I'm going back through basic module theory notes, and I've come across a paragraph explaining that a sub-module of a finitely generated free module may not itself be free. In my course a free module $M$ over $R$ is defined to be one that contains a basis for $M$. A basis must be $R$-linearly independent and any $m\in M$ must be able to be expressed as a finite sum of basis elements.
The lecture notes define the rank of a free module to be the number of elements in a basis for it (which was proved to be the same for any basis). They then state that if N is a submodule of M, then Rank$(N) \leq$ Rank$(M)$. However, they say, a submodule of a free module might not itself be free. They give an example: the module $\mathbb{Z}/ 6\mathbb{Z}$ over $\mathbb{Z}/ 6\mathbb{Z}$ is free; the sets $\{1\}$ and $\{5\}$ are both bases, and I think they are the only ones, for any others wouldn't be linearly independent. They consider the submodule generated by $[2]$, and state that it isn't free because it has a cardinality of 3.
I can see that the submodule isn't free by exhaustively trying the possible subsets as bases, but I can't see what this has to do with its cardinality. I feel like I'm missing something important. Thanks for any help!
A finitely generated $R$-module $M$ is free if $M \cong R^n$ as $R$-modules. If $|M| < |R| < \infty$, then there is no way to have an isomorphism $M \cong R^n$.