In my lecture the professor gave an example of a submodule of a finitely generated module not being free as:

The submodule generated by [2] in Z/6Z. With the Z/6Z considered as a module over itself in the usual way.

Because this submodule is of cardinality 3, it cannot be free. Why is this??

Also, he stated that Z/2Z isn't free as a Z-module. Couldn't [1] form a basis?? Is it because it can't be linearly independent as x[1] for any even x yields 0. or Equivalently the uniqueness of expressions fails, ie. [1] = 1[1] = 3[1] = 5[1] etc?

Does is therefore follow that Z/nZ for any n cannot be free as a Z-module?

Thanks for anyhelp!