Is my proof correct? I have made use of the fact isomorphism preserves order of elements, which I proved couple of exercises back. I am also interested in other ways of proving it. Is there a more explicit way or is this explicit enough?

Problem Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.

Solution Recall that isomorphism preserves order of elements and hence if there exists an isomorphism from $\phi: \mathbb{C}-\{0\} \mapsto \mathbb{R}-\{0\}$, then $x \in \mathbb{C} - \{0\}$, and $\phi(x) \in \mathbb{R} - \{0\}$, then $\vert x \vert = \vert \phi(x) \vert$. Now note that the element $i \in \mathbb{C} - \{0\}$ has order $4$. However, no element in $\mathbb{R}-\{0\}$ has order $4$. Hence, no isomorphism can exist. Hence, the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.


  • $\begingroup$ This would be probably best way... I guess there would be no easier way.... well done! $\endgroup$
    – user87543
    Dec 28 '13 at 15:16
  • $\begingroup$ @DietrichBurde I am working on Dummit and Foote. So as I write down the solution, if I am unsure and have some questions, I ask for verification to make sure my understanding is correct. Is it not allowed on this website? $\endgroup$
    – John Smith
    Dec 28 '13 at 16:49
  • $\begingroup$ Your argument also works with the real numbers under addition. $\endgroup$ May 23 '17 at 14:14

Another way would be. Let's take $i$ and see what happen:
a) $f(i^2)=f(-1)=-1$, using basic properties of a morphism.
b) on the other side, we have $f(i)^2=x^2$ that can't produce -1.

  • 3
    $\begingroup$ How do you know that $f(-1)=-1$? $f$ is just an isomoprhism of multiplicative groups, so it's not clear how it interacts with an additive inverse. Of course, you can argue that $-1$ is the unique element of order $2$, but if you're going to do an order argument, you might as well give the argument in the OP. $\endgroup$ Oct 13 '19 at 10:59

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