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
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  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
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  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
  4. But is every proper subgroup of this finite?
    Perhaps the question was: every proper subgroup of finite index?
    Timothy Murphy, Jun 27, 2003
  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
  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} =

    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
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