$G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ that generates $G$ then is it true that $|S| \ge k$?

Question

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.

Answer 1

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 $k$.

Answer 2

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 $|S|<k$.