[GAP Forum] p-group

[Charles Wright]

Dear Derek, Vivek and friends,

I'm glad that Bettina and Derek have provided information about how to 
use GAP on problems such as this.

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^2), 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.

As ever,

Charley



Derek Holt wrote:
> Dear GAP Forum, Dear Vivek,
> 
> 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 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:=19;;
> 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);
> 
> Derek Holt.
> 
> 
> 
> 
> 
> 
> On Wed, Jan 27, 2010 at 08:06:38PM +0530, Vivek Jain wrote:
>> Dear Forum,
>>
>> I want to know that:
>>
>> "Is it possible using GAP to check that given presentation is 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?
>>
>>
>> with regards
>>
>> Vivek kumar jain
>>
>>
>>
>>
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