Let $G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ that generates $G$ then is it true that $|S| \ge k$ ?
I don't know tensor product, and thus would appreciate an answer that avoids the use of this notion.
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1@PeterFranek : My question is more elementary I suppose and is not concerned with free groups – Sep 02 '16 at 13:55
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@SaunDev The argument still works: By assumption, you have an onto homomorphism $\Bbb Z^{|S|}\to G$ and by tensoring with $\Bbb Q$ this leads to an onto linear map $\Bbb Q^{|S|}\to \Bbb Q^k$ – Hagen von Eitzen Sep 02 '16 at 14:01
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1@HagenvonEitzen : I don't know tensor product , isn't there any elementary way ? – Sep 02 '16 at 14:03
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Without loss we can assume that $G$ is $\mathbb{Z}^k$, and we embed it into $\mathbb{Q}^k$.
If the set $S$ generates $\mathbb{Z}^k$ as an abelian group, then it follows that it generates $\mathbb{Q}^k$ as a rational vector space.
To see this let $(q_1, \dots, q_k) \in \mathbb{Q}^k$. Letting $d$ denote the least common denominator, we have $(dq_1, \dots, dq_k) \in \mathbb{Z}^k$.
Thus, as $S$ generates $\mathbb{Z}^k$ we have $(dq_1, \dots, dq_k) = \sum_{s\in S} z_s s$ with integers $z_s$. It follows that $(q_1, \dots, q_k) = \sum_{s\in S} \frac{z_s}{d} s$
Thus, $S$ contains a vector space basis, that is it has a subset of size $k$.
quid
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thanks ... this technique should be applicable for this http://math.stackexchange.com/questions/1912070/let-s-be-a-generating-subset-of-a-free-abelian-group-of-rank-k-s-t-s-k too right ? – Sep 02 '16 at 17:15
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The answer given by quid above is perfect. Here is another argument. Assume, to the contrary, that $|S| < k$. By definition of $S$ being a generating set for $G$, we have a surjective map of abelian groups: $$ f: \mathbb{Z}^{|S|} \to |G| \cong \mathbb{Z}^{k} $$ We also have a projection map from $\mathbb{Z}^{k}$ that sends a $k$-tuple into its first $|S|$ coordinates: $$ \operatorname{pr}: \mathbb{Z}^k \to \mathbb{Z}^{|S|} $$ Composing this with $f$, we get a map $\varphi = f\circ\operatorname{pr}: \mathbb{Z}^{k}\to \mathbb{Z}^{k}$ which is surjective (as both $f$ and $\operatorname{pr}$ are surjective). It is a theorem that a surjective endomorphism of a finitely-generated module is an isomorphism (there is a slick proof due to Vasconcelos -- see Georges' answer here), thus $\varphi$ is injective. This forces $\operatorname{pr}$ to be injective as well (from $\varphi = f\circ\operatorname{pr}$), and that is a contradiction since $|S|<k$.
