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

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

1. ### HerbGuest

Hi,

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

Thanks for any help~

Herb, Jun 26, 2004

2. ### William ElliotGuest

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

3. ### Abraham BuckinghamGuest

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
4. ### VirgilGuest

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

Virgil, Jul 1, 2004
5. ### Abraham BuckinghamGuest

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
6. ### Brian M. ScottGuest

On 1 Jul 2004 13:41:10 -0700 (Abraham
Buckingham) wrote in

Brian M. Scott, Jul 1, 2004
7. ### VirgilGuest

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
8. ### Abraham BuckinghamGuest

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