Re: Number of abelian groups + exact sequence

In article <1164746807.347890.314240@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
galathaea <galathaea@xxxxxxxxx> wrote:

Arturo Magidin wrote:
In article <1164674298.103037.211910@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
galathaea <galathaea@xxxxxxxxx> wrote:

eugene wrote:
Do you have any ideas for the following problem:

How many up to isomoporphism abelian groups A are there, which could be
included in the following exact sequence: 0 -> Z -> A -> Z/6000Z -> 0 ?

look at the cohomology....

Why would he do that, when A is supposed to be abelian?

A is 2-generated. You can write it as Z^r(+)T, where T is torsion, and
r is nonnegative. You have information about what T and r can be. You
can figure out what A must be. There are a few possibilities, but you
can forget about cohomology and semi-direct products.

i am sorry
but i do not understand why cohomology is not applicable
to the abelian case

using cohomology
you get pretty immediately the importance of

Ext^1(Z, Z/6000Z)

and you start forming pullback squares
and computing which factors work

the identity element of Ext^1(Z, Z/6000Z)
for instance
gives the extension Z (+) Z/6000Z

i thought this was the point
of using baer sums in such calculations
and was the way one atually ends up proving
that one can write it as Z^r (+) T

is this all messed up and incoherent?

i have what i think is the correct solution to eugene's problem
and i could post it if comparison is needed
(i am only being cryptic to not answer homework)

[...]

Ext(Z/6000Z, Z) is isomorphic to Z/6000Z but I don't see that that
helps. The problem is that if 0 -> Z -> A -> Z/6000Z -> 0 and
0 -> Z -> A' -> Z/6000Z -> 0 are representatives of the same element of
Ext(Z/6000Z, Z) then A and A' are isomorphic. But, as far as I know, A
may be the "middle group" of some inequivalent extension. So, I suspect
there are _not_ 6000 A's.

I agree with Arturo and John that the the way to go is to observe that
A is generated by two elements and thus is the sum of two cyclic
groups.

--
Paul Sperry
Columbia, SC (USA)
.

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