(Message # 249: 7338 bytes)
Date: Fri, 27 Oct 95 11:39:27 0100
From: "T.Ward"
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Subject: Summary of replies to question on f.g. abelian groups.
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Summary of replies to question on f.g. abelian groups.
Dear all,
A week or two ago I posted the following question:
>If G is abelian, with finite torsion-free rank, and A is
>an injective homomorphism from G to itself, does it
>follow that the quotient G/AG is finite?
>If so, is there a `formula' for the order? For example,
>if G is Z^d, then the quotient group has order
>given by the modulus of the determinant defining
>A.
>More suggestively, if G is Q^d, (Q the rationals)
>then the order (which is 1) is equal to the product
>over all primes of the p-adic norm of the determinant of
>the rational matrix defining A.
>One would expect that there is a `maximum' size
>associated with A: if A is a k by k rational matrix,
>and G_A is the group Z^k[A;A^{-1}] then for any
>other G with rank k the quotient G/AG should be
>smaller than G_A/AG_A, which can be computed.
In doing so I of course left out a key additional
hypothesis: that G should also be torsion-free.
I have had a number of helpful replies:
From many people the observation that if G is allowed to
have torsion then the quotient may be infinite.
From Avinoam Mann, Hyman Bass and Robin Chapman `elementary'
proofs that if G is abelian, torsion-free, of finite torsion-
free rank, and A:G to G is injective, then G/AG is finite.
There does not seem to be a particularly simple way to establish
the size of the quotient in terms of the group G and the
rational matrix A, but there are simple estimates bounding
the size.
From Ray Mines, an outline of a proof using the notion of
Richman type.
====================================================================
Avinoam Mann wrote:
"I think the answer is yes. Consider G as a subgroup of Q^d, and let x_1,..,x_k
be independent elements of G. Then x_i generate a subgroup H isomorphic to Z^d,
and A is determined by its action on H, i.e. on x_i. If Ax_i = y_i, we can
express y_i as rational combinations of x_i (the x_i are a basis for Q^d over
Q), and thus A determines a rational matrix T, and A in turn is determined by
T, indeed A is rge restriction to G of the lin trans of Q^d given by
multiplication by T, taking x_i as a basis for Q^d. Following A by a
multiplication by some integer n, we get a new endomorphism of G which maps H
into H. The matrix T is multiplied by the diagonal matrix nI, and the product
is an integral matrix, say S. Let B be the new endomorphism.
Now if K is any finitely generated subgroup of Q^d containing H, then |K:BK| is
the modulus of detB. But if |G:BG| is infinite we can find, by taking enough
representatives of cosets of BG in G, a K for which |K:BK| is arbitrarily
large, a contradiction. Thus |G:BG| is finite, and AG contains BG, so |G:AG|
is finite. Following the above procedure more carefully should also give the
value of |G:AG|."
====================================================================
Hyman Bass wrote:
"You did not specify that G was torsion free, nor did you impose
finiteness conditions on the torsion. Taking G to be the direct
sum of a sequence of copies of Q/Z, and A a shift endomorphism,
you get G/AG = Q/Z. If, on the other hand, the elementary p-torsion
in G has finite rank for all primes p, then A induces an isomorphism
on the torsion subgroup T, and one can pass to G/T and so reduce to
the torsion free case.
Then one can argue as Avinoah Mann proposes, with one correction. Let
V = Q^d, the rational vector space spanned by G. After multiplying A
by an integer, we can assume that A leaves invariant the free Z-module
L spanned by a rational basis of V contained in G. Now, the extra
ingredient needed in Avinoah's arguement is the observation that if
K is a finitely generated Z-module containing L, then the A-invariant
submodule generated by K, i.e. the sum of all A^n .K, (n > 0), is still
a finitely generated Z-module. (One needs this because his argument only
applies to K's that are A-invariant.) The finite generation of the
'A-span' of K follows because dK is contained in L for some d > 0, and
so the 'A-span' of K is contained in the A-invariant module (1/d)L.
One more comment: K/AK does not necessarily embed in G/AG, but no matter.
The argument bounds the size of any finitely generated subgroup of G/AG.
The same considerations show that G/AG has p-torsion only for primes p
dividing det A, and that |det A| bounds |G/AG| (always assuming that A
leaves L invariant, as above)."
====================================================================
Ray Mines wrote:
"your question is a good one, and is not all that easy to answer. It
is possible that it is worked out in Dave Arnold's book on torsionfree
goups in the Springer Lecture Notes.
Among people who work in finite rank torsion free goups there is
something known as the Richman Type. Let G be such a group. Let F be
a full-rank free contained in G. Then the quotient G/F is called the
Richman type and is a quasi-isomorphism invariant. Let f be the
monomorphism. Then it is easy to see that G/F is isomorphic to
f(G)/f(F). Since f is a monomorphism f(F) is a full rank free. Thus
G/F and G/f(F) are quasi-isomorphic. Moreover, G/f(G) is isomorphic
to G/f(F) / f(G)/f(F). Now it follows, some how or other, that G/f(G)
is finite.
You need to understand the quasi isomorphism to get the last sentence,
but I am late for my class. So I'm off. If you have more questions,
write."
====================================================================
Many thanks to all who replied. The problem came up in some ergodic
theory calculations, where it has the following form: if X is a
solenoid (a connected, compact, abelian group of finite topological
dimension), and T is an ergodic endomorphism of X, then T has
finitely many periodic points for each period.
Thomas Ward
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Dr. Thomas Ward e-mail: t.ward@uea.ac.uk
School of Mathematics url: http://www.mth.uea.ac.uk/people/tw.html
University of East Anglia phone: (01603)-592848
Norwich home: (01603)-401079
NR6 5LB
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