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When constructing a Free Group $F(S)$ from a set $S$ with only one element, then the resulting Free Group is isomorphic to the additive group of integers.

Now I wonder whether there is an Isomorphism from some abelian Free Group $F(S_R)$ (without further building any Quotient Group from it) to the additive Group of reals and if so, then what would be the Set $S_R$ from which such an abelian Free Group is created?

Thanks for all answers.

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    $\begingroup$ The question is odd: "to construct the additive group of reals as a free group" would mean the reals is a free (abelian) group...but it isn't. $\endgroup$ – DonAntonio Nov 14 '16 at 8:45
  • $\begingroup$ Okay, how should I rephrase it? Or do you mean that there is also no Isomorphism from any abelian Free Group to the additive Group of reals and if so why not? $\endgroup$ – exchange Nov 14 '16 at 8:48
  • $\begingroup$ I'm not sure what you had in mind to begin with, so to "rephrase it" is something I can't understand: what is it there? Now, of course there is a homomorphism from some free abelian group onto $\;\Bbb R\;$ , for example take the free abelian group $\;A\;$ on a set with cardinality the continuum. The existence of the wanted homomorphism $\;A\to\Bbb R\;$ follows at once from the universal property of free abelian groups. $\endgroup$ – DonAntonio Nov 14 '16 at 8:52
  • $\begingroup$ @exchange Your question is a little bit confusing. The point is that any free group with more than one generator is not commutative (the free group generated by one element is the integers, but that is more an exception than a rule). So, unless you admit some sort of quotients by relations in the group (such as $xy=yx$, but possibly others) you will never get the additive group of the real numbers. $\endgroup$ – Darío G Nov 14 '16 at 8:56
  • $\begingroup$ @DonAntonio: I know of course that by definition there is a Homomorphism from any Free Group to any other Group. I wanted to know whether one can construct an Isomorphism without building a Quotient Group and suggested myself that the Free Group had to be abelian! $\endgroup$ – exchange Nov 14 '16 at 9:35
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There is no isomorphism from a free abelian group to the additive group of reals, because the latter is not free.

A free abelian group is reduced, that is, it has no (nonzero) divisible subgroup. This is easily proved once you realize that a free abelian group $F$ is a direct sum of copies of $\mathbb{Z}$. So, given $x\in F$, $x\ne 0$, you can surely find $n>0$ such that $ny\ne x$, for every $y\in F$.

Since the additive group of $\mathbb{R}$ is clearly divisible, it follows this group is not free.

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  • $\begingroup$ It might be worth adding that in some way this is the only reason, since the the reals, being isomporphic to a direct sum of groups isomorphic to $\mathbb{Q}$, are the divisible hull of a free abelian group. $\endgroup$ – quid Nov 14 '16 at 9:55
  • $\begingroup$ @quid There are many other reasons for a group not to be free: $\mathbb{Z}[2^{-1}]$ in not free, but has no divisible subgroup either. $\endgroup$ – egreg Nov 14 '16 at 9:58
  • $\begingroup$ @quid: Could you elaborate on what a divisible hull of a free abelian Group is? Thanks. And also it is very interesting that the reals are isomorphic to a direct sum of groups isomorphic to $\mathbb{Q}$. How to construct that isomorphism? $\endgroup$ – exchange Nov 14 '16 at 10:02
  • $\begingroup$ And also, if this is true that the reals are a direct sum of groups isomorphic to $\mathbb{Q}$ and $\mathbb{Q}$ is countable, that is, there is an Isomorphism from $\mathbb{Z}$ to $\mathbb{Q}$, then why is it not possible to construct an Isomorphism from a direct sum of $\mathbb{Z}$ to $\mathbb{R}$? $\endgroup$ – exchange Nov 14 '16 at 10:04
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    $\begingroup$ @exchange Not really so bad, but it could be phrased in a much better way: “is the additive group of reals a free abelian group”? $\endgroup$ – egreg Nov 14 '16 at 11:41

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