> < ^ From:

> < ^ Subject:

Dear Marco,

Thank you. I didn't think of that. If I understand correctly, then it is

enough to find the subdirect products of all M and N such that M is

self-normalized in G and N is self-normalized in H.

Still, I do not see how that could give me a more efficient way of

calculating the set of self-normalized subgroups of G x H. I could not find

a way to calculate the set of subdirect products of two groups.

Thanks,

Avital.

----- Original Message -----

From: Marco Costantini <costanti@giove.mat.uniroma1.it>

To: "Multiple recipients of list" <GAP-Forum@dcs.st-and.ac.uk>

Sent: Tuesday, October 08, 2002 5:13 AM

Subject: Re: Self-normalized subgroups

Dear Avital and dear gap-forum,

Let G be H x K. Then the self-normalized subgroups of G should be the

subgroups S such that the projections of S onto H and K are

self-normalized in H and K respectively.

Let S be a self-normalized subgroup generated by h_1 x k_1, ...,

h_n x k_n. Then the projection of S onto H is the subgroup S_H of H

generated by h_1, ..., h_n and S_H self-normalized, this can be proved

using (h_i x k_1) ^ (h x k) = (h_i ^ h) x (k_i ^ k) .

The same for the analogous subgroup of K.

Greetings,

Marco

Miles-Receive-Header: reply

> < [top]