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?
Maximal Free Subgroups and Torsion
3
$\begingroup$
abstract-algebra
group-theory
free-groups
1 Answers
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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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0I 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 ;) – 2012-10-05
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0@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. $G=\Bbb Z\times\Bbb Z$). – 2012-10-05
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0That 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. – 2012-10-05
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0http://archive.numdam.org/ARCHIVE/RSMUP/RSMUP_1983__69_/RSMUP_1983__69__153_0/RSMUP_1983__69__153_0.pdf 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? – 2012-10-05
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0@W4cc0 Periodic is another name for torsion. It does not mean free-abelian. (Sorry for the delay, I had to drop my brother off somewhere.) – 2012-10-05
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0I know it xD What i really want to understand is how can we say that $H$ is periodic... I thinked that it comes only from the fact that $T$ is a maximal free subgroup, but after your answers i must conclude that also the fact that $pG=G$ (for infinite primes) is involved. The question is How? xD (sorry for my bad english, I can't really be more clear than this) – 2012-10-05
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0No, my answer sketches a proof-by-contradiction. Suppose $H=G/T$ *isn't* periodic. Then there is an element (coset of $T$) $gT\in H$ that has infinite order. If $gT$ has infinite order in $H$, then $\langle g,T\rangle$ is a free abelian subgroup of $G$ that is even greater (in the set of subgroups ordered by inclusion) than the maximally free-abelian subgroup $T$, which is a contradiction. (Nothing is strictly greater than a maximal element!) Hence $H$ must be periodic. There is nothing like $pG=G$ involed in this argument, and I fail to see why you would think so after reading my answer... – 2012-10-05
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0Cause $⟨g,T⟩$ is a free-abelian group while $T$ is maximal with respect to condition of being only a free group :S What i miss? XD – 2012-10-05
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0I believe $T$ is maximal with respect to being a free-*abelian* group (IOW, I believe the "abelian" part of the adjective was left out in the source because the author felt, evidently wrongly, that it would be unnecessary). As I said, the claim "Let $T$ be a maximal free subgroup of [abelian, torsion-free] $G$; then $0\ne T$ and $H=G/T$ is periodic" is **not true** if we are talking about free groups but not talking about them in the abelian sense (I gave a counterexample even). – 2012-10-05
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0The lemma continues... "If $x$ is a free generator of $T$ [...]" This mean that $T$ is a cyclic infinite group and this, in general, happen when $T$ is a free-group and Abelian; otherwise why $T$ must be generate by only one generator? – 2012-10-05
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0Possibile it means a random generator? Not implicating that $T$ is cyclic? XD If so, i'm an idiot u.u Anyway this conversation make me realise this, so thanks ;) – 2012-10-05
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0It means merely that $x$ is an element of some basis ([As defined here](http://en.wikipedia.org/wiki/Free_abelian_group)) for $T$. – 2012-10-05
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1Note 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*." – 2012-10-05
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0Perfect! Thanks you too! Now it's all clear :D (at least for now) – 2012-10-05