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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.

rover2
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4 Answers4

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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 $f(1) = f(x^n) = f(x)^n = 1$

john
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Dionel Jaime
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  • and this fact automatically shows that they are not isomorphic? – rover2 Nov 16 '17 at 04:12
  • Yes: an iso would preserve orders of elements. – Randall Nov 16 '17 at 04:12
  • so the number of elements of finite order in $\mathbb{R}^$ must equal the number of elements of finite order in $\mathbb{C}^$ in order to be an isomorphism? – rover2 Nov 16 '17 at 04:14
  • Yes, which is enough to get what you want, but much more is true, too. (If $\theta: G \to L$ is an iso of groups, then for each $a \in G$ the order of $a$ and the order of $\theta(a)$ are equal.) – Randall Nov 16 '17 at 04:15
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$\phi(i^2)=\phi(-1)=^*-1$. 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.

Matematleta
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  • As a homomorphism of multiplicative groups, how do we get $\phi(-1) = -\phi(1)$? – Randall Nov 16 '17 at 04:24
  • $\phi(1)=\phi((-1)\cdot (-1))=(\phi (-1))^2\Rightarrow \phi(-1)=\phi(-1))^{-1}$ and so either $\phi(-1)=1$ or $\phi(-1)=-1$. But $\phi $ is an isomorphism so the former is impossible. – Matematleta Nov 16 '17 at 04:33
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    @ChilangoIncomprendido I like this. Thanks. – Randall Nov 16 '17 at 04:34
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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.

AJY
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  • @Qudit Shoot, I'm silly. Editing. – AJY Nov 16 '17 at 04:15
  • "$\mathbb{C}^*$ has elements of order $n$ for all positive $n$" Can you please demonstrate this is true, or post a formula to compute such elements... oh the nth roots of unity? i.e. we solve $z^n = 1 $. Thus $\mathcal{O} (1) = 1, \mathcal{O}(-1) = 2 , \mathcal{O}(-\frac{1}{2} + i\frac{\sqrt 3}{2})= 3$. In general it can be shown that $\mathcal{O} ( \cos(2\pi/n ) + i \sin(2 \pi / n ) ) = n $, for $n $ a positive integer. Note that $\mathcal{O}$ is the symbol for order. – john Oct 27 '21 at 19:48
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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.

Randall
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