In the spirit of challenges, here is a surprizingly simple question about finite modules, that I did not know the answer to until recently. It is so elementary that I suspect it was common knowledge to the ancients and only forgotten as algebra became more sophisticated. We all know that finitely generated modules mimic many properties of finite sets. E.g. a self map of a finite set is bijective if and only if surjective, if and only if injective. Similarly, an endomorphism of a finite vector space is an isomorphism if and only if injective if and only if surjective. But what if the ring is not a field, but just a commutative ring with identity? Then e.g. multiplication by 3 is an injective endomorphism of Z, a finite Z-module, but it is neither surjective nor injective? can you find an example in the other direction? i.e. can you find a surjective endomorphism of a finite R module which is not injective? I claim not. There is an abstract proof for modules which are "noetherian" i.e. not only finitely generated, but in which also every submodule is finitely generated. Namely such modules cannot have a strictly increasing infinite sequence of submodules. But the sequence of kernels of iterates of a surjective endomorphism cannot terminate unless they are all zero. But there is a simpler more elementary proof for al finite modules, without assuming noetherian. can you find it? notice this implies that if two finite R modules are isomorphic, then every surjective homomorphism between them is an isomorphism! Isn't that amazing? Hence if you mod out any finite R module by a non trivial submodule, what you get is never isomorphic to the original module. or if you have two properly nested submodules of a finite free module, the two quotients are never isomorphic. in particular a non trivial quotient of a free module is never free on the same number of generators. thus the rank is well defined. again, in any free rank n module, n generators are always independent, but n independent elements may not generate.