Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
-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
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
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. □

share|cite|improve this answer
thanks. hmmm.... – Seoral May 24 '11 at 5:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.