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Given that there is the additive group $\mathbb Q$ of rational numbers, and the multiplicative group $\mathbb Q^*$ of non-zero rational numbers, prove that $(\mathbb Q,+)$ is not isomorphic to $(\mathbb Q^*,\times)$.

How many methods can you think of and can you provide a complete solution?

I am a self-learner of maths and feel difficult to offer a rigorous proof, but here are my thoughts:

  1. I could try to assume a isomorphism $\theta$ exists between the two groups and prove that $\theta$ cannot exist.

  2. I could try to find some property which should preserve under isomorphism but is satisfied only by one of the groups.

However I could not proceed in either direction, could someone please help?

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    $\begingroup$ Do you have a non trivial element of finite order in the first group ? in the second? $\endgroup$ – Clément Guérin Nov 5 '15 at 10:50
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    $\begingroup$ additive, not addictive... $\endgroup$ – lhf Nov 5 '15 at 10:51
  • $\begingroup$ @Rescy_: It would be more fun to discuss the isomorphism question when the second group is taken to be consisting of only the positive rational numbers (under multiplication). $\endgroup$ – P Vanchinathan Nov 5 '15 at 11:02
  • $\begingroup$ the "property" method is almost always the way to go. $\endgroup$ – hunter Nov 5 '15 at 11:07
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Check for the existence of elements of order $2$.

In the light of this your next question is probably about $(\Bbb Q^*_{>0},\times)$. Then we can resort to the fact that $(\Bbb Q,+)$ is divisible.

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  • $\begingroup$ Great method! Actually my next question is to prove $(\mathbb R,+)$ being not isomorphic to $(\mathbb R^*,\times)$, which can also be solved using your method. Many thanks! $\endgroup$ – Rescy_ Nov 5 '15 at 10:56
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This is true for any field $k$: The equation $x^2=1$ in $k^*$ corresponds to the equation $2x=0$ in $k$; however, they always admit different number of solutions.

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  • $\begingroup$ @lhf If $k$ is of char$=2$, then $2x=0$ always hold, but $x^2=1$ admits only one solution. $\endgroup$ – user148212 Nov 5 '15 at 19:05
  • $\begingroup$ you're right, I misread. $\endgroup$ – lhf Nov 5 '15 at 19:07

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