Let $G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ generating $G$ such that $|S|=k$ , then is it true that the elements of $S$ are linearly independent over $\mathbb Z$ ?
I can see that for a free abelian group of rank $k$ , not every linear independent subset of cardinality $k$ generates $G$ ( ex. $\{2\}$ is linear independent but does not generate $\mathbb Z$ ) and I am asking here whether every generating set of cardinality $k$ is linear independent or not .
Yes, this is true. For every prime number $p$, the quotient $G/pG$ is a $k$-dimensional vector space over $\mathbb Z/p \mathbb Z$ and the image of $S$ forms a basis. Now suppose we have a relation $\sum_{s \in S} a_s s = 0$ in $G$. Then we get a relation in $G/pG$ with coefficients in $\mathbb Z/p \mathbb Z$, hence $a_s \equiv 0 \bmod p$ for all $s \in S$. Since $p$ is an arbitrary prime, we get $a_s = 0$ for all $s \in S$, hence $S$ is linearly independent.
Here is another proof. We have a surjective homomorphism $\varphi: \mathbb Z^k \twoheadrightarrow G$, given by $\varphi(a_1,\ldots,a_k) = \sum_{i = 0}^k a_i s_i$ where $S = \{s_1,\ldots,s_k\}$. Let $K$ be its kernel, so we have a short exact sequence $$0 \to K \to \mathbb Z^k \to G \to 0.$$ As a subgroup of $\mathbb Z^k$, the group $K$ is finitely generated and torsion-free, hence free. Since the rank is additive in short exact sequences, we have $\operatorname{rk}(K) = 0$, hence $K = 0$ and $\varphi$ is an isomorphism.
One can also use the vector space result in another way.
Without loss we assume that $G$ is $\mathbb{Z}^k$ and we embed it into $\mathbb{Q}^k$. (One could also consider a tensor product.)
The set $S$ is then also generating/spanning set for the vector-space $\mathbb{Q}^k$. See https://math.stackexchange.com/a/1912310/ for details.
Since its cardinality is equal to the dimension of the space the set is a vector space basis and in particular linearly independent over the rationals.
However, this is (formally) even more restrictive than being independent over the integers.
