Re: Self Study problem help - Group theory

quasi wrote:
> On Tue, 26 Jul 2005 00:45:04 -0700, quasi <quasi@xxxxxxxx> wrote:
>
> Another variation:
>
> How about using for generators {1/p} where p ranges over all primes
> (including 2).
>
> Is this subgroup proper? Is it cyclic?
>
> quasi

Seems to be a proper subgroup as well considering what I now know about
<1/p, 1/p^2, 1/p^3,...> being a proper non-cyclic subgroup. All the
denominators would be square-free. It's also not cyclic for the same
reasons the collection without 2 would be since if it was cyclic we
wouldn't be able to get to prime reciprocals that aren't in the
denominator of the element that would generate the group.

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. 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?

.

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