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7. Computing with GAP

7.10: "Is it possible using GAP to check that a given presentation defines a nilpotent group of class 2 or not?"
For example $G=\langle a,b,c| a^{p^5}, b^{p^3}, c^{p^2}, [a,b]=a^{p^3}, [a,c]=c^p, [b,c]=b^{p^2} \rangle $ where $p$ is a prime.
Also how can we determine its automorphism group using GAP?

Answer: The question was asked in this form in a GAP Forum letter on Jan. 27, 2010 and was answered in the Forum by three letters on Jan. 28. The answer given here joins the contents of these.

First please note that the 'example' really is not a presentation of a single group, but a family of presentations parametrized by the primes p. GAP has no methods for handling such a family of presentations in one step. However GAP has methods to investigate each such presentation for any given (not too big) prime p. In her reply Bettina Eick recommends:

you can use GAP to investigate your question for any fixed prime p.

For example, the nilpotent quotient algorithm of the NQ package or the NQL package of GAP allows you to determine the largest class-c quotient of a finitely presented group for any positive integer c or even the largest nilpotent quotient (if this exists).

Further, there are methods available in GAP to determine the automorphism group of a finite p-group. Check the AutPGrp package for this purpose.

In your given example, you can implement your considered group G in GAP as function in p:

G := function(p)
    local F, f, r, a, b, c;
    F := FreeGroup(3);
    f := GeneratorsOfGroup(F); a := f[1]; b := f[2]; c := f[3];
    r := [a^(p^5), b^(p^3), c^(p^2),
          Comm(a,b)/a^(p^3),
          Comm(a,c)/c^p,
          Comm(b,c)/b^(p^2) ];
    return F/r;
end;

Then you load the relevant packages

LoadPackage("nq");
LoadPackage("autpgrp");

And then you can do the following (for example for p=3):

gap> H := G(3);
[fp group on the generators [ f1, f2, f3 ]]
gap> K := NilpotentQuotient(H);
Pcp-group with orders [ 27, 9, 3, 9, 3, 3 ]
gap> Length(LowerCentralSeries(K));
3
gap> A := AutomorphismGroupPGroup(K);;
gap> A.size;
14348907

Hence for p=3 your group has class 2 and you can see the size of its automorphism group. Generators and further information on the automorphisms is also stored in A, but is perhaps too long to be displayed here.

In a second letter Derek Holt recommends:

You can use the GAP package KBMAG to prove nilpotency of finitely presented groups, using the method described by Charles Sims in his book of computing in finitely presented groups. This uses the Knuth-Bendix completion algorithm.

This process is described and illustrated in Example 4 (p. 13) of the KBMAG manual. I have successfully verifed that your group below is nilpotent of order p^10 for p=2,3,5,7,11,13,17, and I am trying to do 19.

Of course, since these groups are (apparently) finite, you could try use coset enumeration. This will work for small primes such as 2 and 3, but for larger primes the group order will probably be too large, and I think the Sims algorithm will work better.

You first run NilpotentQuotient (as described in Bettina Eick's reply) to find the maximal nilpotent quotient of your group. The aim is then to prove that the group is actually isomorphic to this quotient. You do this by introducing new generators in the presentation which correspond to the power-commutator generators in the maximal nilpotent quotient. You order the generators so that those at the bottom of the group come first and then use the so-called recursive ordering on strings to run Knuth-Bendix.

Here is the basic GAP code to do this.

LoadPackage("kbmag");
SetInfoLevel(InfoRWS,2);
F:=FreeGroup("j","i","h","g","f","e","d","c","b","a");;
j:=F.1;; i:=F.2;; h:=F.3;; g:=F.4;; f:=F.5;;
e:=F.6;; d:=F.7;; c:=F.8;; b:=F.9;; a:=F.10;;
p:=3;;
rels := [a^p/e, b^p/f, c^p/d, e^p/g, f^p/h, g^p/i, i^p/j,
        j^p, h^p, d^p, Comm(a,b)/i, Comm(a,c)/d, Comm(b,c)/h ];;
G := F/rels;;
R := KBMAGRewritingSystem(G);;
SetOrderingOfKBMAGRewritingSystem(R, "recursive");
MakeConfluent(R);

If successful it will halt with a confluent presentation containing the relations of the power-commutator presentation of the computed maximal nilpotent quotient. You have then proved that these relations hold in the group itself (not just in the nilptent quotient), so you have proved that the group is nilpotent. This consists of 65 reduction equations (or 62 when p=2).

The above works quickly for p=2,3,5,7. For larger primes, it helps to restrict the length of the stored reduction relations, and then re-run after completion. You have to experiment to find the optimal maximal length to store. So, for example, the following works fast for p=17:

p:=17;;
rels := [a^p/e, b^p/f, c^p/d, e^p/g, f^p/h, g^p/i, i^p/j,
        j^p, h^p, d^p, Comm(a,b)/i, Comm(a,c)/d, Comm(b,c)/h ];;
G := F/rels;;
R := KBMAGRewritingSystem(G);;
SetOrderingOfKBMAGRewritingSystem(R, "recursive");
O := OptionsRecordOfKBMAGRewritingSystem(R);
O.maxstoredlen := [40,40];
MakeConfluent(R);
Unbind(O.maxstoredlen);
MakeConfluent(R);

In a final letter Charles Wright gave an elegant proof 'by hand' that in fact all the groups for arbitrary primes are of nilpotency class 2 and order at most p^10, a proof, the clues for which can perhaps be obtained from the results of the GAP computations (which of course can be applied in similar form also to other presentations). He writes:

If all you want to show is that this particular example is class 2 of order (at most) p^10, though, it's easy to do it by hand. We're given that a^b = a^(1+p^3), c^a = c^(1-p) and b^c = b^(1+p^2). Hence, (a^(p^3))^b = (a^b)^(p^3) = a^((1+p^3)p^3) = a^(p^3) and c^(a^(p^3)) = c^((1-p)^(p^3)) = c [what a mess of superscripts!], so a^(p^3) (i.e., [a,b]) is in Z(G). Similarly, a^(b^(p^2)) = a^((1+p^3)^(p^2)) = a and (b^(p^2))^c = (b^c)^(p^2) = b^((1+p^2)p^2) = b^(p^2), so b^(p^2) (i.e., [b,c]) is central. Finally, (c^p)^a = c^((1-p)p) = c^p and b^(c^p) = b^((1+p^2)^p) = b, so c^p (i.e., [a,c]) is central. Thus G/Z(G) is abelian. Since G' has order (at most) p^4 and G/G' has order (at most) p^6, G has order (at most) p^10. In the spirit of Burnside, I'll leave the elimination of "(at most)" to the reader.