[GAP Forum] Outer Actions Giving Rise to Distinct Group Extensions

Jeffrey Rolland rollandj at uwm.edu
Tue Dec 15 20:49:29 GMT 2009


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Hello, all!

My dissertation involves taking a closed smooth manifold with finitely
presented fundamental group Q and creating a cobordant manifold having
fundamental group isomorphic to Q |x S ("Q semi-direct product S"),
where S is a given finitely presented superperfect group. I have created
a procedure for doing this, and am now at the point where I want to make
some examples of my theorem. My advisor, in particular, would like me to
"stack" these cobordisms, so the right-hand side of cobordism (W_i, M_i,
M)_{i+1}) becomes the left-hand side of cobordism (W_{i+1}, M_{i+1},
M)_{i+2}) (cobordism are read left-to-right).

Derek Holt kindly gave me an example of a superperfect group with
infinitely many outer automorphisms, none conjugate to each other. This
 example is S*S*S (free product) where S is SL(2,5) aka the Poincare
group. The outer action is having the last two factors act via
conjugation on the first factor and have the last two factors act
trivially on the last two factors. I have been trying to show that one
can make uncountably infinitely many of these infintely long "stacks" of
cobordisms (the stacks are technically called "pseudo-collars")
interleaving two of these outer actions, using matrices in SL(2,5) which
are not conjugate in GL(2,5); one gets one pseudo-collar for each point
in the unit interval, in fact.

Unfortunately, it is quite difficult to prove that the group extensions
one gets from using the different outer actions are not isomorphic.

I was wondering if there isn't a wiser way of going about this example,
using two distinct outer actions of Z, the integers, on S*S*S (or
possibly some other fintely presented superperfect group) so that it is
obvious, or at least simpler to prove, that the group extensions one
obtains at each step are not isomorphic.

[Note that given one group extension G_1 = Z |x S*S*S, one can get
another group extension by having the copy of Z in G_! act again on
S*S*S in the same way and have S*S*S act trivially on itself, producing
G_2 = (Z |x S*S*S) |x S*S*S, and so on.]

Thanks in advance for any assistance you can provide.

Sincerely,
- -- 
Jeffrey Rolland
<rollandj at uwm.edu>
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