0

Evidently, $\{0\}$ has to be excluded since it has no inverse. My question is reduced to: Is there any total order on complex numbers w/o $0$? From what I sense (but not 100% sure), the lexicographical order (also known as lexical order) is a total order but not linear order.

The answer may be implied by this publication: Levi, F.W. (1942), "Ordered groups.", Proc. Indian Acad. Sci., A16: 256–263.

If this: http://mathworld.wolfram.com/TotallyOrderedSet.html is correct. However, kindly notice that $<$ is a total order for $\mathbb{R} \setminus \{0\}$ and $-1 < 1$ is true but when we multiply both sides by $-1$, we get a rather surprising result $1 < -1$ Hence, it is not a liner order for the multiplication. Any comments? Interpretation of it?

Alex Vong
  • 1,654
Jan Tamariani
  • 7
  • 3
  • It seems that alo groups (abelian linearly ordered groups) are not so popular. Mathematica http://mathworld.wolfram.com/TotallyOrderedSet.html equalizes "total order" with linear order hence R{0} should be alo group by Levi, F.W. (1942), "Ordered groups.", Proc. Indian Acad. Sci., A16: 256–263. Can anyone of you, great mathematicians, confirm it before I self-admit myself in mental institution? :) – Jan Tamariani Dec 04 '18 at 19:23
  • According to Wikipedia, linear order, total order and simple order all have the same meaning. – Alex Vong Dec 22 '18 at 00:37
  • @AlexVong: That is why you say "abelian linearly ordered group" or some other combination of those words. It is not only required to be a linear order, that order is also supposed to be compatible with the group structure. – GEdgar Dec 22 '18 at 01:11
  • @GEdgar Maybe this is exactly what the OP is missing: the linear order on $\mathbb{C} \setminus {0}$ has to be translation-invariant as well – Alex Vong Dec 22 '18 at 01:16
  • No because $(-1)^2 = 1$. – bof Dec 22 '18 at 01:19
  • There isn't a total order on $\mathbb{C}$. – Thomas Davis Dec 22 '18 at 01:46

1 Answers1

1

Firstly, the terminology linear order, total order and simple order all have the same meaning. Therefore, linearly ordered group is sometimes also called totally ordered group or simply ordered group. From now on, I will use the terminology linear order and linearly ordered group.

Secondly, as GEdgar has pointed out, the linear order on an abelian linearly ordered group has to be translation-invariant as well. For example, let $(G, \cdot, <)$ be an abelian linearly ordered group, then we must have $a < b$ implying $ac < bc$ for all $a, b, c \in G$.

The answer to your question is no. It is not possible to equip the abelian group $(\mathbb{C} \setminus \{0\}, \cdot)$ with a translation-invariant linear order $<$.

Firstly, we claim that all abelian linearly ordered group $(G, \cdot, <)$ must be torsion-free. (Recall a torsion-free group is a group where all non-identity elements have infinite order.) Suppose not, say $g^n = 1$ for some $g \ne 1, n \in \mathbb{N}$. By trichotomy, either $g < 1$ or $g > 1$. Say $g > 1$. Multiplying $g$ on both sides gives $$g^2 > g$$ $$g^3 > g^2$$ $$\vdots$$ $$1= g^n > g^{n - 1}$$ Chaining these inequalities gives $1 > g^{n - 1} > \cdots > g^2 > g > 1$ which is a contradiction. The proof is similar for $g < 1$.

Finally, observe that $(\mathbb{C} \setminus \{0\}, \cdot)$ is not torsion-free since $-1$ is of order $2$ (see bof's comment). Therefore, it cannot be an abelian linearly ordered group.

Alex Vong
  • 1,654
  • Thank you for helping me to sort it out. There seem to be even more controversy with complex numbers w/o 0 but the torsion free requirement solves it (i^2=-1). – Jan Tamariani Dec 23 '18 at 01:53