> < ^ Date: Mon, 07 Oct 2002 12:46:11 -0500
> < ^ From: Avital Oliver <olivera@macs.biu.ac.il >
> < ^ Subject: Re: Re: Self-normalized subgroups

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.


----- 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.


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