A question about minimal generating sets. A generating set of a group is termed minimal or irredundant if any proper subset of the generating set, generates a strictly smaller (i.e. proper) subgroup. In other words, no generator can be dropped from the generating set. - http://groupprops.subwiki.org/wiki/Minimal_generating_set
From this I understand that no element of this set can be expressed in terms of others in the same set. Hence, they're all linearly independent. 
On Pg.204 of Herstein's "Topics in Algebra", a line says given any minimal generating set $b_{1},b_{2}\ldots b_{q}$ of M, there must be integers $r_{1},\ldots r_{q}$ such that $r_{1}b_{1}+\ldots r_{q}b{q}=0$ in which not all of $r_{1}b_{1},r_{2}b_{2},\ldots r_{q}b_{q}$ are $0$.
How can this be? 
 A: Let $M=\mathbf{Z}/(2\mathbf{Z})\times\mathbf{Z}/(4\mathbf{Z}).$  Now consider $b_1=(1,1)$ and $b_2=(0,1).$  Then $b_1$ and $b_2$ generate $M,$ and $2b_1+2b_2=0,$ but neither $2b_1$ nor $2b_2$ equals $0.$  The relation $2b_1+2b_2=0$ does not allow us to express either of $b_1$ or $b_2$ in terms of the other since we cannot divide by $2$ here: the relation $b_1+b_2=0$ does not hold.  So the situation that Herstein is describing can indeed occur.
The goal of the Herstein passage is to prove that a finitely generated $\mathbf{Z}$-module is the direct sum of a finite number of cyclic submodules.  That result will follow if there is a generating set of $M$ with the property that any relation of the form $n_1b_1+\ldots+n_qb_q=0$ implies $n_jb_j=0$ for all $j.$  The goal of the proof is to construct a minimal generating set with this property.  By "minimal", Herstein means of smallest cardinality.  This is stronger the definition of minimal that you quote.
I think it might help to carry out the steps of the proof on my $b_1$ and $b_2$ above.  Let $(s_1,s_2)=(2,2),$ so that $s_1b_1+s_2b_2=0.$  It should be clear that in any relation of the form $r_1b_1+r_2b_2=0,$ the coefficient $r_1$ is divisible by $s_1=2.$  Herstein spends a paragraph proving the general analog of this statement.
This example also illustrates that $s_1\mid s_2,$ which is the next thing Herstein proves.  So we let $b_1^*=b_1+m_2b_2$ where $m_2=s_2/s_1=1.$  Observe that $s_1b_1^*=2b_1^*=2b_1+2b_2=0$ and that $\{b_1^*,b_2\}$ generates $M.$  In any relation of the form $r_1b_1^*+r_2b_2=0,$ the coefficient of $b_1^*=(1,2)$ must be even.  Hence $r_1b_1^*=0.$  It follows that $r_2b_2=0$ as well, and therefore that we have found our desired generating set.  In the general case, it might be necessary to repeat the process for $b_2,\ldots,b_q$ to construct the generating set.
A: That Herstein statement is incorrect in general, for example $\mathbb{Z \times Z}$ has minimal generating set $\{(1, 0), (0, 1)\}$ that does not satisfy this property.  I suspect he assumes the group is finite abelian.  Anyway, when $M$ is finite abelian the $b_i$ are not linearly independent, you could simply take $r_i$ to be the order of the element $b_i$.  
For finite groups the key to understanding why minimal generating sets are not linearly independent is that, unlike in linear algebra, we can't always divide by coefficients.  The easy example to keep in mind is that $2$ and $3$ generate all of $\mathbb Z/6$ because $3 - 2 = 1$.  But $\{2, 3\}$ is minimal as a generating set because removing either results in a proper subgroup.  So it could be the case that a multiple of one generator equals a multiple of another generator even though neither generator is itself a multiple of the other.
Edit: After thinking about it, I don't think that lemma from Herstein is correct even for finite abelian groups.  For example $\mathbb Z/2 \times \mathbb Z/2$ is finite abelian with minimal generating set $\{(1, 0), (0, 1)\}$ and
$$r_1(1, 0) + r_2(0, 1) = (0, 0)$$
implies $r_1 = r_2 = 0$ so the statement isn't true.
