On Wed, Apr 20, 2016 at 6:30 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6.
The Chinese remainder theorem lets you express integers mod 10 as pairs of integers mod 2 and integers mod 5; these are all the pairs (0, x) where x is nonzero.
I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example.
These are all the pairs of the form (x, 0) where x is nonzero.
I'm not sure how to enumerate these. {6} and {4,6} are two more.
{6} is {(0,1)}. Its dual is {(1,0)} = {1} {4, 6} is {(0,-1), (0,1)}. Its dual is also {1}, since -1 = 1 mod 2. The answer is all products of subgroups of Z/2Z with those of Z/5Z. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com