Re: Infinite group with finite proper subgroups

In article <6470282.1190140456457.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
jane <jane1806@xxxxxxxxxx> wrote:
I want to prove that the Pruefer group

Z(p^oo) = { exp(2ni*pi / p^m) for n,m in Z^+ } is an example

of an infinite group, all proper subgroup of which are finite.

Could anybody help me to prove that.

Any ideas are welcome.

First, note that you do not lose or gain anything if you allow n=0 or
m=0. Then notice that:

* Every element of Z(p^oo) can be written, uniquely, with m>=0,
n relatively prime to p, and 0 <= n < p^m.


Then let H be a subgroup of Z(p^oo). Consider the set:

S = {m in Z | m>=0, there is n in Z^+, gcd(n,p)=1, 0<=n<p^m, }
{ | exp(2n*pi*i/p^m) in H }

S is nonempty (m=0 must be in there).

Try proving that:

* If m is in S and 0 <= k < m, then k is also in S.

Deduce that either S has a maximum, or else S = Z. Go from there.


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

Arturo Magidin
magidin-at-member-ams-org

.

Relevant Pages

  • Re: Infinite group with finite proper subgroups
    ... jane wrote: ... proper subgroup. ... bounded power of p as a denominator, ...
    (sci.math)
  • Re: Infinite group with finite proper subgroups
    ... Thus, if S contained arbitrary large powers of p as denominators, then it would actually contain the whole Zand wouldn't be proper subgroup. ... But there are only finitely many elements from Z, which have a bounded power of p as a denominator, hence any proper subgroup is finite. ...
    (sci.math)
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