Jupiter

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1. Aug26-12 12:04 AM
LASLESH
Dear Jupiter,

Another way of looking at whether (Q, +) is finitely generated or not could be as follows:
(1) All automorphisms of Q are of the form T(q) = kq (for all q in Q, some k in Q)
ie:Stretchings
(2) Assume that Q is finitely generated ==> Q = < q1 , q2,....qr > having r generators
with the qi in Q.
Now, any non-identity permutation of the generators induces a
non-identity automorphism of Q.
(3) Choose the permutation q1 --> q2, q2 --> q3, q3--->q1.
(4) But (1) implies that there must exist a k in Q such that k.q1 = q2, k.q2 = q3 and
k.q3 = q1 which is valid (in the reals) only for k = 1 (Identity automorphism)

Of course, the more geometric proof involving the LCM & GCD of the denominators and numerators of {q1 , q2, ....qr} enables one to explictly construct a single generator element of S = <q1, q2,.....qr> implying cyclicity of S (as a group) which certainly isn't = Q ( via denseness) !!!