Group theory II

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tomtom2357
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Joined: Tue Jul 27, 2010 8:48 am UTC

Group theory II

Ok, I've given up the idea of classifying all groups of a given order, and I'm turning to something a bit more interesting. I am trying to go through groups by their presentation instead. So I started with groups with one generator, which gave me all the cyclic groups. Then I moved on to two generator groups. I started a case by case analysis, and I got to the group: <a,b|a2=b3=(ab)6=[a,b]2=1>. I am not sure how to calculate the isomorphism type of this group, because it is difficult to find whether two elements are the same. Are there any methods which can easily calculate the group from the presentation?

Edit: Here are the cases I have covered so far:
Spoiler:
Case 1: <a,b|[a,b]>: Z2, ZmZ, ZmZn. I had to cover the abelian case first.
Case 2: <a,b|a2>:
Case 2.1: <a,b|a2, b2>: Dihn, and the infinite dihedral group.
Case 2.2: <a,b|a2, b3>:
Case 2.2.1: <a,b|a2, b3, (ab)m>, m=3, 4, 5: A4, S4, A5
Case 2.2.2: <a,b|a2, b3, (ab)6>: Z2x|Z6, (ZmZnx|Z6) where m/n is a squarefree product of primes not congruent to 2(mod 3).
Case 2.2.3: <a,b|a2, b3, (ab)7>:
Case 2.2.3.1: <a,b|a2, b3, (ab)7, [a,b]m: m<=8: PSL(2,7), PSL(2,13), G1344, G10752
Case 2.2.3.2: <a,b|a2, b3, (ab)7, [a,b]9>: It all depends on the order of [a,b]4b (call it n). If n=1, you get PSL(2,8). If n=2, you get a group of order 64512, and one of order 32256. When n>2, but not divisible by 4, you just get a group of order 504n7. When n is divisible by 4, you get a group of order 1008n7, and one of order 504n7. If the order is infinite, then there is the whole group, and a quotient of index 2.
Last edited by tomtom2357 on Fri Aug 23, 2013 8:24 am UTC, edited 5 times in total.
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skeptical scientist
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Re: Group theory II

tomtom2357 wrote:Note: I already know about the undecidability of the word problem, and of the fact that there are uncountably many 2-generator groups, I'm just working through a few relatively simple cases.

As you point out, there are no general, effective methods to do this. One thing which may work in particular cases is proving some sort of normal form theorem. If you can come up with a set of "normal form" words where every word is equivalent to a unique normal form word, that can be a good way of understanding the group.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

tomtom2357
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Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

That could work, but I'm not even sure that the group is finite. I've made a start on it though, first I proved that every element can be expressed in a way that begins with a (for example ababa begins with a). Then I proved that (ab)3=(ab2)3, which really helped.

Edit: I just proved that the group has order 24. Also figured out that it is (isomorphic to) the wreath product of Z2 and Z3, or Z2A4. Now on to the group: <a,b|a2=b3=(ab)6=[a,b]3=1>.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

antonfire
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Re: Group theory II

When I have to do these things I like to draw the Cayley graph of the group. The Cayley graph of <a, b| a2 = b3 = 1> is a 3-regular tree with every vertex replaced by a triangle. (A truncated 3-regular tree.) Add in the relation (ab)6 = 1 and you get a truncated hexagonal grid. So now it's an extension by Z2 of a finite group, and you could work out exactly what extension. (In fact, it's a semidirect product Z2 x| Z6.) Any power of [a,b] is a shift in the grid, and, together with its conjugates, generates a finite index subgroup of Z2. You should find that adding the relation [a,b]n = 1 gives you a semidirect product of (Zn)2 and Z6, which is a group of order 6n2.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

tomtom2357
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Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Wow, that is cool. Are Z2x|Z6 and Zn2x|Z6 the only possible groups such that a2=b3=(ab)6?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

The group D(l,m,n)=<x,y | x^l,y^m,(xy)^n> is sometimes called a von Dick group or an ordinary triangle group.
http://en.wikipedia.org/wiki/Von_Dyck_g ... yck_groups

You can tell whether the group is finite or not by consider 1/l +1/m +1/n.

In the case l=2, m=3 (which, as you've noticed, is the first important case), we find that the group is finite when n is at most 5 (as you've probably found as well, by your question).

The case n=6 is a borderline case. As antonfire mentioned, you get Z^2 semi Z6 in this case. (It is a wallpaper group.)
I havn't checked carefully, but I would think that there are other groups that the ones you mentioned that appear in this case. (What you are really asking is about quotients of Z^2 semi Z6.)
(Note though that you will only obtain solvable groups of derived length at most 2.)

When you move to n=7, things get really interesting. Many (but not all) finite simple groups admit such a generating set for example. The keyword here is Hurwitz Group.
http://en.wikipedia.org/wiki/(2,3,7)_triangle_group

The point is this: even the cases (2,3,6) and (2,3,7) are quite hard and there is already a lot in the literature about them. In fact, finding whether certain groups admit a (2,3,7) presentation was an important research problem for some time.

tomtom2357
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Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

I thought the triangle group was: <a,b,c|a2=b2=c2=(ab)m=(ac)n=(bc)p=1>, which actually contains one more factor of 2. The rest is ok though. Could you try to find these extra quotients? I don't see how there could be any more quotients that still satisfy the conditions, but there might be a few more.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

(EDIT: I've made some changes to this post.)

Yes, what you describe is the triangle group.
The subgroup of index 2 is often called the "ordinary" triangle group.

(I'm actually just quoting Wikipedia here.)

As for the other quotients, I played around a bit in magma.

(You can too, at http://magma.maths.usyd.edu.au/calc/)

Here is my code:

Spoiler:
for n in [2.. 15] do
n;
G<a,b>:=Group<a,b|a^2,b^3,(a*b)^6, (a,b)^n>;

ID:=IdentifyGroup(G);
G:=SmallGroup(ID,ID);

Nor:= NormalSubgroups(G);

for N in Nor do
Q:=quo<G|N`subgroup>;
if not IsCyclic(Q) then
IdentifyGroup(Q);
end if;
end for;
end for;

For each n between 2 and 15, it builds the group G of order 6n^2.
It then finds all the quotients, and for the non-cyclic ones, it gives their ID in the small group library.
This is the output:
Spoiler:
2
<24, 13>
<12, 3>
3
<54, 5>
<18, 3>
<6, 1>
4
<96, 72>
<24, 13>
<12, 3>
5
<150, 6>
6
<216, 99>
<72, 44>
<54, 5>
<24, 13>
<18, 3>
<12, 3>
<6, 1>
7
<294, 14>
<42, 1>
<42, 1>
8
<384, 595>
<96, 72>
<24, 13>
<12, 3>
9
<486, 37>
<162, 15>
<54, 5>
<18, 3>
<6, 1>
10
<600, 183>
<150, 6>
<24, 13>
<12, 3>
11
<726, 6>
12
<864, 724>
<288, 405>
<216, 99>
<96, 72>
<72, 44>
<54, 5>
<24, 13>
<18, 3>
<12, 3>
<6, 1>
13
<1014, 10>
<78, 1>
<78, 1>
14
<1176, 254>
<168, 49>
<168, 49>
<294, 14>
<24, 13>
<42, 1>
<42, 1>
<12, 3>
15
<1350, 59>
<450, 24>
<150, 6>
<54, 5>
<18, 3>
<6, 1>

For example, for n=2, you get (24,13) and (12,3) (which is G itself and the alternating group of degree 4).

For n=3, you get
<54, 5>
<18, 3>
<6, 1>
which are the original group, Z3xZ3 semi Z6, S3 x Z3, and S3, respectively.
This means that all of these groups admit such a presentation...
Another interesting example is <42,1> for n=7. This is the Frobenius group Z7 semi Z6.

It looks like the number of quotients depends mostly on the prime factorisation of n. If n is a prime, you get few quotients. For example, for n=5 and n=11 then indeed the group is the unique non-cyclic one. If you take n=12 for example, you get many quotients. This might make a full classification a little difficult (but not necessarily impossible).

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Wow, that is a lot of groups. Ok, lets see if we can figure out a classification of all these groups. We can cut out the dihedral groups, because my previous <a,b,a2=b2=1> covers all those. Also, A4, S4, and A5 are covered as well. Now we need to create a list of what groups there are for each n. Since Zn2x|Z6 occurs at every step, I'll call it Gn (this may change if other groups appear a lot too), also, I won't include groups that have already been used, as there is no point, and also I am parameterizing by the order of [a,b] anyway, so it is not right to have groups with [a,b] having less order in that section anyway.

n=2: G2
n=3: G3, Z3Dih3 (I'm sorry, but I really don't like the term S3)
n=4: G4 (interesting, the only new group is the main group, I would have thought that n=4 would contribute a few more groups)
n=5: G5
n=6: G6, <72,44> (not sure what this is, is there a magma command that shows a presentation of the group given the small group library number?)
n=7: G7, Z7x|Z6 (why does magma include this group twice?)
n=8: G8 (interesting, powers of two only contribute the main group)
n=9: G9, <162,15> (interesting, multiples of 3 contribute the main group and some quotient of index 3)
n=10: G10 (at this point I'm starting to form a theory about what groups are possible)
n=11: G11
n=12: G12, <288,405> (ok, I have to know what this group is)
n=13: G13, Z13x|Z6 (at least, I think that is what <78,1> is)
n=14: G14, <168,49>
n=15: G15, <450,24>

Ok, I have a theory, which will have slightly more clarity when I find out exactly what the other groups are. The number of extra groups is 2f(n), where f(n) is the number of prime factors (not including multiplicities), not congruent to 2(mod 3). To go further, I need to know what those groups are. Ok, this might just work, it is not too complicated. Unfortunately, we still need to find what groups are possible if the order of [a,b] is infinite.
Last edited by tomtom2357 on Wed Aug 14, 2013 4:18 am UTC, edited 1 time in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

The reason a group is listed twice is because it appears twice as a quotient.
It's giving all quotients, not up to isomorphism.

If you just write
SmallGroup(72,44);
It will give you something called a polycyclic presentation.
By playing around a bit, it turns out that this group is A4 x S3.

Feel free to play around to figure out what some of the others are...

By the way, if you want to classify all quotients, you could try to classify all normal subgroups N of G_n=(ZnxZn) semi Z6.
I think if N is not contained in ZnxZn then one of the generators collapses so you get a cyclic group.
So you have to consider the normal closure of some element under Z6.
It boils down to some arithmetic I think...

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

I would like to know the presentations of the groups Z3Dih3 and A4Dih3 in terms of a and b. I am working on the isomorphism between the ab presentation (<a,b|a2=b3=(ab)6=[a,b]n) and the xyz presentation (<x,y,z|xn=yn=z6=1,xy=yx,zx=y-1z,zy=xyz>). I have figured out that [a,b] and [a,b2] commute, so we can take x=[a,b], and y=[a,b2].
Last edited by tomtom2357 on Wed Aug 14, 2013 6:00 am UTC, edited 1 time in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

There should be a way to answer most of your questions in an automated way, but my magma-fu is rather weak and I only have so much time on my hand Anyway, I worked out the first one by hand, only took a few minutes.
(See details in next post.)

<a,b|a^2,b^3,(ab)^6, [a,b]^3, abbabababb> is isomorphic to Z3xS3.

You can check this by hand, or run the following code:

n:=3;
G<a,b>:=Group<a,b|a^2,b^3,(a*b)^6, (a,b)^n, a*b^2*a*b*a*b*a*b^2>;
ID:=IdentifyGroup(G);
G:=SmallGroup(ID,ID);
IsIsomorphic(G,DirectProduct(CyclicGroup(3),SymmetricGroup(3)));
Last edited by Lul Thyme on Wed Aug 14, 2013 5:52 am UTC, edited 1 time in total.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

By the way, if you want to get your hands dirty, this is what I suggest.

Set
x=[a,b]=ab^2ab and
y=abab^2.

Then note that these are the generators of the group ZnxZn in Gn.
(x and y commute and have order n.)

In fact, Gn= (<x>x<y>) semi <ab>.
Let H=<x>x<y>

Now you only have to compute how a and b act on x and y by conjugation.
You find:
x^a=x^-1, y^a=y^-1, y^b=x^-1, x^b=x^-1y

Now, if you want to find all the normal subgroups of G contained in H you have to find the orbits of ab on H.

For example, if n=3, you have:
1. the identity,
2. {xy,x^2y^2}
3. The other 6 elements of H.

The groups generated by these orbits are:
1.The identity
2.<xy>
3. H
The respective quotients are:
1. H semi Z6=Gn
2. Z3 semi Z6=Z3 x S3
3.1 semi Z6=Z6.

The one you were interested in was 2, which came from quotienting by the group generated by xy. Thus we have to add xy=ab^2ababab^2 to the relations.

You could do something like this for general n, or try to make it more systematic.

tomtom2357
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Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Interesting, <a,b|a2=b3=(ab)6=[a,b]3=[a,b][a,b2]=1> is Dih3Z3. It all comes down to whether the equation x2-x+1=0 is satisfiable in the field Zp. It is the eigenvalue equation of the matrix transformation of conjugation by ab (think of xmyn as a vector <m,n>, and conjugation by ab acts as a matrix transformation on that vector). This equation is only satisfiable if p=3 or 1(mod 3). Now I know how to calculate the presentations of the extra groups. For example, the extra case for n=6 is <a,b|a2=b3=(ab)6=[a,b]6=[a,b]2[a,b2]2=1>=A4Dih3.

Edit: Also, there are no solutions to the equation in Z. Therefore, there are no non-trivial infinite quotient groups. That's quite nice to know, so we only have to worry about the finite case.

Edit 2: Groups like <a,b|a2=b3=(ab)6=[a,b][a,b2]=1> actually force the order of x, so they are no problem.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Now that we have found all groups with a2=b3=(ab)6=1, the classification is not too hard, all groups in this class are a direct product of two cyclic groups with a semidirect product by Z6. The two cyclic groups have a ratio equal to squarefree products of 3 and primes that are 1(mod 3). It is now time to move on to <a,b|a2=b3=(ab)7>. Apparently this is where is gets really interesting.

Wikipedia wrote:All 26 sporadic groups are quotients of triangle groups, of which 12 are Hurwitz groups (quotients of the (2,3,7) group).

Well that's interesting. Seems like this will be a very hard case to cover. Also, there is a link to a paper that seems to say that the monster group is a (2, 3, 7) group. That is a group of order 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000! This is insane!

Also, I entered the groups <a,b|a2=b3=(ab)7=[a,b]n> into magma for n ranging from 2 to 7, and I got some very interesting results. First, forcing the order of [a,b] to be 2, 3, or 5 collapses the group to the trivial group. Second, whether [a,b] has order 6 or 7, they are both the group <1092,25> (why is there a factor of 13 in this group order?, I want to know an element that has order 13). If the order of [a,b] is 4, then we get the group <168,42> which I strongly suspect is the simple group of order 168. If [a,b] has order 8, then it is a group of order 10752, which magma is unable to classify. The groups seem to be growing a lot faster than in the (ab)6=1 case. I suspect that if the order of [a,b] is high enough, the group will be infinite. In fact, if [a,b] has order 9, then magma is unable to identify the group, so it could even be infinite for n=9!

Edit: I was right! I checked on a different site, and <168,42> is indeed the simple group of order 168. Fortunately, that means that there are no quotients, as there are no normal subgroups other than the trivial group and the whole group. Also, the group of order 1092 is simple as well! That is interesting.

Edit 2: It occurs to me that we can still generate the groups with [a,b], [a,b2], and ab, so we shall still refer to them as x, y, and z respectively. They might come in handy, especially since we know that the order of [a,b] is the same as the order of [a,b2], and ab has order 7. I wonder what the presentation in terms of this is (what yx, zx, and zy are).

Edit 3: I strongly suspect that the group of order 10752 has only the group of order 168 as a quotient.
Last edited by tomtom2357 on Thu Aug 15, 2013 2:59 pm UTC, edited 1 time in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

It might be helpful to you to learn a bit about programming in magma if you want to play around with such things.

Here is a link to the handbook:
http://magma.maths.usyd.edu.au/magma/handbook/

A useful command to get a quick idea of the structure is CompositionFactors.
For example :
CompositionFactors(SmallGroup(168,42));
outputs:
G
| A(1, 7) = L(2, 7)
1

Which immediately that the group is the simple group PSL(2,7);
Similarly, you find that SmallGroup(1092,25) is PSL(2,13);

If we move to n=8, then the group is too big for the SmallGroup library.
The trick is then to translate the group to a permutation group, where there are lots more functions available than for finitely generated groups.
For example

G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7, (a,b)^8>;
Order(G);
P:=PermutationGroup(G);
CompositionFactors(P);

outputs:
10752
G
| A(1, 7) = L(2, 7)
*
| Cyclic(2)
*
| Cyclic(2)
*
| Cyclic(2)
*
| Cyclic(2)
*
| Cyclic(2)
*
| Cyclic(2)
1

So this is a group of order 2^6 with a PSL(2,7) on top.
In fact, I checked and the group of order 2^6 is elementary abelian, so its an extension Z2^6.PSL(2,7) (but it doesn't split as this code shows)
N:=NormalSubgroups(P)`subgroup;
HasComplement(P,N);

As you said, it looks like with n>=9 the groups are infinite.

(2,3,7) groups are known to be very wild and I'm afraid that your quest for a classification will end here. (Although there's no harm in playing around.)

tomtom2357
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Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

So, is the simple group of order 168 is the only quotient of the group of order 10752?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

You should be able to answer these questions yourself by now Try this for example:

G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7, (a,b)^8>;
P:=PermutationGroup(G);
NormalSubgroups(P);
N:=NormalSubgroups(P)`subgroup;
Q:=quo<P|N>;
IdentifyGroup(Q);
CompositionFactors(Q);

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Ok, so we have an extra group of order 1344. Well, that finishes the case [a,b]8=1. It seems that the next group is infinite, but that does not stop us from taking a quotient of that group. I'm not going to give up just when it is getting interesting. I just have to find an element that will allow me to take a quotient by that element and give a finite group. Unless, of course, the whole group is simple, which would be really weird. Is there an easy way to check for simplicity for infinite groups?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

No, it is not simple.
There is a command called LowIndexNormalSubgroups which allows you to find all the normal subgroups of a finitely presented subgroups up to a given index.

There are two things you have to be careful. It can sometimes get pretty slow and the output can be gigantic. This is where it might start to help to actually have access to magma rather than use the online calculator.
Still, in this case, we can manage.

Spoiler:
G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7, (a,b)^9>;
Nor:=LowIndexNormalSubgroups(G,25000);
#Nor;
for i in [1..#Nor] do
N:=Nor[i]`Group;
Q:=PermutationGroup(quo<G|N>);
IdentifyGroup(Q);
CompositionFactors(Q);
end for;

The program above shows that our group had only two normal subgroups of index at most 25000.
One is the group itself, of course, the other is a group of index 504. The quotient group is PSL(2,8).
This took 4 seconds for the online calculator, which is actually pretty quick for 25000. (This means that group has relatively "few" quotients.)
It starts slowing down pretty quickly as you increase the value of 25000.
(In fact, something funny seems to happen: for 32255, it still takes about 4 seconds and no new quotients, but for 32256 it runs out of time. It might be that there is a new quotient there that is creating problems. Havn't tried on other machines yet.)

Another thing you could try is just to add "random" words to the presentation, and see what happens...
(In a sense, this is what you've been doing all along. When <a,b|a^2,b^2> was infinite, you added (ab)^n, until you reached n=6, when it was still infinite, so you added [a,b]^m. Similarly, for n=7, you are adding [a,b]^m until you reached an infinite group at m=9. Time to add another word...)

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Lul Thyme wrote:No, it is not simple.
There is a command called LowIndexNormalSubgroups which allows you to find all the normal subgroups of a finitely presented subgroups up to a given index.

There are two things you have to be careful. It can sometimes get pretty slow and the output can be gigantic. This is where it might start to help to actually have access to magma rather than use the online calculator.
Still, in this case, we can manage.

Spoiler:
G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7, (a,b)^9>;
Nor:=LowIndexNormalSubgroups(G,25000);
#Nor;
for i in [1..#Nor] do
N:=Nor[i]`Group;
Q:=PermutationGroup(quo<G|N>);
IdentifyGroup(Q);
CompositionFactors(Q);
end for;

The program above shows that our group had only two normal subgroups of index at most 25000.
One is the group itself, of course, the other is a group of index 504. The quotient group is PSL(2,8).
This took 4 seconds for the online calculator, which is actually pretty quick for 25000. (This means that group has relatively "few" quotients.)
It starts slowing down pretty quickly as you increase the value of 25000.
(In fact, something funny seems to happen: for 32255, it still takes about 4 seconds and no new quotients, but for 32256 it runs out of time. It might be that there is a new quotient there that is creating problems. Haven't tried on other machines yet.)

Huh, that is interesting, it does the same on my computer. There might be something interesting for 32256. 32256 is 64x504, so we shouldn't be surprised. I wish I knew the presentation of PSL(2,8) in terms of a and b, it could be quite interesting, and maybe it could shed some light on the subgroup of index 32256 (I know there is a new quotient, because of your #Nor command. It said 3, instead of 2. I tried the code below to try to find the normal subgroup of order 64 of the group 32256 such that when you quotient out by it, you get the group of order 504. It returned an error, so could you check my code to see if it is code properly?

Spoiler:
G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7, (a,b)^9>;
Nor:=LowIndexNormalSubgroups(G,32256);
N:=Nor`Group;
LowIndexNormalSubgroups(N,504);

Lul Thyme wrote: Another thing you could try is just to add "random" words to the presentation, and see what happens...
(In a sense, this is what you've been doing all along. When <a,b|a^2,b^3> was infinite, you added (ab)^n, until you reached n=6, when it was still infinite, so you added [a,b]^m. Similarly, for n=7, you are adding [a,b]^m until you reached an infinite group at m=9. Time to add another word...)

This is basically what I have been doing. Unfortunately, I have no idea whether the group I generate from that is the same or different from the original group, because magma cannot process either of them. It hasn't returned any finite groups yet, so I was unable to tell. Also, that was exactly my strategy for defining these groups in the first place. After I got to <a,b|a2,b3> it seemed natural to add the order of ab. Then, after I got to <a,b|a2,b3,(ab)6>, I found the first word that I did not know the order of, which was [a,b], and fixed the order of that. Unfortunately, I am out of ideas about which word to fix the order of. Every word I have come across has finite order, and trying to lower that order results in the group becoming the trivial group.

I have read online that all hurwitz groups have order divisible by 84, that is interesting.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

G<a,b>:=Group<a,b|a^2,b^3,(a*b)^7, (a,b)^9>;
Nor:=LowIndexNormalSubgroups(G,32256);
N:=Nor`Group;
LowIndexNormalSubgroups(N,504);

Remember that N is a normal subgroup of G. It has index 32256, but it is still an infinite group.
LowIndexNormalSubgroups(N,504) is not really what you want to do, you dont want to find normal subgroup of N.

The group you are interested in is actually the quotient, so the first thing you should do is
Q:=quo<G|N>;
Now Q is a group of order 32256 but magma is having trouble dealing with it.
Asking Order(Q); returns the right answer but it takes a while on a decent machine.
On this machine, I was able to get it into a Permutation group (it has degree 72).
So H:=PermutationGroup(Q);
If your machine suceeds with this, you should be home free.
You can do
CompositionFactors(H); to see that it is indeed a group of order 64 at the bottom, with PSL(2,8) on top.
NormalSubgroups(H); shows that you have a unique non-trivial normal subgroup, of order 64.
A few more commands and you find that this group of order 64 is actually Z2^6, and that the extension doesn't split. (Very similar to a situation we encountered earlier).

By the way, it looks like the hard part of my previous code was when I try to transform the group into a permutation group. If you just want to know how many quotients there are, you can skip this and you should be able to go further than 32356. (Just as a trial, on my machine, I went to 50000 in a few seconds, and there are still only 3).

Finally, regarding your question about getting a presentation of PSL(2,8) in terms of a,b. There are a few ways to do this.
The "proper" way is to notice that the normal subgroup N gives you the extra relations. The way magma does it, there will be hundreds if not more relations though. There is a command to rewrite this in a more manageable way....

Anyway a less "proper" way to do things is to cheat a bit. Just start with PSL(2,8), take a random element of order 2, a random element of order 3, check the order of the product, etc... (you'll be suprised how quickly this works), then see what other relations they satisfy. Add those, see if you get the right group.

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Lul Thyme wrote:Finally, regarding your question about getting a presentation of PSL(2,8) in terms of a,b. There are a few ways to do this.
The "proper" way is to notice that the normal subgroup N gives you the extra relations. The way magma does it, there will be hundreds if not more relations though. There is a command to rewrite this in a more manageable way....

Anyway a less "proper" way to do things is to cheat a bit. Just start with PSL(2,8), take a random element of order 2, a random element of order 3, check the order of the product, etc... (you'll be surprised how quickly this works), then see what other relations they satisfy. Add those, see if you get the right group.

In a sense, maths is all about cheating, getting the result using sneaky tricks.

This is so confusing! PSL(2,8) is hard! First I have to scale everything down so that the first non-zero element is one, then I have to use these wacky addition and multiplication rules for F8. I've never done this before, so maybe I'll get better if I have more practice (and less sleep deprivation!). But then, give your "you'll be surprised at how quickly this works", I wonder if every pair of elements of order 2 and 3 work.

Edit: scratch that, I tried [1 2 2 1] (this is a 2x2 matrix) and [0 1 1 1], and they don't satisfy the relation (ab)7=1. Is there an easy general presentation of PSL(2,n) that I can use to figure this out?
Last edited by tomtom2357 on Sat Aug 17, 2013 2:44 pm UTC, edited 2 times in total.
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tomtom2357
Posts: 563
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Re: Group theory II

I found a very interesting relation for the group. ([a,b]4b)2=1. It is nice and short, no more relations are needed, and it could probably generalize to give us bigger groups. I suggest we use that relation in the future. The presentation for the group is now: <a,b|a2, b3, (ab)7, [a,b]9, ([a,b]4b)2>.

I looked at what happened when I change the order of [a,b]4b, and I found that it is the trivial group when the order is odd (I checked this up to 13), when it is 2, the group is the simple group of order 504, as stated above, but when the order is 4, it is a group of order 64512 (extension of PSL(2,8) by Z27), so it obviously contains the group of order 32256 as a subgroup. Also, I used the code you gave to find the number of normal subgroups, and there were only those four up to 500000 (it said it was only accurate up to that point).

I will now move on to the case <a,b|a2, b3, (ab)7, [a,b]9, ([a,b]4b)6>. It has order 1102248(=23397), so there might be a quotient of index 2. It is an extension of PSL(2,8) by Z37. I am starting to see a pattern here. If it holds for all orders of [a,b]4b, then we might just have solved the case <a,b|a2, b3, (ab)7, [a,b]9>! This seems just like the case <a,b|a2, b3, (ab)6>, except now it is seventh powers rather than second powers.

Edit 2: Huh, it says that the group has order 16515072, double what my prediction said it would have. The group <a,b|a2, b3, (ab)7, [a,b]9, ([a,b]4b)10> does have order 39375000 as predicted though. It might be possible that my prediction only works for orders that are double a prime. Magma was unfortunately unable to construct the permutation groups of those two though.

Edit 3: I found out (from mathoverflow) that my prediction was true! The groups <a,b|a2, b3, (ab)7, [a,b]9, ([a,b]4b)2p>, where p is prime, are an extension of PSL(2,8) by Zp7.
Last edited by tomtom2357 on Sun Aug 18, 2013 2:34 am UTC, edited 1 time in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

I see you've taken things to Mathoverflow.
(http://mathoverflow.net/questions/13966 ... itz-groups)
Not a bad idea, though be careful about the way you phrase things.
A big conversation like here is not really appropriate there.
Also, people like a bit of background around questions there.

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

I just wanted to find out whether the group <a,b|a2, b3, (ab)7, [a,b]9, ([a,b]4b)2p> was actually an extension of PSL(2,8) by Zp7, and whether there were any additional quotients of these groups that I need to include. Fortunately, there aren't any, so all we need to do is find what happens when p is not prime. I am asking what happens for the other groups on mathoverflow also. It appears that there is only one group when p is not divisible by 4, and two when p is divisible by 4. Now, what we need to know is: What happens when p is infinite? Are there any quotients other than the whole group?

Edit: It appears that someone has looked at the groups <a,b|a2, b3, (ab)7, [a,b]9> before. Unfortunately, it is not a free paper, so I can't get it.
Last edited by tomtom2357 on Mon Aug 19, 2013 1:07 am UTC, edited 1 time in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

What paper is this?

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

It is at http://journals.cambridge.org/action/di ... id=2056248. I'm not sure how much it covers though.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Thanks for the article! It appears that it actually looks at the finite groups, and it arrives at the same result I did: that when n is not a multiple of 4, then the group has order 504n7, and has no further quotients, but when n is a multiple of 4, the group has order 1008n7, and has a quotient of index 2!
Last edited by tomtom2357 on Mon Aug 19, 2013 6:46 am UTC, edited 1 time in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

In an infinite group?
Not that I know of.

It seems to me that even in a finitely presented group, there could be infinitely many normal subgroups of bounded order... (Don't quote me on this.)

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Since the paper covers all cases where n is finite, we must move on to the case where n is infinite. I'm not sure, but I think that the paper refers to one quotient group, which when translated into my a and b, has the extra relation ([a,b]6(ab)5)2=1. Is there a magma command for testing isomorphisms?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Lul Thyme
Posts: 68
Joined: Sun Apr 13, 2008 1:43 pm UTC

Re: Group theory II

In principle, it's IsIsomorphic(,), as in
IsIsomorphic(SymmetricGroup(3), DihedralGroup(3));

but I'm not sure if it works on infinite groups.

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Actually, it turns out that the group with that presentation is not infinite at all, but it is isomorphic to PSL(2,8). Well, back to square one. BTW, IsIsomorphic does not work on infinite groups. There is another way, SearchForIsomorphism(G,H,m), which searches for an isomorphism in which the images of the generators of G and H have length at most m. Unfortunately, it is not foolproof, as there might be an isomorphism such that the images have length greater than m.

Edit: There actually is a central element of index 2 in the group. If you really want to know what it is, I'll post it, but it is really long. This generates a quotient of index 2, and explains the extra quotients. I don't think that there are any others. Now we can move on to the case <a,b|a2, b3, (ab)7, [a,b]10>! This group is quite interesting, as all groups of this type have elements of order 2, 3, 5, and 7. The first group that I have found has order 34440, and unfortunately, magma was unable to make it a permutation group. However, I strongly suspect that it is the simple group of order 34440, and I am checking this now.

Edit2: Actually, since it is the lowest order of the quotient groups (other than the trivial group), then it must be a simple group. Therefore it is the group PSL(2,41). Now I need to figure out why there is a factor of 41 in the group, and what is the presentation. Also, there may be another small (ish) quotient of order 175560. Magma worked for every number up to that, and stopped working at 175560, also, I only asked for the normal subgroups, and the number of them, not anything fancy like the permutation group. If it exists, it is obviously simple, because 34440 doesn't divide into it, and therefore it is the Janko group J1. Interesting, I didn't think we would get to the sporadic groups so early on.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

DavCrav
Posts: 251
Joined: Tue Aug 12, 2008 3:04 pm UTC
Location: Oxford, UK

Re: Group theory II

Lul Thyme wrote:In an infinite group?
Not that I know of.

It seems to me that even in a finitely presented group, there could be infinitely many normal subgroups of bounded order... (Don't quote me on this.)

This is false. In any finitely generated group, there are only finitely many subgroups of a given index.

Edit: OK, so a big long list of stuff to talk about on this topic. I'll just give a few ideas.

1) The smallest Janko group J1 is a Hurwitz group, so is a quotient of the (2,3,7)-triangle group. On this page:
http://brauer.maths.qmul.ac.uk/Atlas/v3/spor/J1/
we see that standard generators of J1 are its Hurwitz generators. The order of [x,y] in standard generators is 19, but we can run code of the form:

x:=G.1; y:=G.2;
repeat g:=Random(G); until Order(x*(y^g)) eq 7 and (x,y^g)^10 eq G!1;

This quickly finds generators that make it a quotient of your group, so there is such a quotient.

2) It turns out that almost every simple group is a quotient of the group <a,b|a^2=b^3=1>. Exceptions are the Suzuki groups (obviously), Sp_4(2^n) and PSp_4(3^n), and a finite number of other groups, such as Alt(6). The finite list is incomplete, but it's highly unlikely to be completed in the way you are doing it! Hopefully I will have completed this list at some point over the next few months, along with other, similar questions.

Edit 2:

Tried to do your LowIndexNormalSubgroups thing:

> G:=Group<a,b|a^2,b^3,(a*b)^7,(a,b)^10>;
> time X:=LowIndexNormalSubgroups(G,180000);
Time: 168.850
#X;
4
> X`Index;

Magma: Internal error
Please mail this entire run [**WITH THE FOLLOWING LINES**]
to magma-bugs@maths.usyd.edu.au
Version: 2.19-3

You can print the whole input for this session by typing:
%P

Machine type: intel64-linux
Initial seed: 4223475622
Time to this point: 169.03
Memory usage: 184.53MB
Segmentation fault

> X`Index;
175560

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

Wait, so do you have magma (the real version not the online calculator)?

Also, for 1) Thanks for proving that the Janko group J1 is in fact a quotient of the group <a,b|a2, b3, (ab)7, (a,b)10>. Could you write the presentation of J1 (in terms of a and b) please (also, the presentation of PSL(2,41)?

for 2) I know that most simple groups are a quotient of the modular group, I am just trying to use certain parameterizations (such as the orders of ab and [a,b]), to make it easier, and to shed some light on how certain groups can be presented. For example, I now know that PSL(2,7) and PSL(2,8) are the only simple hurwitz groups such that the order of [a,b] is 9 or less. Now I want to know what groups are quotients of <a,b|a2, b3, (ab)7, (a,b)10>.

Edit: Actually, I found the presentation of PSL(2,41) on my own. Just add the relation ((a,b)4b)^5 and you get PSL(2,41). It is interesting, when the order of (a,b)4b is 1, 2, 3, 4, or 6, it is the trivial group, when the order is 5, it is PSL(2,41), and when it is 7, 8, or 9, it is an infinite group. The infinite groups do not contain either PSL(2,41) or J1 as quotients (or any other group of order at most 500000), so I suspect that they are simple. When the order is 11 however, the group does contain J1. It is interesting, (a,b)4b seems to be of significant importance in defining these groups. Or maybe, it is just that it is the simplest element that has infinite order in the group.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

DavCrav
Posts: 251
Joined: Tue Aug 12, 2008 3:04 pm UTC
Location: Oxford, UK

Re: Group theory II

tomtom2357 wrote:Wait, so do you have magma (the real version not the online calculator)?

Yes. I am a professional mathematician, so I have this sort of access, along with my own server to run this stuff on.

tomtom2357 wrote:Also, for 1) Thanks for proving that the Janko group J1 is in fact a quotient of the group <a,b|a2, b3, (ab)7, (a,b)10>. Could you write the presentation of J1 (in terms of a and b) please (also, the presentation of PSL(2,41)?

I had a go at this. I had earlier checked that ([x,y]^4y)^11 is not sufficient to make the group finite...

Magma wrote:> G:=Group<a,b|a^2,b^3,(a*b)^7,(a,b)^10>;
> time X:=LowIndexNormalSubgroups(G,180000);
Time: 172.890
> G1:=G/X`Group;

>> G1:=G/X`Group;
^
Runtime error in `: Field 'Group' does not exist in this record

Magma: Internal error
Please mail this entire run [**WITH THE FOLLOWING LINES**]
to magma-bugs@maths.usyd.edu.au
Version: 2.19-3

You can print the whole input for this session by typing:
%P

Machine type: intel64-linux
Initial seed: 239760821
Time to this point: 1754.62
Memory usage: 288.53MB
Segmentation fault
> X`Index;

Magma: Internal error
Please mail this entire run [**WITH THE FOLLOWING LINES**]
to magma-bugs@maths.usyd.edu.au
Version: 2.19-3

You can print the whole input for this session by typing:
%P

Machine type: intel64-linux
Initial seed: 239760821
Time to this point: 1754.63
Memory usage: 288.53MB
Segmentation fault
Magma: Fatal Error: Magma: Irrecoverable error; exiting

I think it's easier to work on a different problem, given how annoying Magma seems to be about this. How about proving that PSL(5,q) is a quotient of the modular group for q even?

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

DavCrav wrote:
tomtom2357 wrote:Wait, so do you have magma (the real version not the online calculator)?

Yes. I am a professional mathematician, so I have this sort of access, along with my own server to run this stuff on.

Cool. Can you check whether the group <a,b|a2, b3, (ab)7, [a,b]10, ([a,b]4b)7> is infinite? I just found out that when I ask for the order, and magma says 0, it does not mean infinite, it means it does not know. Therefore the online magma calculator is of no use to this problem.

Also, is there a way to check for simplicity in an infinite group? I could not find any normal subgroups of this group (magma could check up to index 500000), so I suspect that it is simple.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

DavCrav
Posts: 251
Joined: Tue Aug 12, 2008 3:04 pm UTC
Location: Oxford, UK

Re: Group theory II

tomtom2357 wrote:Cool. Can you check whether the group <a,b|a2, b3, (ab)7, [a,b]10, ([a,b]4b)7> is infinite? I just found out that when I ask for the order, and magma says 0, it does not mean infinite, it means it does not know. Therefore the online magma calculator is of no use to this problem.

Also, is there a way to check for simplicity in an infinite group? I could not find any normal subgroups of this group (magma could check up to index 500000), so I suspect that it is simple.

Magma says 0 if the group exceeds the ability of the Order function to check. The process uses an algorithm called coset enumeration. Increasing the memory is unlikely to help you here, and in particular Magma cannot prove that a group is infinite unless it has an infinite abelian quotient. Since this group is perfect, that isn't going to happen.

There is in general no way to check for simplicity of an infinite group. There is in general no way of checking the infinitude of an infinite group, so you're already in bad shape. 500000 is a pathetically small number though: since it's perfect, all quotients are going to be simple or bigger, and so 100 trillion might be a rough indication.

Very few examples of finitely presented, simple groups are known. It's difficult to write one down, and very difficult to prove it's simple. My guess would be 1) it isn't simple, and 2) if it is, it isn't provable.

Edit: You appear to be very motivated, but perhaps asking the wrong sort of questions. Do you have access to university-level algebraists you could ask for direction towards more interesting areas?

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

Re: Group theory II

I wonder if we could find a central element, like one that was found with the group <a,b|a2, b3, (ab)7, [a,b]9>. Is there any hope of that?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.