Prove that $\mathbb{R}^*$ is not isomorphic to $\mathbb{C}^*$.

Question

Prove that $\mathbb{R}^*$, the multiplicative group of non-zero real numbers is not isomorphic to $\mathbb{C}^*$ the multiplicative group of non-zero complex numbers.

I am unsure how to go about this proof. I know in order to be isomorphic, the function must be a bijective homomorphism. In order for the function to be a homomorphism, for $a,b\in\mathbb{R}^*$, $\phi(ab)=\phi(a)\phi(b)$ for multiplicative groups.

Answer 1

Count the element of finite order. $\mathbb{R}^*$ has 2 elements of finite order, namely, $1, -1$.

$\mathbb{C}^*$ has $1,-1,i,-i$ and some more, but this is all you need.

It is true that elements of finite order map to elements of finite order in an isomorphism. To see this note that if x is of finite order $n$ in some group $G$, and $f$ is an isomorphism to another group $f(1) = f(x^n) = f(x)^n = 1$

Answer 2

Note that $\mathbb{C}^*$ has elements of order $n$ for all positive $n$, which is not true of $\mathbb{R}^*$. Prove that this means they aren't isomorphic groups.

Answer 3

To the contrary, let $\theta: \mathbb{C}^* \to \mathbb{R}^*$ be an iso. Note that $-i$ has order 4 in $\mathbb{C}^*$. The same is then true for $\theta(-i)$ in $\mathbb{R}^*$. But if $x$ is real with $x^4=1$ then $x= \pm 1$, but neither of these has order 4, a contradiction.

Answer 4

$\phi(i^2)=\phi(-1)=^*-1$. On the other hand, $\phi(i)=x\Rightarrow (\phi(i))^2=x^2$ so we get $x^2=-1$ and as $x\in \mathbb R$ this is a contradiction.

$^*$ is true because $ \phi(1)=\phi((-1)\cdot (-1))=(\phi (-1))^2\Rightarrow \phi(-1)=\phi(-1))^{-1}$ and so either $\phi(-1)=1$ or $\phi(-1)=-1$. But $\phi $ is an isomorphism so the former is impossible.