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Prove that the groups \(\displaystyle{\left ( R,+ \right )}\) , \(\displaystyle{\left ( R-\left \{ 0 \right \},\cdot \right )}\) are never isomorphic, where \(\displaystyle{\left ( R,+,\cdot \right )}\) is a field.

Nice question! Suppose \(\varphi: (R,+) \to (R \setminus \{0\},\cdot)\) is an isomorphism. We must have \(\varphi(0)=1\). There is an \(x \in R\) with \(\varphi(x)=-1\). Then \(\varphi(2x) = \varphi(x)^2 = 1 = \varphi(0)\). So we must have \(2x = 0\) and so either \(x=0\) or \(2_R = 0\). In both cases we get that \(R\) has characteristic \(2\). Now observe that \(1 = \varphi(0) = \varphi(1+1) = \varphi(1)^2.\) So (as we have characteristic \(2\)) we get \( (\varphi(1)-1)^2 = 0\) giving (since \(R\) is a field) \(\varphi(1)=1\). But since \(1_R \neq 0_R\) then \(\varphi\) is not injective, a contradiction.

Thank you Demetres Christofides for your nice and short solution. Although ,the part you write that \(\displaystyle{R}\) has characteristic
\(\displaystyle{2}\) wasn't so clear to me.Maybe it's a matter of symbolism.Anyway i explained it like this , if \(\displaystyle{x=0}\) then \(\displaystyle{\varphi \left ( 0 \right )=-1}\) so
\(\displaystyle{1=-1}\) and \(\displaystyle{r=-r}\) for every \(\displaystyle{r \in R}\) and \(\displaystyle{R}\) has characteristic \(\displaystyle{2}\), or
\(\displaystyle{x\neq 0}\) and so there is \(\displaystyle{x^{-1}}\) and then \(\displaystyle{xx^{-1}+xx^{-1}=0}\) so \(\displaystyle{1=-1}\) ,again
\(\displaystyle{R}\) has characteristic 2 .

Hi Alkes. I get that the characteristic is equal to \(2\) in exactly the same way as you do. In fact, for \(x \neq 0\) we can be quicker. Since we are in a field we know that \(2x=0\) and \(x\neq 0\) imply that \(2=0\). (We can take this for granted, but the proof is the one that you give.)