[GAP Forum] groups of order p^n and exponent p
Siddhartha Sarkar
sidhu at mri.ernet.in
Fri Dec 3 10:09:44 GMT 2004
Dear Prof. Stefan Kohl,
The result you mentioned about the number of isomorphism types
of groups of order p^n, is probably by C.Sims London Math.
Soc.(3)15(1965) 151-166. Is there any similar result if one fixes the
exponent. Specially, is it known in case exponent p? Kindly reply.
Thanks to both of you and Prof. Bettina Eick for the response
of my earlier mail(s); it was really of help. Sorry for including a long
mail last time. It was a mistake.
regards,
Siddhartha Sarkar
> ------------------------------
>
> Message: 5
> Date: Wed, 24 Nov 2004 17:59:40 +0100
> From: Stefan Kohl <kohl at mathematik.uni-stuttgart.de>
> Subject: Re: [GAP Forum] commutator relations in p-group
> To: GAP Forum <forum at gap-system.org>
> Message-ID: <41A4BDFC.6060308 at mathematik.uni-stuttgart.de>
> Content-Type: text/plain; charset=us-ascii; format=flowed
>
> Dear Forum,
>
> Siddhartha Sarkar wrote:
>
> > I need to know the classification of p-groups satisfy the following
> > commutator relation:
> >
> > (1) say the group G is minimally generated by d elements
> > x_1, x_2,...,x_d.
> > (2) if d is even, the relation is [x_1, x_2]...[x_{d-1},x_d] = 1
> > (3) for d odd, the relation is [x_1, x_2]...[x_d,x] = 1
> > for some element x in G.
>
> Your condition is relatively weak --
>
> A pc-presentation of a p-group of order p^n requires n(n+1)/2 relations.
> Out of these you prescribe only one.
>
> Given that there are asymptotically p^((2/27)n^3) isomorphism types
> of groups of order p^n, it seems likely that there also is a huge
> number of p-groups which satisfy your relation, and that it does not
> make sense to attempt to classify them in general.
>
> Bettina Eick has already suggested you off-list to use NQ or ANUPQ --
> maybe these packages help you in investigating your problem.
>
> By the way: when replying to Forum digests, please make sure that
> you only cite the relevant part. Thanks!
>
> Best wishes,
>
> Stefan Kohl
>
More information about the Forum
mailing list