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?

  • 2
    $\begingroup$ 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
  • 4
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.