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Honestly, I have been thinking on this problem for hours but couldn't find a way:

Let $G$ is torsion-free group and $X$ is a maximal independent subset, then $G/\langle X\rangle$ is torsion.

I know:

  1. The main problem is to show that any $g+\langle X\rangle\in \frac{G}{\langle X\rangle}$ has a finite order.

  2. $\langle X\rangle=\sum_{x\in X}\langle x\rangle$.

  3. If the group $G$ is torsion-free and $x\in G$, then the equation $nx=y$ has a unique solution in $G$.

Thanks for any hint.

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I'm guessing by the tag that $G$ is abelian, and by independent you mean linearly independent (over $\mathbb{Z}$). –  Colin McQuillan Oct 9 '12 at 21:25
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1 Answer

up vote 3 down vote accepted

Let $G$ be an abelian torsion-free group, and let $X$ be a maximal linearly independent set. By maximality of $X$, for any $x\in G\setminus X$, there is a linear relation $$ cx + c_1x_1 + \dots + c_kx_k=0$$ for some $k\geq 0$, some integers $c,c_1,\dots,c_k$ not all zero, and some $x_1,\dots,x_k\in X$. Note that since $X$ is independent we cannot have $c=0$. But $cx \in \langle X\rangle$.

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In fact, you say: since for any $x\in G-X$, $G=<X,x>$ then the identity element, $0$, can be written as above linear relation? Right? –  B. S. Oct 10 '12 at 6:49
    
That's not what he said at all! It is not necessarily true that $G = \langle X,x \rangle$ for any $x \in G \setminus X$. –  Derek Holt Oct 10 '12 at 8:06
    
@BabakSorouh: the point is that if $X\cup\{x\}$ were a linearly independent set, then it would be a linearly independent set strictly larger than $X$, which contradicts the maximality of $X$. So $X\cup\{x\}$ is not linearly independent. –  Colin McQuillan Oct 10 '12 at 9:19
    
@ColinMcQuillan: Thanks. Honestly, I pointed in an improper way my question in the first comment. I wanted to be sure of what you noted me nicely in the last comment. Thanks again. –  B. S. Oct 10 '12 at 9:28
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