In fact I think that it is true that every finite abelian group is isomorphic to a subgroup of the group of units of Z/(n) for n a product of distinct primes. To see this note that if n is a product of distinct primes p_1,...,p_k then the group of units is a product C_{p_1-1}xC_{p_2-1} x...xC_{p_k-1} where C_{p-1} is a cyclic group of order p - 1. Then it suffices to show that for any sequence q_1,...,q_s of prime powers there are distinct prime p_1,...,p_s such that q_i divides p_i -1 for all i. Then then C_{p_i-1} will have a subgroup of order q_i , etc.. This follows from Dirichlet's theorem_on_arithmetic_progressions <https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions> , Namely since for any one of the q's, gcd(q,1) = 1, there are infinitely many primes p of the form p =q * k +1 Hence q divides p -1. And since there are infinitely many such primes we can chose a different prime p_i for each q_i. Of course here I use the fact that every finite abelian group is the product of cyclic groups of prime power order. If I recall correctly this is a trick that Everett Dade told me about over half century ago. On Sun, Apr 24, 2016 at 11:06 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
4 14 16 26
Note that that's isomorphic to the other one of the two abstract groups of order 4, the one that isn't a cyclic group.
Inspired by that, here's an absurdly bold conjecture with remarkably little evidence: *Every* finite abelian group is isomorphic to some multiplicative group mod N. Someone prove me wrong.
After I get my program to find all such groups through at least N=1000 working, I'll check and see if it appears to be true. Of course whatever I find will constitute neither a proof or a disproof, but it may inspire insight. In support of that, is there any online database of small abstract abelian groups? I see little point in reinventing that wheel. Thanks.
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