Consider the following problem from Hungerford algebra:
A subset $X$ of an abelian group $F$ is said to be linearly independent if $n_1x_1 + \ldots + n_kx_k = 0 $ always implies $n_i = 0$ for all $i$ (where $n\in \mathbb{Z}$ and $x_1, \ldots , x_k$ are distinct elements of $X$).
Prove that
(a) If $F$ is free abelian, it is not true that every linearly independent subset of $F$ may be extended to a basis of $F$.
(b) If $F$ is free abelian, it is not true that every generating set of $F$ contains a basis of $F$. However, if $F$ is also finitely generated by $n$ elements, $F$ has rank $m\leq n $.
I think a is false, as let a linearly independent subset $\{x_1 , \ldots , x_k \}$, then it can be extended to basis by simply adding the basis elements which are not in span of $\{x_1, \ldots ,x_k \}$.
For b, I am clear with the part that which is after however but couldn't find reasoning of before however part.
(a) Let $F= \Bbb Z, X= \{ 2 \}$.
(b) Let $F=\Bbb Z, X=\{ 2, 3 \}$.
Adding elements that are not in the span of $X$ may destroy linear independence.