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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.

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marked as duplicate by Batominovski, Community Nov 16 '17 at 4:24

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    $\begingroup$ Count elements of order 4. $\endgroup$ – Randall Nov 16 '17 at 4:10
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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.

Elements of finte 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 $f(1) = f(x^n) = f(x)^n = 1$

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  • $\begingroup$ and this fact automatically shows that they are not isomorphic? $\endgroup$ – rover2 Nov 16 '17 at 4:12
  • $\begingroup$ Yes: an iso would preserve orders of elements. $\endgroup$ – Randall Nov 16 '17 at 4:12
  • $\begingroup$ 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? $\endgroup$ – rover2 Nov 16 '17 at 4:14
  • $\begingroup$ 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.) $\endgroup$ – Randall Nov 16 '17 at 4:15
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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.

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  • $\begingroup$ As a homomorphism of multiplicative groups, how do we get $\phi(-1) = -\phi(1)$? $\endgroup$ – Randall Nov 16 '17 at 4:24
  • $\begingroup$ $\phi(1)=\phi((-1)\cdot (-1))=(\phi (-1))^2\Rightarrow \phi(-1)=\phi(-1))^{-1}$ and so either $\phi(-1)=1$ or $\phi(-1)=-1$. But $\phi $ is an isomorphism so the former is impossible. $\endgroup$ – Matematleta Nov 16 '17 at 4:33
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    $\begingroup$ @ChilangoIncomprendido I like this. Thanks. $\endgroup$ – Randall Nov 16 '17 at 4:34
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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.

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  • $\begingroup$ @Qudit Shoot, I'm silly. Editing. $\endgroup$ – AJY Nov 16 '17 at 4:15
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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.

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