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Let $G$ be a non-trivial torsion-free Abelian group. If $T$ is a maximal free subgroup of $G$, then $G\over T$ is periodic?
How can we prove this?

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Show that if a coset $gT$ has infinite order in the factor group then $\langle g,T\rangle$ is a free-abelian subgroup of $G$ with proper subgroup $T$, contrary to the hypothesis on maximality of $T$.

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I was trying that way too... But (i'm not an expert on that) free groups are the same of free-abelian groups? I know that a free group is an free-abelian group only when this is an infinite Cyclic group... So i'm stuck in prove that $⟨g,T⟩$ is Cyclic cause if the subgroup $T$ and the subgroup generates by $g$ have no common items, then $⟨g,T⟩$ is direct product. I'm wrong? Thanks for Helping me ;) – W4cc0 Oct 5 '12 at 21:39
@W4cc0 Free groups are generally nonabelian, since the letters (and thus all of the other nonidentity elements) are not allowed to commute. Indeed a free group is only abelian if it is rank $1$, i.e. the infinite cyclic group. Free abelian groups on the other hand, while still generated by a set of letters without any torsion allowed, have commutativity relations imposed so that everything commutes. I assumed the exercise was speaking about free-abelian subgroups (due to $G$ being declared abelian from the get-go). The claim isn't really true otherwise (see e.g. $G=\Bbb Z\times\Bbb Z$). – anon Oct 5 '12 at 21:41
That is, $\Bbb Z\times\Bbb Z$ has maximally free (not free in the abelian sense) subgroup $T=\Bbb Z\times 1$ but the factor group is not periodic. | Back on topic, $\langle g,T\rangle=\langle g\rangle\times T$ as an internal direct product is precisely what you want in order to establish free-abelianness. – anon Oct 5 '12 at 21:42… Here's my real problem (Lemma 2, Paragraph 3) When it says that $H$ is periodic it refers to free-abelian group? I think no. Maybe make uses of the fact that $pG=G$ for infinite primes? – W4cc0 Oct 5 '12 at 21:47
Note that Lang (in Algebra) uses the terminology "free generator" in the way that I indicated: For a set $S$, "[We shall] call $F_{ab}(S)$ the free abelian group generated by $S$. We call elements of $S$ its free generators." – peoplepower Oct 5 '12 at 22:47

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