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Is the following true?

$G$ is a torsion free abelian group of rank $>n$.

Let $S$ be be a subgroup of $G$ generated by $s_1,s_2, \cdots ,s_n$.

(1) If $m_1s_1+m_2s_2+ ...+m_ns_n$ is linearly independent modulo $p$ for every prime $p$, then $S$ is linearly independent. That is, if $m_1s_1+m_2s_2+ ...+m_ns_n=0$ implies $m_i =0$ (mod $p)$ for all $i$ and every prime $p$, then $m_1s_1+m_2s_2+ ...+m_ns_n=0$ implies $m_i=0$ for all $i$.

(2) Free abelian groups are never divisible.

How do I go about proving statement (2) above? Any hints will greatly be appreciated. thanks

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What is $G$ supposed to be? (The second statement follows directly from the universal property of free abelian groups. If the first statement means what I think it means, you'll have an easier time proving the contrapositive.) –  Qiaochu Yuan May 23 '11 at 12:11
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-1: I can't parse condition (1), even if $G$ is meant to be an abelian group: I don't what a "linearly independent abelian group" is. As for (2): can you prove that $\mathbb{Z}$ is not divisible? Can you reduce to this case? Finally, the title is just about maximally unhelpful. –  Pete L. Clark May 23 '11 at 13:19
    
Sorry for the unhelpful title and unclear question that i have let you all suffer on. I was in a hurry to go out and forgot to state the question clearly. Now i have edited it. Please take a look at it. –  Seoral May 23 '11 at 15:01
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you are still not stating the condition correctly. You mean to say that if $\{ s_1, ... s_n \}$ is linearly independent $\bmod p$ for all $p$ then it is linearly independent in $G$. (But now that you've stated the rest of it in so much detail, shouldn't it be obvious? If the condition holds for all $p$ then the $m_i$ are integers divisible by all primes...) –  Qiaochu Yuan May 23 '11 at 15:24
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1 Answer 1

up vote 2 down vote accepted

Part 1:

Definition: Suppose G is an abelian group, and S is a subset of G generating the group H. For each prime p, the group H/pH is a vector space over the field Z/pZ of p elements. Consider the set S + pH = { s + pH : s in S } in the group H/pH. We say that S is linearly independent mod p if S + pH is a linearly independent set in the vector space H/pH over Z/pZ.

Proposition: Suppose G is an abelian group, and S is a subset of G that is linearly independent mod p for every prime p. Then S is a basis of a free abelian subgroup of G.

Proof: Suppose Σ as s = 0 for some integers as. Then reducing the equation mod p produces a linear dependence relation in the vector space H/pH. Hence each as ≡ 0 mod p, by the definition of linearly independent. In particular, as is an integer divisible by every prime p. The only such integer is 0, and as = 0 for all s in S. This means that S is "independent", as in, it is a basis of the free abelian subgroup it spans. □

As far as I can tell, you don't need to assume G is torsion-free (but, the subgroup generated by S is torsion-free), and you don't need to assume S is finite. Note that no subgroup can possibly by linearly independent mod p according to my definition. You want a subset S, not a subgroup, even though you really are quite interested in the subgroup H it generates.


Part 2:

This is sort of a restatement of the previous part.

Proposition: Let G be a free abelian group with basis S. Then no element of S is divisible by any prime number p.

Proof: Suppose x in S is divisible by p, that is, suppose there is some y in G such that py = x. Write y as a linear combination of S: y = Σ as s. Then x = py = Σ (pas) s. Since S is a basis, the coefficients are uniquely determined and so we get 1 = pax and 0 = pas for sx, s in S. These are just equations of integers, and canceling p one gets ax is the "integer" 1/p, a contradiction. □

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thanks. hmmm.... –  Seoral May 24 '11 at 5:17
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