I think Keith is asking for "The number of subsets of the multiplicative semigroup Z/nZ which are representations of some group." I would have thought that "The number of multiplicative subgroups of the ring Z/nZ" was perfectly clear as well, but the thread so far makes me think others would disagree. Representation theory in general is perfectly happy to think about group representations over finite rings (and modules, i.e. "vector spaces over rings"). If you have a set of things and a binary operation and the things transform like a group, then congratulations, you're the proud owner of a group representation, whether or not e.g. the image of the group's identity element is actually the identity element of the larger set that you're mapping into. Ed Clark's reference, Stanley Payne's "Multiplicative Subgroups of Z/nZ" ( http://math.ucdenver.edu/~spayne/classnotes/subgroup.ps), starts by showing that the maximal multiplicative subgroups are exactly of the form bU, where U is the group of units in Z/nZ, and b ranges over all elements where ab=n and gcd(a,b)=1. The nicely-annotated bibliography of that paper probably has things worth reading. --Michael On Thu, Apr 21, 2016 at 4:34 AM, Joerg Arndt <arndt@jjj.de> wrote:
* Keith F. Lynch <kfl@KeithLynch.net> [Apr 21. 2016 08:21]:
Allan Wechsler <acwacw@gmail.com> wrote:
Modulo 10, only 1, 3, 7, and 9 have reciprocals. These four form a cyclic group under multiplication, generated by 3 (or 7). This group has only two subgroups, {1} and {1,9}. These are the three groups Dan enumerated. ....
Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6. I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example. I'm not sure how to enumerate these. {6} and {4,6} are two more.
Right.
Dan Asimov <dasimov@earthlink.net> wrote:
I think it would also fair to say {0} is a group under multiplication modulo 10.
Right. And that's all eight of them.
The following modulo confusion on my side.
Mod N, the identity elements are those numbers I such that I^2 = I mod N. For N=10, those are 0, 1, 5, and 6. The numbers of identity elements mod N appear to be given by A034444. Mod 30, there are 8 of them, and 28 groups.
A group G has one neutral ("identity") element e by definition: a * e = a = e * a for all a \in G
Not sure how to name your I, "projection" springs to mind.
Every number 0...N-1 "belongs to" at most one identity element. For N=10, 0 belongs to 0; 1,3,5,7 belong to 1; 5 belongs to 5; and 2,4,6,8 belong to 6. That's all 10 numbers. They each appear in at least one group.
Usually the multiplicative group mod n consists of all u, 1<=u<n, such that gcd(u,n)=1 (the "units"). There are phi(n) units.
For these groups, just the neutral element e satisfies e * e = e.
Cf. A000688
You seem to have something different in mind. I'd like to warn about using existing terminology with non-standard meanings.
Mod N, the number of numbers that *don't* appear in any multiplicative group appears to be given by A055654. It appears to be always 0 for square-free N, and never 0 if N is not square free. Those groupless numbers are always divisors of N.
I think the order of any of these groups always divides phi(N).
The order of any subgroup of G is a divisor of the order of G. As the order of the multiplicative group mod N is phi(N), so your statement follows.
Except for very small N, you'll always have {0}, {1}, {1, N-1}, and the phi(N) numbers relatively prime to N. The last may have other subgroups. If M divides N, I think you'll always also have, for each group in M, the same group in N only with each term multiplied by N/M. For instance {2,4,6,8} where N=10 is an echo of {1,2,3,4} where N=5. And {5} where N=10 is an echo of {1} where N=2. So there's probably some simple recurrence relation which can be used to find how many multiplicative groups there are mod N.
How many are there?
1,2,3,3,4,6,5,6,5,8,5,10,7,10,14,9,6,10,7,14,17,10,5,24,7,14,7,17,7,28,9,12,17,12,24,17,...
Not in OEIS. I plan to add it. This discussion thread makes it obvious that I need help in phrasing my description of what this is about, so someone please help. Thanks.
First step: define everything.
Best regards, jj
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