# Classification of Groups of Order (p^l)(q^m)(r^n)

1. Mar 11, 2007

### Tom Mattson

Staff Emeritus
Hello,

I'm taking a seminar in group theory, and the main focus of the course is the classification of low-order groups (by classification, I mean finding all of the groups of a given order). I've got the hang of classification of (some) groups G that are products of powers of 2 distinct primes p,q (that is, the order of G is plqm). I'm having trouble when a power of a third prime r is thrown into the mix.

Can anyone point me to any online sources that treat this sort of thing? Specifically, it would be very helpful to find something for p=2, q=3, and r=5.

Thanks,

Tom

2. Mar 12, 2007

### matt grime

You're not going to get one in general, is my gut feeling. The point for p^aq^b, is that we know these are soluble (Burnside). This fails for the products of three primes (|A_5|=4.3.5).

In the case of 2.3.5, at least the group is soluble (any group of order 59 or less is). You can apply Sylow to show one of the p-subgroups is normal (can't remember which), then do something with the semi-direct products of the two smaller sized subgroups.

Of course things like Magma, have small subgroup databases - though I don't know how they are generated. And there huge numbers of groups of given sizes (there are 267 groups of order 64 for instance). Up to what do you wish to classify them? How explicit do you want to be and how much time do you have?

http://www.math.usf.edu/~eclark/algctlg/small_groups.html

might be useful (google for 'groups of order 30').

Last edited: Mar 12, 2007
3. Mar 12, 2007

### Tom Mattson

Staff Emeritus
I want to classify them up to isomorphism. In the seminar I am taking our objective is to classify all groups of order 64 or less.

I'll check it out, thanks.

4. Mar 12, 2007

### matt grime

I meant how explicit do you want to be in your description of them. Do you want to say that all groups of order 6 are semi-direct products of C_3 with C_2, and there are precisely 2 of them or do you want to write them out with generators and relations? As I said, there are 267 groups of order 64. That is not an easy thing to write down.

5. Mar 12, 2007

### Tom Mattson

Staff Emeritus
So far it's been very simple. We've not had to cite an isomorphism class that isn't one of the few groups with which we are already familiar ($\mathbb{Z}_n$, the permutations $S_n$, alternating group $A_n$, the dihedrals $D_n$, and the quaternions $Q_n$). At the moment I'm trying to find the isomorphism classes of groups of order 66, and I don't even know if I have to go beyond that list.

6. Mar 12, 2007

### matt grime

As I said, Magma, and presumably GAP have complete lists of 'small groups' should you need a reference. GAP is free. Magma has all groups up to order 2048 or something. I would imagine GAP can match it or get very close trying.

EDIT: I checked and it goes up to 2000 in GAP with the exception of groups of order 1024 (Magma also omits this order - I presume there just to many of them)

EDIT EDIT: Yes, there are over 460,000,000 million groups of order 1024.

EDIT EDIT EDIT: There are 4 groups of order 66. All polycyclic. They are

S_3 x C_11
D_22 x C_3
one of type 3:2+11:2 (no, I don't understand the lingo)
C_66

EDIT EDIT EDIT EDIT: That oddly described 3rd one is a group with 3 generators, x,y,z and relations

x^2=e, xy^2=yx, zx = xz^10, y^3=e zy=yz, z^11=e,

so it's some extension of C_33 by C_2 (y,z generate a copy of C_33 in there), conjugation by x sends y to y^-1 and z to z^-1.

Last edited: Mar 12, 2007
7. Mar 12, 2007

### matt grime

Here is a (now better since it omits order 1 and order p a prime) list of

Order Number of groups of that order
4 2
6 2
8 5
9 2
10 2
12 5
14 2
15 1
16 14
18 5
20 5
21 2
22 2
24 15
25 2
26 2
27 5
28 4
30 4
32 51
33 1
34 2
35 1
36 14
38 2
39 2
40 14
42 6
44 4
45 2
46 2
48 52
49 2
50 5
51 1
52 5
54 15
55 2
56 13
57 2
58 2
60 13
62 2
63 4
64 267
65 1
66 4
68 5
69 1
70 4
72 50
74 2
75 3
76 4
77 1
78 6
80 52
81 15
82 2
84 15
85 1
86 2
87 1
88 12
90 10
91 1
92 4
93 2
94 2
95 1
96 231
98 5
99 2
100 16

Last edited: Mar 12, 2007
8. Mar 12, 2007

### mathwonk

order 66, can you find 4 such groups? using Z/6, Z/11 and S(3), one can form 2 direct products and 2 semi direct products.

also every such group has (by sylow) a normal subgroup of order 11 and one of order 3. to show there are no other groups, it would suffice to show every such group has a subgroup of order 6. but probably since the subgroup of order 3 is normal, the element of order 2 should generate a subgroup of order 6 with it.

i think that does it for 66.