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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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    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.)2012-03-01

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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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    @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.2012-03-01
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$(\mathbb{R}, +)$ 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.2012-03-01