Self Study problem help - Group theory

I am studying the third edition of 'Abstract Algebra' by Dummit and
Foote on my own and have come across a question which I not certain I
have solved correctly, or rigorously. As always even though this is
self study I would ask that you give as little help as possible and
avoid giving me an answer, only guideance on how to improve my proof,
or where the reasoning went wrong rather then an explicit solution.

The problem is (paraphased) 'Exhibit a proper subgroup of the rationals
under addition which is not cyclic'.

I had trouble at first, but I think I have a good answer, I'm just not
certain how rigorous it is. This text defined a cyclic group to be a
group that's generated by a single element.

So, consider X which is generated by the infinite collection <1/p_1,
1/p_2, ...> where p_n is a prime greater then 2. Now note that adding
any two two together gives the product (p_i + p_k)/(p_i * p_k). So, all
the elements generated by this collection will have an integer for a
numerator and the denominator will decompose into powers of the primes
p_n listed in the numerators of X. Now, no matter what the numerator
is, when we reduce the fraction we can only divide out by the primes in
the bottom, and none of them are 2, so no matter what we can never
generate 1/2. (I'm not certain if my reasoning is solid here, I keep
thinking somehow I can create 1/2 by a clever combination of additions
and subtractions on the rationals I have). Now this means we have a
proper subset, since we chose to generate the group we know that it's
closed.

Now assume that X is cyclic. This would mean some element, well call it
k = (v / [(p_a)^x*(p_b)^y*...(p_c)^z] would generate X. But, since k^m
for all integers m would only contain powers of a finite number of
prime denominators there would exist elements of X that k would not
construct. This contradicts the assumtion that X is cyclic. This part
of the "proof" relies on the same logic about the denominator being a
product of primes so I feel if the first part fails this part will to.

Thanks in advance, Abe Buckingham.

.

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