I learnt that $(\mathbb{R},\times) < (\mathbb{C},\times)$, Which means the first is a subgroup of the second one. But in the first group inequality is defined, while it's not in the latter. This got me thinking, is there anything about a group which tells you which symbols ($=,\neq,<,\geq,\cdots$) are defined within a group? Shouldn't these be explicitly stated when defining a group, so you should have $(\mathbb{G},\circ,=)$, with a group, an operation, and something to compare elements within the group?
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@user1729 No, it was defined to be $(x^2)^{-1}$. Was that definition wrong? – Frank Vel Jan 21 '15 at 13:55
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How you define inequality in the first group? – Ofir Schnabel Jan 21 '15 at 13:55
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1Sorry, I didn't read your first sentence carefully enough. They do commute, because $x^{-1}x^{-1}\ast x^2=1$, so $(x^{-1})^2=(x^2)^{-1}$. – user1729 Jan 21 '15 at 13:56
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@OfirSchnabel well I know you can use $\geq$ between real numbers, but that this doesn't work for complex numbers. Or should I not apply my thinking this way? – Frank Vel Jan 21 '15 at 13:58
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1But $\leq $ is for additive group of $\mathbb{R}$. Anyway the answer is no. Yow do not consider additional structre. – Ofir Schnabel Jan 21 '15 at 13:59
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@user1729 Thanks I also realised it's a special case of $(x\circ y)^{-1} = y^{-1}\circ x^{-1}$ where $x=y$ – Frank Vel Jan 21 '15 at 14:15
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I've removed the first question, as it was answered in the comments while both answers address the second, more interesting, question. In general, @fvel, please ask just one question at a time. It makes everything neater and tidier. – user1729 Jan 22 '15 at 09:36
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- When talking about $\mathbb R$ and $\mathbb C$ as multiplicative groups, we only discuss the structure of multiplication, and not anything else. Thus any other structure of $\mathbb R$ or $\mathbb C$ is irrelevant.
In particular, having an order is not part of the structure of being a group. So the answer is no - there is nothing in the definition of group which tells you which other structures you might want to define on that set.
Fredrik Meyer
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@GPerez Thanks, it's hard to unlearn familiar things when learning new unfamiliar things... – Frank Vel Jan 21 '15 at 15:20
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Inequality. There is always a notion of "inequality" in groups. In a group $G$ two elements $g, h\in G$ are equal if and only if $gh^{-1}=1$. Therefore, I do not understand your claim that inequality is not defined on $\mathbb{C}$ but is on $\mathbb{R}$†.
However, what is true is that there exist groups where it is undecidable if a combination of elements (a word) $g_1g_2\cdots g_n$ is equal to the trivial element. The key term is "the word problem for groups", and it is known to be insoluble in general. I wrote a lengthy expository math.SE answer on this topic here.
Order. There is a notion of "order" in a group. However, you have to be careful, because groups are not necessarily commutative. We talk about "left-ordered groups", "right-ordered groups" and "bi-ordered groups".
- A group $G$ is left-ordered if there exists a total order $<$ on the set $G$ such that if $h<k$ then $gh<gk$ for all $g\in G$.
- A group $G$ is right-ordered if there exists a total order $<$ on the set $G$ such that if $h<k$ then $hg<kg$ for all $g\in G$.
- A group $G$ is bi-ordered if it is both left and right ordered.
However, a group under a given operation is either (for example) left-orderable or not left-orderable. That is, it is not something we "give" to a group, but rather something we may or may not be able to find. A nice illustration of this idea (but it is about fields rather than groups) can be found in this math.SE answer of mine: the complex numbers $\mathbb{C}$ can be ordered in such a way that addition is respected, and (omitting $0$) in such a way that multiplication is respected, but not in such a way that both are respected. However, the reals $\mathbb{R}$ can be ordered in a way which respects both addition and multiplication.
† Having thought about this for a bit, I think both $\mathbb{R}$ and $\mathbb{C}$ should have insoluble word problem. So inequality does not "work" in these groups. Because of this thought I asked a related question here.
