Showing that $(\mathbb{R} \setminus \{ 0 \}, \, \times) \not \cong (\mathbb{C} \setminus \{ 0 \}, \, \times)$

Question

I'm trying to show that $(\mathbb{R} \setminus \{ 0 \}, \, \times) \not \cong (\mathbb{C} \setminus \{ 0 \}, \, \times)$ as follows: note that there exists an element (namely $i$) in $\mathbb{C} \setminus \{ 0 \}$ that has order $4$, but no element of $\mathbb{R} \setminus \{ 0 \}$ has order $4$.

I have the intuition for why this is true -- the only element in $\mathbb{R} \setminus \{ 0 \}$ that has order not equal to $1$ or $\infty$ is $-1$, which has order $2$.

Can anyone think of a way to show rigorously that no element in $\mathbb{R} \setminus \{ 0 \}$ has order $4$?

Suppose that $x\neq 1,0$. If $|x|>1$; $x^4>|x|>1$. If $|x|<1$, $x^4<|x|<1$. — Pedro, Sep 17 '14 at 01:17
Hint: $x^4 - 1 = (x^2 - 1)(x^2 + 1)$. If $x$ has order $4$, it must be a root of which factor? — Viktor Vaughn, Sep 17 '14 at 01:21

score 2 · Answer 1 · answered Sep 17 '14 at 01:17

2

Hint: how many solutions does $x^{4}-1 = 0$ have in $\mathbb{R}$? What are those solutions, and their respective orders in $\mathbb{R}^{\times}$?

answered Sep 17 '14 at 01:17

Alex Wertheim

20,278

score 0 · Answer 2 · answered Sep 17 '14 at 01:28

0

You could verify that if $|x|>1$ then $|x^n|>1$, so $x$ does not have finite order. Similarly if $|x|<1$, then $|x^n|<1$, so again $x$ doesn't have finite order. So your intuition is correct. The only finite order elements are $\pm1$.

answered Sep 17 '14 at 01:28

Cheerful Parsnip

27,278

Showing that $(\mathbb{R} \setminus \{ 0 \}, \, \times) \not \cong (\mathbb{C} \setminus \{ 0 \}, \, \times)$

2 Answers2