An additional property that holds when N is even but not a multiple of 4: Each maximal subgroup consists solely of either even numbers or odd numbers, never both. Tom Tom Karzes writes:
Actually I think I can strengthen that: As long as N is even, but not a multiple of 4, than I believe this still holds. E.g., it holds for 18, etc.
Tom
Tom Karzes writes:
If N is even, but contains no squares (i.e. each prime factor occurs only once), then N/2 is idempotent under multiplication mod N, and it follows that (e + N/2) is idempotent iff e is idempotent. In this case, I believe the maximal subgroups for e and (e + N/2) are isomorphic to each other, so all the subgroups repeat (under isomorphism) starting at N/2. Does this sound right?
Tom
Warren D Smith writes:
From: Thane Plambeck <tplambeck@gmail.com> In every finite commutative monoid M, you get its maximal subgroups by....
(1) Computing first its set of idempotents X (ie those elements x such that x^2 =x). (2) Then, for each x in X, compute the subset S(x) of y's such that x divides y and y divides x.
Each set S(x) is a maximal subgroup of M with identity the idempotent x.
For the particular case of the integers modulo 30, I got these idempotents and maximal subgroups
In[1164]:= idempotents = Select[elements, mult[#, #] == # &] Out[1164]= {0, 1, 6, 10, 15, 16, 21, 25}
Ie, there are 8 idempotents. So there will be 8 maximal subgroups. Here they are.
In[1173]:= Map[allMutuals, idempotents]
Out[1173]= {{0}, {1, 7, 11, 13, 17, 19, 23, 29}, {6, 12, 18, 24}, {10, 20}, {15}, {2, 4, 8, 14, 16, 22, 26, 28}, {3, 9, 21, 27}, {5, 25}}
Listing all the subgroups is more work, but they're easily found via this divide and conquer technique.
--Mod p^e, with p=prime, the only idempotents are 0 and 1. (Proof: consider the final digit in the radix-p representation of the idempotent...) Therefore mod N, where N has known factorization into K prime powers, there will be 2^K idempotents, and hence 2^K maximal subgroups.
Also, the largest subgroup will arise from the idempotent (1,1,1,...,1)=1 as identity, and it will have cardinality=EulerTotient(N) and consist of all residues relatively prime to N.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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