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.
Interesting suggestion.
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.
There's a simple decomposition theorem for finite abelian groups. Consider any choice E: {2, 3, 5, 7,...} —> {0, 1, 2, 3, ...} of a nonnegative integer e(p) for each prime p, *such that* e(p) > 0 *for only finitely many primes* p. Then such e are in one-to-one correspondence with finite abelian groups, up to group isomorphism. Namely, E <—> Z/2^e(2) + Z/3^e(3) + Z/5^e(5) + Z/7^e(7) + ... where + denotes direct sum. For more see https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Classificatio... (and ignore the Z^k direct summands, since that article more generally addresses the classification of finitely *generated* abelian groups, which is only infinitesimally more complicated). —Dan