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How can we prove that $\mathbb{Z}$ is not isomorphic to $\mathbb{Q}$?

Both of them have a countable number of elements, so cardinality doesn't help. $0$ is the identity and $-x$ is the inverse of $x$ in both case. What to use then?

Kunal
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2 Answers2

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Hint:

Note that $\mathbb Z$ is cyclic.

Can you prove that $\mathbb Q$ isn't?

Ludolila
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$\mathbb{Q}/\mathbb{Z}$ contains elements of arbitrary finite order. This is not true for $\mathbb{Z}/n\mathbb{Z}$ for any $n$.

  • This seems to me too much like assuming what you want to prove, or at least the argument is incomplete. Isn't it at least conceivable that there is some very strange embedding of $\Bbb Z$ inside of $\Bbb Q$ so that $\Bbb Q /\Bbb Z$ would have surprising properties, such as containing elements of arbitrary finite order? There might be many copies of $\Bbb Z$ inside of $\Bbb Q$, and it matters which one you use when you take the quotient. – MJD Jan 12 '14 at 15:16
  • I think your argument is that if $Q\cong Z$ via some isomorphism $\phi$, then for each $nZ\subseteq Z$ we can find a corresponding subgroup $\phi(nZ) = \overline{nZ}\subseteq Q$, and since the $nZ$ exhaust the subgroups of $Z$, one of the $\overline{nZ}$ will have to be equal to the canonical copy of $Z$ inside $Q$. But since we know none of them are, because the elements of $Z/nZ$ have bounded order, no such isomorphism can exist. Is that what you intended? – MJD Jan 12 '14 at 15:30
  • I see now! Thanks for your patience. – MJD Jan 12 '14 at 15:41