$G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ linearly independent over $\mathbb Z$ , then is it true that $|S| \le k$?

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Let $G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ linearly independent over $\mathbb Z$ , then is it true that $|S| \le k$ ?

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Yes, this is true. Introducing an independent generating set, we can just as well assume we are talking about $\mathbb{Z}^k$ and we can consider embedded in $\mathbb{Q}^k$.

Thus, a set $S$ with $|S| \gt k$ will be dependent over the rationals (I assume you know the result for vector spaces). Take this relation and multiply by an integer to clear denominators. You then have a dependence over the integers.

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Consider $G\otimes_ZQ$ it is a $Q$-vector space of dimension $k$. Let $ZS$ the subgroup of $G$ generated by $S$. The dimension of $ZS\otimes_ZQ$ is the cardinality of $S$, which is inferior to $k$ as a subvector space of a $k$-dimensional vector space.