For showing that two groups (Q,+) and (R,+) are not isomorphic, can we use the fact that Q/Z is a torsion qroup while R/Z is a mixed one wherein, Z is group of integers? :)
Is being torsion group effective here?
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0@Myself: There's no doubt the method can work, but the point is to notice the small details. Many a proof has slipped on the small details, after all. (-: – 2011-04-30
1 Answers
You are essentially trying to use the fact that if $f\colon G\to K$ is an isomorphism, and $N\triangleleft G$, then $G/N$ is isomorphic to $K/f(N)$. But you have no warrant for assuming that the image of $\mathbb{Z}$ under a putative isomorphism $f\colon \mathbb{Q}\to\mathbb{R}$ would necessarily be equal to $\mathbb{Z}$. In fact, you cannot make such an assertion, because it is false even in $\mathbb{Q}$: there are isomorphisms from $\mathbb{Q}$ to itself as an abelian group that does not map $\mathbb{Z}$ to $\mathbb{Z}$ (e.g., $q\mapsto \frac{q}{2}$).
So your argument cannot work as written. Now, if you could show that for any infinite cyclic subgroup $C$ of $\mathbb{R}$ you have $\mathbb{R}/C$ not torsion, then you'd be done. But the cardinality argument is by far the simplest here.
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1Thanks. What I had not consider was really what Steve D and you pointed above. Your three last sentences were really essential. I just thought the concept of torsion could help me considering this simple elementary problem. :) – 2011-04-29