I need help in describing the equivalence classes of covering spaces
of the torus, using the theorem that the equivalence classes of
covering spaces of a topological space are in one-to-one
correspondence with the subgroups of the fundamental group of the
topological space.  I know that the fundamental group of the torus is
Z^2, that the universal cover is the R^2, and that the torus can be
n-covered by a connected spiraled tube.  Can you help me?

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Clarification of Question by topologystudent-ga on 24 Jun 2004 11:06 PDT

http://planetmath.org/encyclopedia/ClassificationOfCoveringSpaces.html 
This site describes the equivalence relation for covering spaces and
the theorem that I am using.  Basically, for covering spaces A (by
homomorphism p) and B (by q), base space C, A and B are equivalent iff
there exist a homeomorphism f between them such that the diagram
commutes.
                   A     f--->    B
                     |           |
                   p  |         |  q
                       \|/    \|/
                           C

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Clarification of Question by topologystudent-ga on 24 Jun 2004 11:10 PDT

http://planetmath.org/encyclopedia/ClassificationOfCoveringSpaces.html 
This site describes the equivalence relation for covering spaces and
the theorem that I am using.  Basically, for covering spaces A (by p)
and B (by q), base space C, A and B are equivalent iff there exist a
homeomorphism f between them such that the diagram commutes.
Using this equivalence, the number of preimages of a point in C is
invariant, so a 2-fold spiral and a 3-fold spiral are not equivalent.
                   A     f--->    B
                     |           |
                   p  |         |  q
                       \|/    \|/
                           C

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Clarification of Question by topologystudent-ga on 02 Jul 2004 03:58 PDT

Can you give tell me any way to cover the torus, other than a
connected, spiraled tube, a torus, or R^2?

Googling '"subgroups of z^2"'
http://google.com/search?q=%22subgroups+of+z%5E2%22
yields the following result on the first page of results:
http://abel.math.harvard.edu/~fcale/assig/Ideals2.pdf

This page won't load (perhaps it's been deleted) but Google's cache of it
http://google.com/search?q=cache:abel.math.harvard.edu/~fcale/assig/Ideals2.pdf
indicates that
    Up to isomorphism, the only subgroups of Z^2 are the trivial group,
    free groups on one generator,(which are isomorphic to Z^1) and free
    groups on two generators, which are isomorphic to Z^2.

What equivalence relation are you using?

I asked what equivalnce relation your using for your equivalence
classes, because if it's homeomorphism, then, of course, all the
n-fold spirals are homeomorphic.  I don't recall precisely the
statement of the theorem you mention.

Clarification of what I wrote earier:

Of course, when I note that Z is the only group other than the trivial
group and z^2 which has an isomorphic copy of itself as a subgroup of
Z^2, that doesn't mean that there's only one copy of Z that's a
subgroup of Z^2.  No.  Rather, (1,0) (the point in Z^2 whose first
coordinate is 1 and whose second is 0) generates a Z, (2,3) generates
a Z, etc.  Some of these are subgroups of one another (e.g., <(6,0)>
is a subgroup of <(3,0)>) but others are not.