, 36 = 4 9 = 2^2 3^2 has 2×2 ways. one from column A, one from column B. This is the first time we have more than one entry in more than one column, so to speak. (In these terms, something of the form p^2 q has two entries in column A but only one in column B.)

To be specific, we use the partitions of 2 twice, once for p =2 and once for p = 3.

Then the 4 possibilities are
C4 x C9
C4 x C3 x C3
C2 x C2 x C9
C2 x C2 x C3 x C3.

Oh, and C4 x C9 ~ C36.

Finally, 40 = 2^3 5 is of the form p^3 q with p = 2, q = 5:
C2 x C2 x C2 x C5
C4 x C2 x C5
C8 x C5 ~ C40.

Thru 50?

44 = 2^2 11, 45 = 3^2 5, and 49 = 7^2 are all of the form p^2 q or p^2, and therefore have 2 distinct abelian groups.

48? It’s embarrassing, but it was faster to copy and execute the following code than to do this one in my head.

Ah, yes, that’s of the form p^4 q and has as many abelian groups as p^4, namely 5.

To be specific,

so the 5 possibilities are

C16 x C3 ~ C48
C8 x C2 x C3
C4 x C4 x C3
C4 x C2 x C2 x C3
C2 x C2 x C2 x C2 x C3.

Next?

52 = 2^2 13, of the form p^2 q, and 60…

is of the form p^2 q r. (Two from column A, but 1 each from columns B and C.)

54? This is 2*27 = 2*3^3, so it has the same number as any p^3, namely 3.

56?

is of the form p^3 q, hence has 3, just like p^3. Still, I find it convenient to let Mathematica display them, old hat though they be:

C8 x C7 ~ C56
C4 x C2 x C7
C2 x C2 x C2 x C7.

Though 70?

63 = 3 21 = 3^2 7, and 68 is…

Both are of the form p^2 q.

64 = 2^6. We need the partitions of 6:

Yup: 11 of them.

Specialize to p = 2…

C64
C32 x C2
C16 x C4
C16 x C4 x C4
C8 x C8
C8 x C4 x C2
C8 x C2 x C2 x C2
C4 x C4 x C4
C4 x C4 x C2 x C2
C4 x C2 x C2 x C2 x C2
C2 x C2 x C2 x C2 x C2 x C2.

Through 80:

75 = 3 25 = 3 5^2 is of the form p^2 q; and 76

is also of the form p^2 q.

72?

Ah, three choices from column A and two from column B. This should be interesting. For the 2^3 we get:

and for the 3^2 we get:

The 6 possibilities, then, are
C8 x C9 ~ C72
C8 x C3 x C3
C4 x C2 x C9
C4 x C2 x C3 x C3
C2 x C2 x C2 x C9
C2 x C2 x C2 x C3 x C3.

And 80?

That’s of the form p^4 q, hence the same number as p^4… which I’ll let Mathematica display for me, again.

So we get
C16 x C5 ~ C80
C8 x C2 x C5
C4 x C4 x C5
C4 x c2 x C2 x C5
C2 x C2 x C2 x C2 x C5.

We’re almost done; through 90:

81 = 3^4… has 5 just as 16 and 80 do.

84 = 4 21 = 2^2 3 7 and 90 = 9 10 = 2 3^2 5 are of the form p^2 q r, hence have 2.

88 = 8 11 is of the form p^3 q, hence has 3.

Oh, C90 ~ C2 x C9 x C5.

The last ten:

92?

OK, of the form p^2 q, hence 2 possibilities.

98?

Of course, 99 = 3^2 11. That’s why 92, 98, and 99 have two possibilities: they’re each of the form p^2 q.

And 100 = 2^2 5^2 has 2 x 2:

To be specific,

C4 x C25
C2 x C2 x C25
C4 x C5 x C5
C2 x C2 x C25.

And which one is C100? C4 x C25.

And 96?

OK, that’s the 7 different possibilities for p^5, like 32.

Let me try a larger example. What if we wanted two choices from each of columns A, B, and C? The smallest such group would be of order

2^2 3^2 5^2

Get the partitions of 2 in general…

but specialize to p = 2, 3, 5:

There are 8 possibilities. Right?

Right. We take one possibility from each “column”:

C4 x C9 x C25 ~ C900
C4 x C9 x C5 x C5
C4 x C3 x C3 x C25
C4 x C3 x C3 x C5 x C5
C2 x C2 x C9 x C25
C2 x C2 x C9 x C5 x C5
C2 x C2 x C3 x C3 x C25
C2 x C2 x C3 x C3 x C5 x C5

C12 ~ C3 x C4

we saw that

C2 x C2 x C3 ~ C2 x C6.

Is there an easier way to get other such relationships? That is, can we collapse the list of cyclic groups?

Yes, we can do it “by hand”, but there’s a simple tabular way to do it. Let’s start with p^2 q r… the smallest possibility is 2^2 3 5

One of the possibilities is 4 3 5, so I write C4 x C3 x C5. It’s isomorphic to C60 because 3,4,5 are pairwise relatively prime. (We could also write this as C12xC5 and C4 x C15, but I’ll sette for the maximal collapsing, one factor C60.)

For the other possibility, C2 x C2 x C3 x C5, I write

So I could write C30 x C2.

What the heck?

Focus on the formatted table. I have two factors of 2, so my first column has two 2s, and that dictates the number of rows. I have only one factor each of 3 and 5, so I pad out the columns with 1s. And I have 3 columns, because I have three distinct primes.

Once I have the formatted table, I multiply all the entries in each row.

Let me be clear. This array is specifically for C2 x C2 x C3 x C5. If I had been looking at C4 x C3 x C5, I would have only one row, three columns:

Let’s try another example… with 2 3^2 5, which is order…

The easy possibility – one row! – is C2 x C9 x C5 ~ C90. For the other, C2 x C3 x C3 x C9, the presence of two 3s says I need 2 rows, the three distinct primes says I need 3 columns:

That is, C2 x C3 x C3 x C5 is isomorphic to C30 x C3.

Incidentally, I believe this construction is finding what are called the invariant factors of the abelian group. That is, the invariant factors of C2 x C3 x C3 x C5 are 3 and 30.

My specification is a vague on one point. What if we have, say, C2 x C3 x C4? That can be written in two equivalent forms, as

C6 x C4
C2 x C12.

But are the invariant factors 4 and 6, or 2 and 12?

2 and 12.

All I’m sure of is that any invariant factor must divide the next larger one – and 4 does not divide 6, while 2 does divide 12.

Actually, let me try to illustrate the terminology. Recall the groups of order 900:

C4 x C9 x C25 ~ C900
C4 x C9 x C5 x C5
C4 x C3 x C3 x C25
C4 x C3 x C3 x C5 x C5
C2 x C2 x C9 x C25
C2 x C2 x C9 x C5 x C5
C2 x C2 x C3 x C3 x C25
C2 x C2 x C3 x C3 x C5 x C5.

For C900 ~ C4 x C9 x C25, the elementary divisors are 4, 9, 25. The invariant factor is 900.

For C4 x C9 x C5 x C5, the elementary divisors are 4, 5, 5, 9; the invariant factors are 180 and 5. Let’s use the table method: I need two rows because I have two 5s, and 3 columns because I have three underlying primes.

Returning to three equivalent versions of one abelian group of order 24…

C2 x C3 x C4. This displays the elementary divisors (powers of primes).

C2 x C12. This displays the invariant factors.

C6 x C4. This is just another equivalent form, distinguished by no name that I know of.

And I think that’s enough for now. We can figure out how many abelian groups there are of order n… we can write them out in terms of elementary divisors… or in terms of invariant factors.

Oh, I should tantalize you. Earlier, we have formed the direct products of cyclic groups with non-abelian groups: C2 x D3 ~ D6 but C2 x D4 is not isomorphic to D8. We could also take direct products of non-abelian groups. In both cases, the resulting products are non-abelian.

We could also form something called a semi-direct product, to get more non-abelian groups. It turns out, for example, that D3 is a semi-direct product of C2 and C3.

But determining the structure of non-abelian groups requires what are called the Sylow theorems – and they should be just the beginning. (If they were enough, the classification of finite simple groups would have been done by 1900.) Instead of just finding ways to combine smaller groups into bigger ones, we find ways to determine possible structures of a given order.