Re: Self Study problem help - Group theory

On 26 Jul 2005 07:25:27 -0700, abe.buckingham@xxxxxxxxx wrote:

>...
>So that considered, it seems that to get a set of generators for Q we'd
>need to take all the P_n of the form {1/p, 1/p^2,...} where p is the
>nth prime and finally the genators would be <P_1, P_2, P_3,..> for all
>P_n for natural n. We could exclude a finite number of these generators
>since if we take out say 1/(p^m) then p/(p^(m+1)) constructed by taking
>p repetitions of 1/(p^(m+1)) and canceling a p to get 1/(p^m).
>
>This generates all of Q since we can always construct 1/x for any x
>since we can finitely generate any composite denominator using the
>linear combination arguement and the fact that primes are always
>relatively prime to each other. naturally we can then construct all
>numbers of the form y/x with y and n elements of the naturals and x not
>equal to 0. That's all of Q so seems to work ok.
>
>Now clearly since removing only a finite number of those generators
>will still construct all of Q, then we must remove a infinite
>collection in order to attain a proper subgroup. It seems that we can
>remove the 'tail' of any P_n which would remove all reciprocal powers
>of that prime above a certain power, therefore we'd be unable to
>construct numbers with those prime powers in the denominator. So we can
>remove an entire P_i, just the tail of it, or the tail of all of them
>as long as we're left with an infinite number of generators and get a
>proper subgroup of Q.

Well analyzed.

>If we allow the removal of any collection from
>these P_n's even if they leave a finite collection of generators then
>it would seem we could describe every possible subgroup of Q. Is this
>correct or am I missing something important?

Well the basic idea is right but the last statement might be
overstating it a little bit since you could have a subgroup with no
numerator = 1, for example <2/3>.

But I don't think you've missed anything important -- it's clear that
for any actual problem involving additive subgroups of Q you have the
experience, understanding and power to resolve it.

Let me recommend a little gem of a book by my favorite author:

Infinite Abelian Groups
by Irving Kaplansky

This book would be perfect for you -- clear, elegant presentation,
challenging problems, just at the right level.

But before leaving this problem, let me point out some ideas which are
potentially applicable to other groups besides the additive group of
Q.

In an arbitrary group G (not necessarily abelian), consider an
ascending chain H1, H2, H3, ... of subgroups of G. Let H be the union
(set-theoretic union) of H1, H2, H3, ...

Is it clear that H is a subgroup of G? In general, is it always true
that the union of an arbitrary (even uncountable) ascending chain of
subgroups is a subgroup?

Now suppose you have an, infinite strictly ascending chain H1, H2, H3,
.... , and let H be the union of the chain. Can H be finitely
generated?

Now look again the subgroup <1/p, 1/p^2, 1/p^3, ...>. Can it be
expressed as the union of an infinite strictly ascending chain of
subgroups?

This doesn't really make any of the proofs you gave any simpler, it
just gives a different way of looking at it.

quasi
.

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