Date: Fri, 4 Nov 94 00:10:11 CST
From: rusin (Dave Rusin)
To: magidin@yam.berkeley.edu
Subject: Re: Homomorphism to abelian group
Newsgroups: sci.math
In article <39c8r0$okm@agate.berkeley.edu> you write:
>
>Since G/T(G) is free, the exact sequence
You're assuming of course that G is finitely generated. What if
G/T(G) = Q (additive group of rationals) for instance?
dave
==============================================================================
Date: Fri, 4 Nov 1994 08:10:29 -0800 (PST)
From: Arturo Magidin
Subject: Re: Homomorphism to abelian group
To: Dave Rusin
On Fri, 4 Nov 1994, Dave Rusin wrote:
> In article <39c8r0$okm@agate.berkeley.edu> you write:
> >
> >Since G/T(G) is free, the exact sequence
>
> You're assuming of course that G is finitely generated. What if
> G/T(G) = Q (additive group of rationals) for instance?
> dave
>
Oops. You're right. I assumed it was finitely generated.
I'll post a correction. Thanks.
[sig deleted]
==============================================================================
Date: Fri, 4 Nov 94 00:16:00 CST
From: rusin (Dave Rusin)
To: frisinv@alleg.edu
Subject: Re: Homomorphism to abelian group
Newsgroups: sci.math
In article <39bkr5$2vf@mustang.alleg.edu>,
Count Zero wrote:
>I ran into this problem while I was working a something and for some
>reason just could not solve it proof, counter-example or otherwise.
>
>Given abelian groups G and X and a nonzero homomorphism from X to G/T(G)
>where T(G) is the torsion subgroup of G, is there a nonzero homomorphism
>from X to G?
>
>Thanks.
A posted solution began,
>Since G/T(G) is free, the exact sequence
Is that what you intended -- that G/T be free? It must be torsion free,
but the example of G/T = additive group of the rationals shows that
it need not be. Do you know G is finitely generated?
dave
==============================================================================
From: Vincent Frisina
Date: Sat, 5 Nov 94 17:40:48 -0500
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: Homomorphism to abelian group
>Is that what you intended -- that G/T be free? It must be torsion
>free, but the example of G/T = additive group of the rationals
>shows that it need not be. Do you know G is finitely generated?
I had not intended that G be finitely generated (and thus G/T(G)
free). In fact, I had asked this orginally hoping that I would not
have to limit myself only to FG abelian groups in some work I'm
doing.
Vincent Frisina
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Homomorphism to abelian group
Date: 7 Nov 1994 21:05:13 GMT
In article <39bkr5$2vf@mustang.alleg.edu>,
Count Zero wrote:
>
>Given abelian groups G and X and a nonzero homomorphism from X to G/T(G)
>where T(G) is the torsion subgroup of G, is there a nonzero homomorphism
>from X to G?
I assume you mean, "Is there a homomorphism from X to G _lifting_
the original homomorphism?". The answer is no (!) (It has already been noted
that the answer is "yes" if G is finitely generated.)
Where else would one turn in this case but to Kaplansky's "Infinite Abelian
Groups"? As has been noted in this group, it suffices to look at the
special case where X=G and the map G-> G/T(G) is the canonical one. Then
the question is, "Is every abelian group the sum of its torsion subgroup
and a (torsion-free) complement?" Kaplansky refers to Baer who in 1936 showed:
A torsion group T has the property that:
T is a summand of every abelian group G with T(G)=T
if and only if T = B + D where B has bounded torsion and D is divisible.
Addressing the other half of the group, Kaplansky quotes Griffith ca. 1968
who showed:
A torsion-free group F has the property that
F is a summand of every abelian group G with G/T(G)=F
if an only if F is free.
If I understand Baer's example, a non-split group would be the direct
product of the groups Z/(2^n) (n=1,2,3,...). The torsion subgroup is
the set of sequences of elements in these groups for which the order of
the terms in the sequence stays bounded.
The existence of a guaranteed splitting is equivalent to the vanishing of
the cohomology group H^2(G/T, T), but inasmuch as it is not possible
(as far as I know!) to "classify" all torsion-free groups it is unlikely
one could hope to compute their cohomology. (The coefficient group T is
much easier to deal with).
I would have to say that as a rule of thumb anything you would want to
know about abelian groups is either trivial or true only for some
smaller class of groups (finitely generated, countable, etc.)
dave
==============================================================================
From: x151509@rubc.rz.ruhr-uni-bochum.de (Joerg Winkelmann)
Newsgroups: sci.math
Subject: Re: Homomorphism to abelian group
Date: 8 Nov 1994 16:06:39 GMT
Count Zero (frisinv@alleg.edu) wrote:
: I ran into this problem while I was working a something and for some
: reason just could not solve it proof, counter-example or otherwise.
: Given abelian groups G and X and a nonzero homomorphism from X to G/T(G)
: where T(G) is the torsion subgroup of G, is there a nonzero homomorphism
: from X to G?
As has been pointed out before, the answer is positive if G is finitely
generated.
However, in general the answer is negative.
Example:
Let G be the direct product over all primes p of the finite groups Z/pZ.
In other words, an element g in G may be regarded as a map which
to each prime p associates an element g(p) in Z/pZ.
Now the torsion subgroup T(G) is simply the group of all g in G such
that g(p)=0 for all except finitely many p.
Given a prime q, an element g is q-divisible
(i.e. there exists an element h such that h^q=g)
if and only if g(q)=0.
It follows that there exists no element g<>0 in G which is
q-divisible for all primes q simultaneously.
On the other hand, for every g in G and every prime q there exists
an element t in T(G) such that g+t is q-divisible:
It suffices to define t by t(p)=0 for p<>q and t(q)=g(q).
It follows that every element in G/T(G) is divisble through all
primes.
This implies that there exists no non-zero group homomorphism from
G/T(G) to G.
In particular, X=G/T(G) is a group which admits a non-zero homomorphism
to G/T(G) (the identity for example) but there does not exists
any non-zero homomorphism from X=G/T(G) to G.
Joerg Winkelmann
: Vince Frisina
==============================================================================
From: fred@noether.math.fau.edu (Fred Richman)
Newsgroups: sci.math
Subject: Re: Homomorphism to abelian group
Date: 9 Nov 94 13:52:41
In article <39m4q9$2f9@mp.cs.niu.edu> rusin@washington.math.niu.edu (Dave Rusin) writes:
In article <39bkr5$2vf@mustang.alleg.edu>,
Count Zero wrote:
>
>Given abelian groups G and X and a nonzero homomorphism from X to G/T(G)
>where T(G) is the torsion subgroup of G, is there a nonzero homomorphism
>from X to G?
I assume you mean, "Is there a homomorphism from X to G _lifting_
the original homomorphism?".
Why would you assume that? If G is the product of one cyclic group of each
prime order, then there is a nonzero homomorphism from the additive group
Q of rational numbers to G/T(G), but none from Q to G.
Fred
==============================================================================
From: rgep@emu.pmms.cam.ac.uk (Richard Pinch)
Newsgroups: sci.math
Subject: Re: Homomorphism to abelian group
Date: 9 Nov 1994 19:57:45 GMT
In article
fred@noether.math.fau.edu writes:
>In article <39m4q9$2f9@mp.cs.niu.edu>
>rusin@washington.math.niu.edu (Dave Rusin) writes:
>
> In article <39bkr5$2vf@mustang.alleg.edu>,
> Count Zero wrote:
> >
> >Given abelian groups G and X and a nonzero homomorphism from X to G/T(G)
> >where T(G) is the torsion subgroup of G, is there a nonzero homomorphism
> >from X to G?
>
> I assume you mean, "Is there a homomorphism from X to G _lifting_
> the original homomorphism?".
>
>Why would you assume that? If G is the product of one cyclic group of each
>prime order, then there is a nonzero homomorphism from the additive group
>Q of rational numbers to G/T(G), but none from Q to G.
>
No there isn't. T(G) = G in this case, so G/T(G) = 0, so there isn't a
non-zero homomorphism from anything to G/T(G).
Richard Pinch; Queens' College, Cambridge
==============================================================================
From: mareg@csv.warwick.ac.uk (Dr D F Holt)
Newsgroups: sci.math
Subject: Re: Homomorphism to abelian group
Date: 10 Nov 1994 09:22:52 -0000
In article <39r9jp$cpr@lyra.csx.cam.ac.uk>,
rgep@emu.pmms.cam.ac.uk (Richard Pinch) writes:
>In article
>fred@noether.math.fau.edu writes:
>>In article <39m4q9$2f9@mp.cs.niu.edu>
>>rusin@washington.math.niu.edu (Dave Rusin) writes:
>>
>> In article <39bkr5$2vf@mustang.alleg.edu>,
>> Count Zero wrote:
>> >
>> >Given abelian groups G and X and a nonzero homomorphism from X to G/T(G)
>> >where T(G) is the torsion subgroup of G, is there a nonzero homomorphism
>> >from X to G?
>>
>> I assume you mean, "Is there a homomorphism from X to G _lifting_
>> the original homomorphism?".
>>
>>Why would you assume that? If G is the product of one cyclic group of each
>>prime order, then there is a nonzero homomorphism from the additive group
>>Q of rational numbers to G/T(G), but none from Q to G.
>>
>
>No there isn't. T(G) = G in this case, so G/T(G) = 0, so there isn't a
>non-zero homomorphism from anything to G/T(G).
>
>Richard Pinch; Queens' College, Cambridge
I presume, by "product" he meant the unrestricted product.
Let x_p denote a generator of the cyclic group of order p.
Then the element g = (x_2,x_3,x_5,x_7,...,x_p,....) lies in G - T(G).
Let Q denote the additive group of rationals. Then we can define a
homomorphism from Q to G/T(G) by mapping 1 to g+T(G).
This works, essentially because, for any n>0, x_p has an n-th root for all but
finitely many p, and so g+T(G) has an n-th root.
But G itself is not a divisible group, so there is no nonzero homomorphism
from Q to G (there is no possible image for 1).
Derek Holt.
==============================================================================
From: rgep@pmms.cam.ac.uk (Richard Pinch)
Newsgroups: sci.math
Subject: Re: Homomorphism to abelian group
Date: 10 Nov 1994 11:05:32 GMT
In article <39sopc$6rt@parmesan.csv.warwick.ac.uk>, mareg@csv.warwick.ac.uk (Dr D F Holt) writes:
[deleted - it's the above post -- djr]
Quite right! I shouldn't post hasty responses after a long day!
Richard Pinch
==============================================================================