> < ^ Date: Wed, 09 Oct 2002 09:31:25 +1000
^ From: L G Kovacs <kovacs@maths.anu.edu.au >
> < ^ Subject: Re: Self-normalized subgroups

Dear Gap-Forum,

To describe the subgroups K of a direct product G x H, one needs
to know not only the subgroups U of G and the subgroups X of H
(so one can form all possible K = U x X ), but also all
isomorphisms f : U/V -> X/Y between (nontrivial) sections of G
and sections of H (so one can form all possible K = { (u,x) |
f((uV) = xY } ). [This result goes back to Goursat, but was
rediscovered by Remak, and may be found in many texts.]

How can one tell whether the K so defined is self-normalizing? If
K = U x X, the test is simply that U and X are to be
self-normalizing (in G and H, respectively). When K is formed
from an f, the answer is more difficult. Part of it is that U/V
should have trivial centralizer (I mean, if an element g of G is
such that all commutators [u,g] lie in V then g must lie in V:
in particular, U/V must have trivial centre), and of course the
same must hold for X/Y. The other part is much harder to check: f
must not be the restriction of any isomorphism
f* : U*/V -> X*/Y such that U is a proper normal subgroup of U*.
The necessity of these conditions is obvious, and it is not hard to
show that together they are also sufficient, but I have not seen them
stated anywhere in the literature.

Unfortunately, this theory only underlines the difficulty of finding
a good algorithm.

Laci Kovacs

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