# Is being torsion group effective here?

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? :)

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Sure, but why not make a cardinality argument? –  t.b. Apr 29 '11 at 12:51
Expanding Theo's comment: an isomorphism is first of all a bijection and there are no bijections between $\mathbb Q$ and $\mathbb R$. –  lhf Apr 29 '11 at 12:55
Actually to make this kind of argument, you must show Z gets mapped to Z under any such isomorphism, which isn't true. –  user641 Apr 29 '11 at 12:57
@Steve: that's not necessary. Given a prospective isomorphism $\phi : \mathbb{Q} \to \mathbb{R}$ you just need to consider the image of $\mathbb{Z}$, which is certainly a group isomorphic to $\mathbb{Z}$. –  Qiaochu Yuan Apr 29 '11 at 13:00
@Qiaochu: Yes, but it need not be $\mathbb{Z}$ itself. So she would have to argue that $\mathbb{R}/C$ is not torsion for any infinite cyclic subgroup $C$ of $\mathbb{R}$. –  Arturo Magidin Apr 29 '11 at 13:23

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.