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In this question, it is shown that $(\mathbb{R}, +)$ is not a free group. But my question is: if it is not a free group, exactly what relations is it subject to?

My other question is: are there sets other than $\mathbb{R}$ that generate $(\mathbb{R}, +)$?

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For one thing, it is subject to $aba^{-1}b^{-1}=e$. – Alex Becker Mar 1 '12 at 2:32
Guess: Generators: $(g_x)_{0 < x < \varepsilon}$ (with $\varepsilon > 0$). Relations: $g_x+g_y=g_{x+y}$ whenever $0 < x,y,x+y < \varepsilon$. (In the realm of abelian groups.) – Pierre-Yves Gaillard Mar 1 '12 at 5:54
up vote 3 down vote accepted

To see that $(\mathbb{R},+)$ is not a free (abelian) group, note that $\mathbb{R}$ is divisible: given any $r\in\mathbb{R}$ and any $n\gt 0$, there exists $s\in\mathbb{R}$ such that $ns = r$.

But a free (abelian) group is never divisible: if $F$ is a free abelian group, and $x$ is a basis element, then there is no element of $F$ such that $2y=x$. If there were, we would be able to write $y = k_1x_1+\cdots + k_mx_m$ for some distinct basis elements $x_1,\ldots,x_m$ and integers $k_i$. Then $x = 2k_1x_1+\cdots 2k_mx_m$, so then there is at most one nonzero $k_i$, say $k_1$, $x_1=x$, and $2k_1 = 1$, which is impossible. Thus, free abelian groups cannot be divisible, and therefore free groups cannot be divisible (since quotients of divisible groups are divisible, and a free abelian group is a quotient of a free group).

So $\mathbb{R}$ is subject to infinitely many relations, some of which tell you any two elements commute (it satisfies the identity $x+y-x-y$); others that tell you that certain elements are "half" other elements, and so on.

Explicitly producing generating sets is not hard: any cofinite set generates $\mathbb{R}$ as an abelian group; the set of reals on $[0,1)$ generate $\mathbb{R}$.

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Maybe I should post this as a separate question, but doesn't the Steinhaus theorem imply that any set of positive Lebesgue measure generates $(\mathbb{R},+)$? If $A$ is such a set, then $A+A$ contains an interval and this interval generates $\mathbb{R}$ similarly to your last sentence. Are there generating sets of measure zero? – dls Mar 1 '12 at 6:04
Thanks, it makes sense to me now. I'm getting the impression that the idea of generators and relations is in general not straightforward and sort of difficult - would you say this is the case? It would be nice for someone else to say so because then it'll be easier to just accept as being the way it is. – Alex Petzke Mar 1 '12 at 16:09
@AlexPetzke: There are good things about generators and relations, and bad things. For one, there is no algorithm which, given even a finite presentation and a word in the generators, will tell you whether the word represents the trivial element or not (this is called the "Word Problem for Groups"). For uncountably generated groups, such as $\mathbb{R}$, things just get way too nasty to be useful. For finitely, or countably generated groups, they can be extremely useful. – Arturo Magidin Mar 1 '12 at 16:14

$(\mathbb{R}, +)$ is* a $\mathbb{Q}$-vector space whose dimension is the cardinality of $|\mathbb{R}|$. So the main things you need to think about to answer whatever questions you have are:

  • Linear algebra over $\mathbb{Q}$ (e.g. to understand it has** a basis over $\mathbb{Q}$)
  • The group structure of $(\mathbb{Q}, +)$ (e.g. to understand exactly how it fails to be a free abelian group)

*: The vector space structure is uniquely determined from the group structure.

**: This invokes the axiom of choice, I think

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Yes, ($**$) does indeed invoke the Axiom of choice. – Arturo Magidin Mar 1 '12 at 3:26

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