infinite abelian group with finite proper subgroups

Discussion in 'Undergraduate Math' started by Jon Gerard, Jun 26, 2003.

  1. Jon Gerard

    Jon Gerard Guest

    Let G be an infinite abelian group such that every proper subgroup of
    G is finite. Show that G is isomorphic to Z(p^infty), where

    Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }

    I've been looking at this one for a week and nothing. Can somebody
    help! It's driving me crazy.
     
    Jon Gerard, Jun 26, 2003
    #1
    1. Advertisements

  2. You say you have gotten nothing, so some comments, which may help
    getting you started; I haven't thought it through in detail, so I
    could be wrong in some things. But it seems right to me:

    First, note that G is a torsion group; as such, it can be decomposed
    into a direct sum of its p-parts, that is

    G = G_2 \oplus G_3 \oplus G_5 \oplus ...

    where G_p = {g in G; g^{p^m}=e for some m>0}.

    Now, let's consider only the non-trivial p-parts; so we have

    G = G_{p_1} \oplus G_{p_2} \oplus ...

    Note that there cannot be an infinite number of non-trivial p-parts;
    if there were, then the subgroup G_{p_2}\oplus ... would be a proper
    infinite subgroup. So

    G= G_{p_1} \oplus G_{p_2} \oplus ... \oplus G_{p_m};

    since G is infinite, at least one of the G_{p_i} is infinite. But
    then, G_{p_i} itself is an infinite subgroup of G, nontrivial. Hence
    G=G_{p_i}. That is, G is an infinite p-group.

    Now consider the quotients G/pG, pG/p^2G, p^2G/p^3G,...
    these quotients are vector spaces over F_p, hence they have a
    well-defined dimension. Prove that the dimensions are non-increasing,
    and that all of them are finite and nonzero. From that, you should be
    able to conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G is
    cyclic; and then that (b) all of them are cyclic; using the fact that
    G does not have infinite proper subgroups.

    Once you have that all of them are cyclic, it should not be hard to
    write down an isomorphism.

    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Jun 26, 2003
    #2
    1. Advertisements

  3. Jon Gerard

    J. Woodward Guest

    Hmmm...I think I follow you, but I still don't see the actual isomorphism...


     
    J. Woodward, Jun 27, 2003
    #3
  4. But is every proper subgroup of this finite?
    Perhaps the question was: every proper subgroup of finite index?
     
    Timothy Murphy, Jun 27, 2003
    #4
  5. Well, assuming everything I said was correct, and you have shown that
    G is a p-group, and moreover, that each of p^iG/p^{i+1}G is cyclic,
    what does that mean? Can you show that G/p^iG is cyclic for all i>0?

    Now, take G/pG. This is cyclic of order p, say with generator
    x+pG. Since G/p^2G is also cyclic, you should be able to see that it
    is generated by an element which, when multiplied by p, is equal to
    x. Call it x/p. Then you need to find one which when multiplied by p
    is x/p; call it x/p^2. Etc.

    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Jun 27, 2003
    #5
  6. Probably because the final two paragraphs are complete bollocks... All
    the quotients given will be trivial.

    I screwed up; I used the p-powers subgroups instead of the subgroups
    of elements of exponent the p-powers...

    Here's a better way to proceed; we know that G is an infinite
    group. First, consider, pG. If pG were trivial, then G would be of
    exponent p, hence an infinite dimensional vector space over F_p, which
    clearly has infinite proper subgroups; so pG is nontrivial. If pG
    is finite, then G/pG is an infinite dimensional vector space over F_p,
    and you can lift an infinite proper subgroup to G. So pG is infinite,
    hence pG=G. So G is divisible (being a p-group and p-divisible).

    At this point you could invoke classification theorems for divisible abelian
    groups, which will tell you that G must be a bunch of copies of
    Z_p^{infty}, and thereby conclude that it must be equal to one of
    them. But here's a way to proceed without invoking those
    classification theorems:

    For each n>0, let G[p^n] = {g in G : g^{p^n}=e}, the subgroup of
    elements of epxonent p^n.

    G[p] is nontrivial; let x in G[p] be different from e1. Since G is
    p-divisible, there is an element x_2 such that px_2 = x; an element
    x_3 such that px_3=x_2; ...; an element x_{n+1} such that px_{n+1} =
    x_n.

    Let H = < x, x_2, x_3,...>

    It is now easy to verify that H is nontrivial and infinite, so H=G. It
    is now also easy to given an isomorphism from Z_{p^{infty}} to G: sent
    1/p + Q/Z to x, and 1/p^n +Q/Z to x_n, n>1.

    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Jun 28, 2003
    #6
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.