Date: Wed, 7 Aug 1996 10:14:46 -0500
From: Hyman Bass
Subject: Jenkinson problem
Dear Oliver,
If I understand your question correctly, I think that the answer is "No."
Here is what seems to be the simplest kind of counterexample.
Let G be any group. Let u, v:G ---> G be maps such that u(v(a)) = a for all
a in G. In G x G consider the sets (graphs of u and v),
H = { (a, u(a) ) | a in G }
and
H' = { (v(b), b ) | b in G }.
The relation,
(*) u o v = Id
is equivalent to the inclusion of H in H'. This does not entail H = H' unless
(*) implies that
(*') v o u = Id.
This would be the case if, for example, u and v were homomorphisms, and G
satisfied some kind of finiteness (e.g. noetherian) conditions. Otherwise,
there are easy examples showing that (*) does not imply (*'), even for G =
Z,
when we can take u(a) = 2a, and v(b) = [b/2].
[You claimed in your message that the w_i in your notation would have to be
group theoretic "G-words" in the free variables, and so H and H' would have
to be subgroups when G is abelian; if true, this would be an *assumption*.
It does not follow from your first description, as the previous example
above shows.]
Best wishes,
Hy Bass
--
Subject: Re: Jenkinson problem
To: Hyman Bass
Date: Thu, 8 Aug 1996 13:46:34 +0100 (MET DST)
Cc: group-pub-forum@maths.bath.ac.uk
Dear Hy and Geoff,
Apologies for the ambiguity - yes, I did mean that the non-freely varying
coordinates should be 'G-words' in the freely varying ones, rather than
just having _some_ functional dependence.
I wanted them to be generalisations of linear (rather than non-linear)
vector subspaces.
Hy wrote:
>
> Let G be any group. Let u, v:G ---> G be maps such that u(v(a)) = a for all
> a in G. In G x G consider the sets (graphs of u and v),
>
> H = { (a, u(a) ) | a in G }
> and
> H' = { (v(b), b ) | b in G }.
> The relation,
>
> (*) u o v = Id
>
> is equivalent to the inclusion of H in H'. This does not entail H = H' unless
> (*) implies that
>
> (*') v o u = Id.
>
> This would be the case if, for example, u and v were homomorphisms, and G
> satisfied some kind of finiteness (e.g. noetherian) conditions.
Okay, thankyou.
Giovanni Cutolo emailed me directly with an explanation along similar lines.
He showed me that it works if G is locally (residually finite). With his
permission I'll forward his message to GPF.
Am I right in thinking that Noetherian implies residually finite?
I'd still be interested if anyone has any examples of groups which are _not_
locally (residually finite) but for which the property still holds.
Also I'd be interested if anyone knows a class of groups for which the
property does _not_ hold.
Cheers,
Oliver
----
Date: Wed, 7 Aug 1996 11:57:48 -0600 (CST)
From: Giovanni Cutolo
To: Oliver Jenkinson
Subject: Re: An elementary problem?
Dear Oliver,
I have only a partial answer to your query, but hope it may help.
The arguments below give a list of group classes in which the property you
are interested in holds.
Firstly, your definition of `m-parameter subsets' can be slightly
generalised as follows, just to avoid the lack of symmetry between the
sets that you have called H and H'.
Fix a free group F of rank m, with basis x_1, x_2, ... , x_m. Let
w=(w_1,...,w_n) an n-tuple of elements (words) of F, such that
w_1,...,w_n generate F (thus n> or = m).
Then the `m-parameter subset' defined by w relative to a group G,
let us call it S(w,G),
is the image of the mapping w* from G^m to G^n defined by
w*(g)=(w_1(g),...,w_n(g)) for all g in G^m.
Let me fix some further notation.
Since the w_i's generate F, there are words w^_1, ..., w^_m on n symbols
such that w^_i(w)=x_i for all i in {1,..., m} (the elements x_i form the
fixed basis of F). Call w^ is the mapping from G^n to G^m defined
by w^(h)=(w^_1(h),...,w^_n(h)) for all h in G^n. Of course w*w^ is the
identity map of G^n (in particular, this shows that w* is injective).
Take now another n-tuple v=(v_1,..,v_n) of generators of F. Define
S(v,G), v*, v^ similarly as for w.
It is straightforward to see that S(w,G) is contained in S(v,G) if and
only if w*=w*v^v*. (Indeed, given any g in G^m, if w*(g)=v*(h) for some h
in G^m then h=v*v^(h)=v^(w*(g)); hence w*(g) belongs to S(v,G) if and only
if v*(v^(w*(g)))=w*(g). )
Thus, fixed w and v as above, the class of groups G such that S(w,G) is
contained in S(v,G) is defined by a set of equations between values of
words, hence it is a variety, let me call it V(w,v).
Let X(w,v) be the class of groups G such that S(w,G) is contained in
S(v,G) if and only if the reverse incusion holds (the groups you are
interested in form the intersection of all these X(w,v) ).
As you remark, all finite groups belong to X(w,v) (this still holds with
my definitions, as w* and v* are injective). Moreover every group G which
is either:
- locally in X(w,v); or
- residually in X(w,v)
still belongs to X(w,v). (If you are not a group theorist: locally-X means
that every finite subset of G is contained in a subgroup which satisfies
X, residually-X means that the intersection of all normal subgroups N of G
such that G/N satisfies X is the identity subgroup.) This is immediate
form the fact that V(w,v) and V(v,w) are varieties.
Thus, for instance, all groups which are locally(resdually finite) are in
X(w,v) for any choice of w and v, and so have the property you are
interested in. Some examples of groups which are locally (residually
finite) are:
- all abelian groups,
- more generally: all metabelian groups,
- all locally (polycyclic-by-finite) groups,
- all groups of matrices over finitely generated integral domains,
- all free groups,
but many more relevant classes of groups are locally (residually finite).
So, your property holds for all those groups. What happens in general is,
in my feeling, a much tougher question. If I happen to think something
about that I'll let you know. Please feel free of dropping a note to me if
some more references are needed.
Best regards,
Giovanni Cutolo
********************************************************************
Giovanni Cutolo
Dipartimento di Matematica e Applicazioni `R. Caccioppoli'
Universita` degli Studi di Napoli `Federico II'
Compl. Universitario Monte S. Angelo - Via Cintia
I-80126, Napoli - ITALY
tel. +39 81 675695 Fax: +39 81 7662106
e-mail: cutolo@matna2.dma.unina.it cutolo@ds.cised.unina.it
cutolo@matna1.dma.unina.it
********************************************************************
---------------------
From: Jim Howie
Date: Thu, 8 Aug 1996 14:08:35 +0100
To: omj@maths.warwick.ac.uk
Subject: Re: Jenkinson problem
Cc: group-pub-forum@maths.bath.ac.uk
X-Sun-Charset: US-ASCII
Oliver,
I have been following the correspondence on your problem with interest.
I suspect that the answer is till "no", even if your funcions are given
by G-words. I do not have an actual proof, though.
My proposed way of constructing a counterexample would go something
like this. Let U, V be two group theoretic words in two variables,
ie elements of the free group of rank 2. Let
H = { (a, b, U(a,b), U(b,a) ), a, b \in G }
and
H' = { (V(a,b), V(b,a), a, b), a, b \in G }.
Then the condition that H be contained in H' trnaslates to
a = V(U(a,b),U(b,a)) \forall a, b \in G (*),
while the condition that H' be contained in H translates to
a = U(V(a,b),V(b,a)) \forall a, b \in G (**).
Now I haven't yet specified what G is. Take G to be some free
group in the variety specified by the identity (*). Then in
general
[Jim then tried to kill this when Giovanni's note arrived, but
it escaped into the world. Hence the mysterious, tantalizing ending]
--------------------_
Date: Thu, 8 Aug 96 14:15:23 BST
From: Geoff Smith
cc: group-pub-forum@maths.bath.ac.uk
Subject: Re: An elementary problem? (fwd)
Oliver Jenkinson wrote:
Am I right in thinking that Noetherian implies residually finite?
-
No; Olschanskii proved that if p is prime and bigger than 10^40
there os an infinite p-group in which every proper subgroup
has order p. Such a group in Noetherian but not residually finite.
Geoff Smith
Ref:
A. Yu. Olschanskii
Groups of Bounded Period with Subgroups of Prime Order
Algebra and Logic 21 (1982) 369-418
--------------------