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Is my proof for the question in the title correct?

Note that $\mathbb{R} - \{0\} \subset \mathbb{C} - \{0\}$, since every real number is a complex number. Therefore, since any $\phi: \mathbb{R} - \{0\} \to \mathbb{C} - \{0\}$ would map from a set to another set of bigger cardinality, no such $\phi$ could be surjective. Thus, no isomorphism exists between the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$.

Kurtland Chua
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    The two sets in question have the same cardinality, so there is a surjective map from one to the other. You have to use a different argument. Namely, you have to find something which is true of one of the groups, but not the other. – Michael Albanese Jan 08 '16 at 02:03
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    The argument about cardinality doesn't work because both groups are equicardinal. Is there a cyclic subgroup of order $3$ in $\mathbb{R}^\times$? How about $\mathbb{C}^\times$? – Batominovski Jan 08 '16 at 02:04

4 Answers4

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Lemma: If $X\cong Y$ via $f:X\to Y$, then $x$ and $f(x)$ have the same order.

Hint: What's the order of $i$? Does $\mathbb{R}-0$ have an element with this order?

The Substitute
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If $f$ is an isomorphism between $(C-0,\times)$ and $(R-0,\times)$, $f(i)^2=-1$ impossible

Tsemo Aristide
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    You need to say more about why an isomorphism would send $-1$ to $-1$. (The answer is that in both groups $-1$ is the unique element of order exactly $2$.) – Qiaochu Yuan Jan 10 '16 at 20:38
  • Nobody said that $f(-1)$ has to be $-1$, this is not the argument used here. – Tsemo Aristide Jan 10 '16 at 20:43
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    Okay, then maybe the argument is that $f(i)$ must be an element of order $4$, but if so then you should say that. – Qiaochu Yuan Jan 10 '16 at 20:46
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    I continue to have no idea why you feel the need to be so rude to me. You claim to demonstrate a contradiction but don't say what that contradiction is. There are a few ways to get a contradiction from here but you should spell out which one you intend. I think I'm making a reasonable request for a clarification here and I don't know why you insist on responding with insults. – Qiaochu Yuan Jan 10 '16 at 20:49
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    You have that $f(i)^2 = f(i^2) = f(-1)$, but you haven't provided an argument for why $f(-1) = -1$ (and in fact you claim that you don't use this fact). If you don't think your last comment was rude then why did you delete it? – Qiaochu Yuan Jan 10 '16 at 20:53
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Generally, $$t(R^*,\cdot)=\{+1,-1\}\neq t(C^*,\cdot)=\{\exp(2ki\pi/n)|k,n\in \mathbb{Z}\}$$ $t(G)$ means the torsion subgroup.

Mikasa
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Fixed $p \not= 2$ prime, take all the complex $p$-th roots of $1$. It is a subgroup of order $p$, and every element has thus order $p$ (exept 1). On the contrary there no elements of order $p$ in the real numbers, and no subgroup of order $p$.

Maffred
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