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I have seen this answer, see here.

To prove multiplicative groups $\mathbb{R}−\{0\}$ and $\mathbb{C}−\{0\}$ are not isomorphic, they have used the fact that if $\varphi:G \to H$ is an isomorphism then for all $x \in G$.

We have $|x|=|\varphi(x)|$. So let there is isomorphism $\varphi$ between the multiplicative groups $\mathbb{R}−\{0\}$ and $\mathbb{C}−\{0\}$. Now $i \in \mathbb{C}−\{0\}$ has order $4$ as $i^4=1 so |i|=4$. But there is no element of order $4$ in $\mathbb{R}−\{0\}$. So there is no isomorphism between these multiplicative group. And in that answer the proof ends here.

But I think we have to proof the statement that there doesn't exist element of order $4$ in $\mathbb{R}−\{0\}$. So suppose there does exist element $x \in \mathbb{R}−\{0\}$ such that $|x|=4$. Now $x^4=1$. Now as $x \in \mathbb{R}−\{0\}$ so we can treat this $x$ as $4-th$ root of unity, so the roots are $1, \alpha,{\alpha}^2,{\alpha}^3$ where $\alpha=e^{\frac{2\pi i}{4}}$. But $\alpha,,{\alpha}^3 \notin \mathbb{R}-\{0\}$ and $|1|=1\neq 4$ and $|{\alpha}^2|=|-1|=2$. So none of the roots have order $4$. So there doesn't exist element of order $4$ in $\mathbb{R}−\{0\}$.

Is it correct now?

Shaun
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Alexander
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    $\alpha^2=-1\in\mathbb R$ but it has order $2$. An easier way to solver $x^4=1$ is just to factor $x^4-1=(x^2+1)(x^2-1)=0$ and $x^2+1>0$ over $\mathbb R$, hence $x^2=1$, $x=\pm 1$. – Just a user Aug 02 '22 at 04:38
  • @Justauser oh okay thanks. now i have edited that$ |{\alpha}^2|=2$ – Alexander Aug 02 '22 at 04:42
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    $C_n={z\in \Bbb{C}\setminus{0}: z^n=1}$ cyclic group of order $n$ but only elements of $\Bbb{R}\setminus {0}$ of finite orders are $\pm 1$ – Sourav Ghosh Aug 02 '22 at 04:42
  • @Alexander Now it looks good, but I still prefer the factorization. – Just a user Aug 02 '22 at 04:43
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    Just a comment: it can be confusing to use the notation $|x|$ to denote the order of an element when that element is a real/complex number, since it's easy to confuse it with the norm. – Captain Lama Aug 02 '22 at 13:23

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