2Z x 3Z x 5Z x ... is NOT countable ?

Discussion in 'Undergraduate Math' started by Herb, Jun 26, 2004.

  1. Herb

    Herb Guest

    Hi,

    I wonder how to prove 2Z x 3Z x 5Z x ...(The direct product of Z's
    prime proper subgroups) is NOT countable.

    If someone can help me, please post reply.

    Thanks for any help~
     
    Herb, Jun 26, 2004
    #1
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  2. You know B = 2Z x 2Z x 2Z x ... is uncountable because
    |B| = 2^N = number of subsets of the integers

    Another way to see this is to map B onto all the binary numbers
    in [0,1]. Your ring is even larger because 3Z has more elements
    than 2Z, 5Z has more elements than 2Z, etc.
     
    William Elliot, Jun 26, 2004
    #2
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  3. Represent each element of this group as an ordered sequence and
    diagonalize? Since the order of each prime group in the product
    increases and you can find and each group has order greater then one
    you should be able to construct and unlistable element.
     
    Abraham Buckingham, Jul 1, 2004
    #3
  4. Herb

    Virgil Guest

    Isn't pZ, for any prime p, isomorphic to Z as an additive Abelian group?

    If so, then 2Z x 3Z x 5Z x ... is isomorphic to Z x Z x Z x ... which
    is, in turn, isomprphic to Z^Z, the set of all functions from Z to Z
    under pointwise addition.
     
    Virgil, Jul 1, 2004
    #4
  5. If I'm intepreting the question correctly (I'm not clear on what prime
    proper subgroup means) then pZ is a finite group of order p which
    means the groups are cyclic, and therefore isomorphic to an additive
    Abelian group. If it means the group generated by the powers of p it
    would still be isomorphic to an infinite additive Abelian group - so
    either way I think you're right. As for the rest of your comments I'm
    really not sure - could you describe the isomorphism fomr 2Z x 3Z x 5Z
    x.... to Z x Z x Z x.... for me? Could you also explain what is ment
    by the pointwise addition from Z to Z as well? I understand what is
    ment by the set of all functions from Z to Z. All of my knowledge of
    group theory comes from 'Elements of Abstract Algebra' by Allan Clark
    and I must admit I'm not very strong on the matieral at this point.
    Does diagonalizing fail for some reason that I'm overlooking?
     
    Abraham Buckingham, Jul 1, 2004
    #5
  6. On 1 Jul 2004 13:41:10 -0700 (Abraham
    Buckingham) wrote in
     
    Brian M. Scott, Jul 1, 2004
    #6
  7. Herb

    Virgil Guest

    I was under the impression that pZ is the (infinite) set of all integers
    multiples of p, and that Z/p or Z/pZ, or something like that, is the
    (finite) set of residue classes mod p.

    However, for the desired proof, it is only necessary that each "factor"
    in the infinite product contain at least two members to be able to
    construct an analogue of Cantor's "diagonal" proof showing
    uncountability (any countable "list" of members of the direct product
    must be incomplete, because we can construct, a la Cantor, an element of
    the direct product not in it).
     
    Virgil, Jul 2, 2004
    #7
  8. Ah thank you So this is a group under regular addition since pm + pn =
    p(m+n) and the isomorphism becomes obvious. Thanks for your help. :)
     
    Abraham Buckingham, Jul 3, 2004
    #8
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