My professor gave us this as an example of a module (I guess my professor were comparing the linear independence in a vector space to that in a module):
In a vector space $V$ over a field $k,$ we have if $v \in V, v \neq 0.$ Then $$cv = 0 \textbf{ for } c \in k \Leftrightarrow c = 0.$$ So, $$\{c \in k| cv = 0 \} = \{0\}.$$
$G \cong \mathbb Z_n$ is a $\mathbb Z$-module. $g \in G, g \neq 0, \{m \in \mathbb Z| mg = 0\} \supset n\mathbb Z$ and if $\langle g \rangle = \mathbb Z_n$ then $\{m \in \mathbb Z| mg = 0\} \supset n\mathbb Z.$
And I am for my life do not understand the idea my professor wanted to convey. Could anyone help me in trying to understand my professor mind please?
This example shows that while a single nonzero vector $v$ is always linearly independent ($cv=0\implies c=0$), this doesn't hold for modules over rings.
Indeed, no single element of the $\Bbb Z$-module $\Bbb Z_n$ can be linearly independent, since we have $nx=0$ for every $x\in\Bbb Z_n$.
The last part should say that if $x\in\Bbb Z_n$ generates the whole module $\Bbb Z_n$ (that is, (any representative of) $x$ is coprime to $n$), then $$\{m\in\Bbb Z:mx=0\}\ =\ n\,\Bbb Z\,.$$