Multiplicative Group of Complex Numbers is not Isomorphic to Multiplicative Group of Real Numbers

Theorem

Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.

Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.

Then $\struct {\C_{\ne 0}, \times}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.

Aiming for a contradiction, suppose $\struct {\C_{\ne 0}, \times}$ is isomorphic to $\struct {\R_{\ne 0}, \times}$.

Let $\phi: \C_{\ne 0} \to \R_{\ne 0}$ be an isomorphism.

Note that $\order i = 4$ in $\struct {\C_{\ne 0}, \times}$.

$\order {\map \phi i} = 4$ in $\struct {\R_{\ne 0}, \times}$

Then $\paren {\map \phi i}^4 = 1$.

So $\paren {\map \phi i}^2 = \pm 1$.

In $\R$ we must have $\paren {\map \phi i}^2 \ge 0$.

Hence $\paren {\map \phi i}^2 = 1$.

This gives $\map \phi i = \pm 1$.

However $\order 1 = 1$ and $\order {-1} = 2$ in $\struct {\R_{\ne 0}, \times}$.

Thus there is no order $4$ element in $\struct {\R_{\ne 0}, \times}$, and thus $\map \phi i$ does not exist.

Therefore $\phi$ is not an isomorphism.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.9$