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Let A be the group of positive rational numbers under multiplication. Prove that A is isomorphic to a subgroup B such that A does not equal B.

So I know how prove isomorphisms.(define mapping, prove 1-1, prove onto, and prove operation-preserving). However, I am thrown by the fact that the subgroup does not equal A. Is it that the order is not equal? Thanks in advance for your help.

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  • $\begingroup$ I think you may be able to map 2 to 3 and 3 to 2. $\endgroup$
    – fleablood
    Commented Oct 18, 2015 at 23:09
  • $\begingroup$ $x \mapsto x^2$. $\endgroup$
    – Rob Arthan
    Commented Oct 18, 2015 at 23:32
  • $\begingroup$ @RobArthan the mapping $x \mapsto x^2 $ is not one-to-one nor onto. So it is not an isomorphism candidate that works. $\endgroup$
    – user277658
    Commented Oct 19, 2015 at 3:44
  • $\begingroup$ $x \mapsto x^2$ is one-to-one on $A$ (since the elements of $A$ are positive) and you don't want it to be onto (you want the image $B$ not to be equal to $A$). $\endgroup$
    – Rob Arthan
    Commented Oct 19, 2015 at 8:04
  • $\begingroup$ @RobArthan If it is not onto, how is it an isomorphism? $\endgroup$
    – user277658
    Commented Oct 19, 2015 at 17:05

1 Answer 1

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Hint: Start out by listing some subgroups of A: What's the smallest subgroup containing 1 and 2? 1 and 3? 1 and 2 and 3? Can any of these 3 be isomorphic to A? Can you find a subgroup that is a better potential candidate?

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  • $\begingroup$ I don't see how your suggested approach will help. $\endgroup$
    – Rob Arthan
    Commented Oct 18, 2015 at 23:34
  • $\begingroup$ The 3 examples can't be isomorphic because they're finitely generated. A good subgroup to look at would be one that isn't finitely generated. $\endgroup$
    – mathematician
    Commented Oct 20, 2015 at 9:52
  • $\begingroup$ That's a bit of a roundabout hint, but I see how it might help. Note that the OP's question is about $\Bbb{Q}_{+}$ under multiplication so $1$ is in any subgroup. $\endgroup$
    – Rob Arthan
    Commented Oct 20, 2015 at 23:08

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