In my lecture the professor gave an example of a submodule of a finitely generated module not being free as:
The submodule generated by  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  form a basis?? Is it because it can't be linearly independent as x for any even x yields 0. or Equivalently the uniqueness of expressions fails, ie.  = 1 = 3 = 5 etc?
Does is therefore follow that Z/nZ for any n cannot be free as a Z-module?
Thanks for anyhelp!