Re: Self Study problem help - Group theory

Ok let me see if I can get it. I did mean to say that (a/b) + (c/d) had
odd denominator naturally, that word seems to be giving me trouble
today.

<k> definitely does not contain 1/k^2, it's all the integer multiples
of k. I see where I caused the confusion by saying k^m when I didn't
mean expodentiation but rather just repeated applications of the group
operation, and should have written k*m instead to be less ambiguous.
I'm simply accustomed to the text using the multiplicative notation for
group operations so I continued with that absent-mindedly without
considering my audience can't read my mind - sorry about that.

To answer your new question I can see now how taking the reciprocal of
any two distinct primes would have been enough. Just one prime would
make a cyclic group (obvious from the definition) so I'd need two. If I
took say <1/2, 1/3> I could then take (1/3) + (1/3) + (1/3) to get 1,
and - (1/2) - (1/3) = -5/6 therefore 1-(5/6) and make 1/6. Since the
denominator is always at most 6 since I can't generate powers of 2 or 3
then 1/6 generates 1/3 and 1/2 it generates the group and it's cyclic,
therefore does not meet the requirements.

So any abstracting I can see that given two primes p and q then I can
construct all numbers of the form (n*p + m*q)/(p*q) for integers n and
m and since p and q are relatively prime there exists an n and m such
that n*p + m*q = 1 and therefore 1/(p*q) can be constructed and
generates the group. since p/(p*q) = 1/q and q/(p*q) = 1/p.

Now consider a finite collection of primes. We can proceed by induction
(?) since we know that for any two we can construct 1/(p*q) which
generates that group. Given another prime called say, r we can
construct a linear combination of n*(p*q) + m*r = 1 since r and p*q are
relatively prime meaning we can construct 1/(p*q*r) and so on with
until we exaust the primes on our list since each prime will be
relatively prime to the product of the previous primes.

So I'm stuck taking an infinite collection of primes? Seems I would
need to rely on the infinitude of primes in order to ensure that no
element could generate the rest of them. I could choose a 'sparse'
collection of primes though, say every billionth prime and the
arguement would hold? This way I can always choose a prime that isn't
in the finite number that would be in the denominator of the would be
generator. Or have I missed something important?

.

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