I wrote: ----- Just to mention that, although the classification of finite abelian groups really isn't that hard, Dirichlet's theorem on primes in arithmetic progressions, on the other hand, is a pioneering and deep theorem that introduces several new ideas and was probably the first proof in analytic number theory. So I wonder if the existence of any finite group up to isomorphism as a multiplicative subsemigroup of at least one ring Z/N (N in Z+) can be proved by more elementary methods than an appeal to Dirichlet. ----- after Andy wrote:
On Apr 24, 2016, at 9:54 PM, Andy Latto <andy.latto@pobox.com> wrote:
But the new question of whether every abelian group arises in this way is much easier. If we want to find the cyclic group of order M, we need to find a prime power p^n such that M divides phi(p^n). But this is easy, by Dirichlet's theorem. We can even require n = 1. We're then looking for a prime p such that M divides p-1, that is p must be of the form k * M +1. There are infinitely many such primes, so we can find as many cyclic groups of any desired order as we like, Since any finite abelian group is the direct product of cyclic groups, we can obtain any abelian group in this way.
But a colleague kindly pointed out that Andy's proof uses only the case of an arithmetic sequence = 1 mod M, which is among a number of cases of Dirichlet's theorem that can be proved without recourse to the advanced techniques of characters and analytic number theory, but instead by generalizing Euclid's proof of the infinitude of primes: This is proved in some generality in this lovely paper, Prime numbers in certain arithmetic progressions by M. Ram Murty and Nithum Thain: http://projecteuclid.org/download/pdf_1/euclid.facm/1229442627 <http://projecteuclid.org/download/pdf_1/euclid.facm/1229442627> in Theorem 6, or more specifically see the remark after the Corollary to Theorem 5. —Dan