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How many groups and semigroups and rings and the like - abstract algebra

To try to get a clue as to the asymptotic behavior of all these binary-algebra counts, I decided to compare them to the asymptotic behavior of the general number of groupoids. I looked for something with form

n^(c*n^2) / n!

where I tried to find c.

For quasigroups and loops, it is roughly 1/2.

For semigroups, it is roughly 1/4.

For commutative semigroups, it it roughly 3/16.

For monoids, both in general and commutative, it is roughly 1/8.

-

For groups, however, c tends to 0, and their rate of increase is much less. The counts of them jump around quite a lot, and those counts are very sensitive to the exponents of the order's prime factors. So it's hard to make a simple statement.
 
The prime-power-order groups or p-groups have many more elements than those of their "neighbors".

For powers of 2: (exponent, order, # abelian groups, # groups
  • 0, 1, 1, 1
  • 1, 2, 1, 1
  • 2, 4, 2, 2
  • 3, 8, 3, 5
  • 4, 16, 5, 14
  • 5, 32, 7, 51
  • 6, 64, 11, 267
  • 7, 128, 15, 2328
  • 8, 256, 22, 56092
  • 9, 512, 30, 10494213
  • 10, 1024, 42, 49487365422
Mathematica doesn't have a count for 2^(11) = 2048.
 
I tried the  Higman–Sims asymptotic formula for it, and it is, for order p^n for prime p, p^( (2/27)*n^3 + O(n^(8/3)) )

That greatly overestimates how many 2-groups there are. I found a power c from 2^(n^c) and from n = 2 to 10 it is
0, 0.766784, 0.964395, 1.0784, 1.16478, 1.24084, 1.32654, 1.43337, 1.55055

It becomes approximately linear after 3 and 4, slowing down a bit until 5 and 6 and then speeding up a bit.

For primes p greater than 2, Mma has counts for orders up to p^7, and for prime powers in the order from 0 to 7, these are the dominant powers of p in the counts:
0, 0, 0, 0, 0, 1, 2, 5

That's too few terms to make a convincing identification as a sequence in OEIS.


Can we go beyond p-groups? There is a simple way of doing so. Some groups are direct product of p-groups for different primes. These groups are "nilpotent" ones, along with the p-groups themselves. Nilpotency means that if one repeats an operation on a binary algebra's set, one will end up at a single element. For semigroups, one uses the operation, but for groups, if one tries, one ends up with the original group again. So for groups, one defines a "commutator":

[A,B] = group generated by all [a,b] = a*b*inv(a)*inv(b) for a in A and b in B

One can repeat this operation in various ways.

The "derived series" is a series of form G(1) = [G,G] with G(k+1) = [G(k),G(k)]. If it ends at the identity group, the group is "solvable". Every abelian group is solvable, and the smallest non-solvable group is the A(5), the group of even permutations of length 5 and order 60. This group and its non-solvability appears in the proof that general quintic equations cannot be solved with nth roots. The connection is in rearranging the roots of the equation, and using the result that nth roots imply that the group of rearrangements must be solvable.

The "lower central series" is a series of form G(1) = [G,G] and G(k+1) = [G,G(k)]. if it ends at the identity group, then the group is nilpotent.

The "upper central series" goes in reverse. Starting with the identity group, G(k+1) has all b that satisfy [a,b] in G(k) for all a in G. The member after the identity group is the "center" of the group. If this sequence ends at the original group, then the group is nilpotent.


Needless to say, there are plenty of non-nilpotent groups, so I have only limited success in estimating how many groups there are.
 
Classification and enumeration of finite semigroups - Andreas Distler's PhD thesis - uses nilpotent semigroups to estimate how many there are.

Nilpotency for semigroups is defined more simply than for groups: use the semigroup operation. One gets a power series:
S(1) = S*S = S^(2)
S(k+1) = S*S(k) = S^(k+1)
If it ends at one element, then the semigroup is nilpotent. The first power of S that gives nilpotency is the degree of nilpotency.

The null semigroup: for constant a and x, y in S, x*y = a. It has degree 2. Only the trivial semigroup has degree 1.

The order-2 semigroups have power series:
1: 12 - 1 - 1 - 1 - 1 ...
2ab, 3, 4: 12 - 12 - 12 - 12 ...

I'll write them for short as
1: 12 - 1
2ab, 3, 4: 12
with the last member being the one that repeats endlessly.


The order-3 semigroups have power series:
1, 4, 5, 10, 12ab, 13, 14, 15ab, 16ab, 17ab, 18ab: 123
2, 9, 11ab: 123 - 23
3: 123 - 23 - 3
6: 123 - 13
7: 123 - 3

Only one of them is nilpotent with degree 3.

In general, nil-3 groups can be constructed with sets A and B satisfying
A*A <= B, A*B = B*A = B*B = {z}, the power series' endpoint.

The A*A operation is completely arbitrary, like the general groupoid operation, and AD used that result to calculate how many nil-3 semigroups there are. This was a nontrivial task, since it was necessary to divide out isomorphisms, like for groupoids. So we have these binary-operation algebras for which we have formulas for numbers of counts:
  • Groupoids: (general, self-converse, commutative) * (general, with identity)
  • Semigroups: nilpotent with degree 3
  • Groups: abelian, nilpotent (using p-groups)
 
Taking a power series on a semigroup produces a chain of subsemigroups. The null semigroup makes the trivial semigroup in one step, and those that are groups make themselves.

1, 4, 5, 10, 12ab, 13, 14, 15ab, 16ab, 17ab, 18ab: Makes itself
2: Makes 23 - 23.32 - Z2
3: Makes 23 - 33.33 (2-element null semigroup) - then 3 (trivial semigroup) - a nil-3 semigroup
6. Makes 13 - 31.13 - Z2
7. Makes 3 (trivial semigroup) - is the 3-element null semigroup
8. Makes 23 - 23.33 - boolean
9. Makes 23 - 22.23 - boolean
11a. Makes 23 - 22.33 - left zero
11b. Makes 23 - 23.23 - right zero


Turning to monoids, I note that every finite one can be expressed as the combination of a group G and a semigroup S. G*G = G and S*S <= S, and g*S = permutation of S and S*g = permutation on S for every g in G.

The endpoint of a semigroup-like power series is G + (S subset).

An easy kind of monoid is the "trivial action" one, where g*S = S*g = (unchanged order of S) for all g in G. So one can make a lower limit on the number of monoids by finding the number of trivial-action ones. I've done that, and it's very close to the numbers of monoids themselves.

So having a nontrivial action makes a strong constraint on the semigroup part. That's g*S = (changed order of S) and likewise for S*g, both for some g in G.

If the group part has order 1 (the identity group), then the monoid is a trivial-action monad (TAM), since the only group element is the identity. Likewise, for total order n, if the group part has order n-1 or n, the the monoid is a TAM because the semigroup part has only one element or none, giving only one possible action.
 
For groups, there is an additional kind of decomposition that some of them may have. Consider subgroup H of group G.

That group has "left cosets" x*H for x in G and also "right cosets" H*x for x in G. They are all disjoint and the same size as H. That makes the order of H evenly divide the order of G. This is Lagrange's theorem, and it makes a big constraint on possible subgroups of a group. A constraint that does not exist for semigroups or monoids more generally.

The two sets of cosets need not be equal, but when they are, they are called plain cosets. They are equal when x.H.inv(x) = H for all x in G, though the elements in H may be rearranged. In that case, H is a "normal subgroup". One can then construct an operation table for cosets C1*C2 = C3 and one finds that it has the group properties. Thus, G and H have a "quotient group".

If a group has no normal subgroups, it is a "simple group". An obvious one is Z(p), the cyclic group of prime order p. Since a prime number has no divisors other than 1 and itself, Lagrange's theorem shows that this group has no nontrivial subgroups. The trivial ones are itself and the identity group.

Some decades ago, mathematicians succeeded in finding all the simple groups. Their proof too some 10,000 pages in a large number of articles scattered over mathematics journals. They found several infinite families, including the Z(p) family, and 26 "sporadic groups", outside of those families. Most of them are huge, and the largest one is called the monster group.
 
Having considered unary-operator and binary-operator algebras, the next one is a ternary-operator one. But there has been very little done on that.

OEIS has A091510 - OEIS - Number of nonisomorphic algebras with a ternary operation (3-d groupoids) with n elements.

The number increases asymptotically as n^(n^3)/n! -- super fast. By comparison, ordinary (2D) groupoids are at n^(n^2)/n! and 1D functions as 0.442 * 2.956^n / sqrt(n)

For order-2 algebras, a unary operator has 4 possibilities with 3 distinct, a binary operator has 16 possibilities with 10 distinct, and a ternary operator has 256 possibilities with 136 distinct.


Ternary quasigroups — page one.

Quasigroups? Defined like binary-operator ones: for a, b, c, one can find x, y, z: a*b*x = c, a*y*b = c, z*a*b = c

These are related to Latin cubes, a 3D generalization of Latin squares.

Associativity? A ternary version: (a*b*c)*d*e = a*(b*c*d)*e = a*b*(c*d*e)

Power-associativity is a weaker version where all the input variables are equal.

Commutativity? a*b*c = b*a*c = c*b*a = a*c*b = b*c*a = c*a*b

Identity? Use a pair of values.
 
I'll now turn to pairs of binary algebras, as generalizations of addition and multiplication taken together.

The usual starting point for this is the ring. It has a set of values and analogs of addition and multiplication with these properties:
  • Addition forms an abelian (commutative) group
  • Multiplication forms a semigroup (associative only)
  • Multiplication is distributive over addition:a*(b+c) = a*b+a*c and (b+c)*a = b*a+c*a.
About the second property, common variations are to add the commutative property and/or an identity, making multiplication a monoid.

The additive identity is usually called 0, and when it is present, the multiplicative identity is usually called 1. Some mathematicians reserve "ring" for those rings with 1, sometimes calling a ring without 1 a "rng" (no i, pronounced "rung").

One can do a + a + a + ... for some nonzero a. The largest number n that produces 0 is the "characteristic" n and if 1 is present, it generates Z(n), the ring of integers modulo n. Otherwise, the ring has characteristic 0 and if 1 is present, it generates Z, the ring of integers.

Distributivity and the additive identity together yield an interesting result.

a*(b+0) = a*b + a*0 = a*b -> a*0 = 0
(b+0)*a = b*a + 0*a = b*a -> 0*a = 0

So 0 is a zero of multiplication, an absorber or annihilator.

If nonzero a and b make 0 when multiplied, then a and b are a pair of "zero divisors". If a ring with no zero divisors, if a*b = 0 implies at least one of a and b being zero, then the ring is called a "domain", and if commutative an "integral domain".

The ring Z(n) is a domain if n is a prime number, but not otherwise. That is because if n = n1*n2, both n1 and n2 being greater than 1, then n1*n2 = 0 mod n, making n1 and n2 zero divisors in Z(n).

Z(prime) is not only a domain but also an integral domain.

If all the ring elements but 0 form a group under multiplication, then the ring is a field. All the finite fields are known. Their orders are p^m for prime p, and there is only one field for each order, GF(p^m). It has additive group Z(p)^m and multiplicative group Z(p^m-1). For m = 1, the field is Z(p), while for m > 1, the construction is more complicated.

The elements are polynomials in some variable x, polynomials with degree at most m-1 and with coefficients in Z(p). Addition proceeds as with polynomial addition in general, and multiplication is polynomial multiplication modulo some "primitive polynomial", a polynomial that cannot be factored in GF(p^m).

To see how it works, I will construct GF(4). GF(2) is {0, 1} with addition and multiplication modulo 2. GF(4) is thus {0, 1, x, x+1}, and the primitive polynomial is x^2 + x + 1. The possible factors of a degree-2 polynomial are x and x+1, and their products are x^2, x^2+x, and (x+1)^2 = x^2 + 2x + 1 = x^ + 1 (arithmetic is modulo 2 here). The remaining polynomial is x^2 + x + 1. For larger fields, there will be more than one primitive polynomial, and one can choose whatever one wants of them.

GF(2^m) can easily be done fast with typical computer hardware. Each bit in sequence corresponds to a power of the variable. Addition is xor, and multiplication can be done by shift and add, then subtracting out shifted copies of the primitive polynomial.
 
If a ring has multiplicative inverses, it is called a "division ring": a*inv(a) = inv(a)*a = 1 for all nonzero a.

Every finite division ring is also a field -- its multiplication is commutative. Also, every finite domain is also a finite integral domain and a finite field.

Here is the multiplication table for GF(4):

0 * (anything) = 0
1 * (anything) = (that anything)
x * x = x + 1
x * (x+1) = 1
(x+1) * (x + 1) = x
{1, x, x+1} form group Z(3)

Turning from fields to rings, we find a very interesting result. Every finite ring can be expressed as a direct product of rings with prime-power orders:

R(n) = product of R(p^m) over p, m in factorization of n (prime p to power m)

So we only need to do prime-power rings (p-rings?).

A related result is the  Multiplicative group of integers modulo n, what I'll call Zx(n). These groups' elements are all numbers from 1 to (n-1) that are relatively prime to n. They factorize as

Zx(n) = product of Zx(p^m) over p, m in factorization of n

For p > 2, the group is Z(p^(m-1)*(p-1))

For p = 2, the group is, for m = 1, the identity group, and for m > 1, Z(2) * Z(2^(m-2))

That is also true for the additive group of integers modulo n,  Cyclic group, Z(n). From the Chinese remainder theorem,

Z(n) = product of Z(p^m) over p, m in factorization of n

As I'd mentioned earlier, this is also true of nilpotent groups - they are direct products of prime-power groups.
 
So we must concern ourselves with prime-power rings.

For generators a, b, c, ...the elements are na*a + nb*b + nc*c + ... where na, nb, nc, ... are in Z(p^m(a)) where m is the power associated with generator a. If the ring has 1, then it is also a generator.

For power 1, the additive group is Z(p), and the two kinds of rings are the zero ring and the ring Z(p), equal to GF(p). They are both commutative, and one has an identity and the other one doesn't. I'll list them as {# - id - cmt, - id + cmt, + id - cmt, +id +cmt} where id = identity and cmt = commutative property. Thus,
{0, 1, 0, 1}

For power 2, the additive groups are Z(p^2) and Z(p)*Z(p), and there are 11 kinds of rings.

For Z(p^2), there are three rings: the zero ring, Z(p^2), and a third ring with generator a satisfying a^2 = p*a.
{0, 2, 0, 1}

For Z(p)*Z(p), there are three composite rings, Z(p)*Z(p), (zero)*Z(p), and (zero)*(zero)
{0, 2, 0, 1}

There are two rings with identity (generators 1, a): a^2 = c0 + c1*a (primitive polynomial for GF(p^2)) and a^2 = 0. There is a commutative ring without identity, a^2 = b, other products zero, and two non-commutative ones without idenitty, a^2 = a, a*b = b, others zero and a^2 = a, b*a = b, others zero.
{2, 1, 0, 2}

Both reducible and irreducible for Z(p)*Z(p): {2, 3, 0, 3}
All for p^2: {2, 5, 0, 4}

I won't describe the rings for higher prime powers.

For prime power 3,

Irreducible:
Z(p^3): {0, 3, 0, 1}
Z(p)*Z(p^2): {(6,3+2p), (6,7), 0, (2,3)}
Z(p)*Z(p*Z(p): {(7,6+p), 3, 1, 3}
The ()'s contain (value for p = 2, value for p > 2)

All:
Z(p^3): {0, 3, 0, 1}
Z(p)*Z(p^2): {(6,3+2p), (11,12), 0, (3,4)}
Z(p)*Z(p)*Z(p): {(11,10+p), 10, 1, 6}
Total: {(17,13+3p), (24,25), 1, (10,11)}
Grand total: (52, 50+3p)

General formulas get more complicated for higher powers of primes, so I'll only do 2 and 3, what I've seen numbers for.

All:
For 2^3: {17, 24, 1, 10} = 52
For 3^3: {22, 25, 1, 11} = 59
For 2^4: {215, 125, 13, 37} = 390
For 3^4: {>338, >114, 18, 39} = (509)
For 2^5: {>826576, >3566, 99, 109} = (>830350)

Some of these calculations are incomplete, but it is evident that the number of rings increases rapidly with increasing prime power.
 
I must note what "self-converse" means. A groupoid is self-converse if it is isomorphic to one with a transposed operation table. It will be that if it is commutative, but it need not be commutative, as this example shows:

332.332.113

Do 123 -> 213, interchanging 1 and 2. Once one does that, one finds

331.331.223

the transpose.

Here are the smallest orders where one sees that effect.
  • Groupoid: 3
  • Unital groupoid: 4 (identity)
  • Quasigroup: 4 (division)
  • Loop: 5 (division, identity)
  • Semigroup: 4 (associativity)
  • Monoid: 5 (associativity, identity)
  • Group: ? (associativity, identity, division (makes inverses) )
So adding properties not only makes fewer of these entities, it makes this property appear later with increasing order.
 
Let's see where the "lower central series" gets us. For groupoid (G,*) this is the series G, G*G, G*G*G, ... (powers 1, 2, 3, ...)

If division is present, then that means that G*G = G.

This property states that for every a and b in G, there is a unique x and y such that a*x = b and y*a = b. The two quotients, x and y, need not be distinct. It is easy to show that it gives G*G = G.

Likewise, if a left identity or a right identity is present, then that means that G*G = G. That is because, for left identity el and right identity er, el*a = a and a*er = a for all a in G. A plain identity is a two-sided identity, both a left one and a right one.

So it's only for groupoids and semigroups that G*G can be a proper subset of G, and thus that this series can reach a single element, making it nilpotent. If it does that, then that element is a zero of the groupoid. But a groupoid can have a zero without being nilpotent. Like if it has both an identity and a zero. Also, this series may terminate at a subgroupoid with some number of elements between 1 and the original groupoid's order.
 
I'll now consider homomorphisms, functions that take a groupoid's elements and make a new groupoid: f(a) = a', where a is in the source domain and a' in the receiver domain or codomain or range.

The source set is partitioned by the function into subsets, each of which maps onto one receiver element that is distinct from all the others. f(A) = a, f(B) = b, with b != a, etc.

f(A*B) = a*b = c = f(C) means that A*B must be a subset of C.

An isomorphism is a homomorphism that has one source element for every receiver one, and an automorphism is an isomorphism of the source onto itself.

There are two trivial homomorphisms, the identity automorphism, f(a) = a with the same operation, and the null homomorphism, f(a) = z, with z*z = z.

I'm mentioning them because they are necessary for relating different algebraic structures. Like {0,1} with addition modulo 2, and {1,-1} with multiplication. They are isomorphic, 0 <-> 1 and 1 <-> -1, even if their content is not identical. A non-isomorphic homomorphism can also be useful for finding the structure of an algebraic structure.


For instance, for the unary case, consider homomorphism F on a set with unary function f. F(f(a)) = f'(F(a))

Let F(a) = a'. Consider f'(a') = a'. Then F((repeated f)(a)) = a'

More generally, for source subset A for F, f(A) must be a subset of some source subset, whether A or some other one.
 
Let's consider some special cases. If there is a left identity, then it and some other elements, EL, will map onto some left identity.

f(EL*A) = el'*f(A) = f(A)

That means that EL*A is a subset of A. But EL includes a left identity, el, and el*A = A. This means that EL*A = A. Likewise, for a right identity, A*ER = A, and for a two-sided identity, E*A = A*E = A. For a left zero, ZL*A = ZL, a right zero, A*ZR = ZR, and a two-sided zero, Z*A = A*Z = Z.


In the theory of quasigroups, one can relax the notion of homomorphism to produce "homotopy", a triplet of mapping functions f, g, h:
f(a)*g(b) = h(a*b)

If all three functions are bijections, then the homotopy is an "isotopy", much like a homomorphism being an isomorphism. If an isotopy is from the original set to itself, then it is an "autotopy", much like an isomorphism being an automorphism.

An autotopy on a Latin square is a permutation of its rows, a permutation of its columns, and a permutation of its symbols.


Turning to monoids, one has f(group part) = group part and f(semigroup part) = semigroup part, except for the null homomorphism: f(everything) = z, with z*z = z.


Homomorphisms of groups have some interesting properties. The elements that map onto the identity are called the kernel, and they form a normal subgroup. The result group of a homomorphism is the quotient group for the kernel.
 
Here are some monoids with nontrivial group actions on the semigroup parts. Their minimum order is 4, 2 each for the group and semigroup parts.

Commutative:
1 2 3 4
2 1 4 3
3 4 3 4
4 3 4 3

Non-Commutative:
1 2 3 4
2 1 3 4
3 4 3 4
4 3 3 4

1 2 3 4
2 1 4 3
3 3 3 3
4 4 4 4

The number of monoids for each order: 1, 2, 7, 35, 228, 2237, 31559, 1668997
Without trivial-action ones: 0, 0, 0, 3, 8, 97, 766, 10360

For commutative monoids: 1, 2, 5, 18, 76, 403, 2555
Without trivial-action ones: 0, 0, 0, 1, 2, 18, 82

As I mentioned earlier, a "group action" is a permutation on symbols. A set of symbols that stays related is an "orbit". The permutations of symbols in each orbit is a quotient group of the original group, with the kernel being the elements of the original group that map onto the identity permutation. So one can use group theory to determine what permutations are possible for some group.

There is a kind of group action that comes from the group itself. Doing the operation with some element causes a permutation of the elements. It can be shown that they are all in the same orbit.

I'll show some examples.

Z2 -- identity 1 2, flip 2 1
Z3 -- identity 1 2 3, rotation 1: 2 3 1, rotation 2: 3 1 2
Z4 -- identity 1 2 3 4, rot 1: 2 3 4 1, rot2: 3 4 1 2, rot3: 4 1 2 3
Also identity 1 2, rot 1: 2 1, rot 2: 1 2, rot 3: 2 1 -- because identity, rot 2 is a normal subgroup of Z4, with quotient group (0: even rotations, 1: odd rotations).
Etc.
 
Z2*Z2 -- identity: 1 2 3 4, flip 1: 2 1 4 3, flip 2: 3 4 1 2, flip both: 4 3 2 1
and
identity and flip 2: 1 2, flip 1 and flip both: 2 1
and
identity and flip 1: 1 2, flip 2 and flip both: 2 1


Turning from group-like algebras to rings, let's see what we can find out about their structure.

They can have nontrivial zero divisors, pairs of nonzero members that give zero when multiplied together. For the ring Z4, 2*2 = 0 mod 4. So 2 is a zero divisor

For non-commutative multiplication, left and right zero divisors may be distinct. But for finite rings, all left ones are right ones and vice versa, thus making them two-sided zero divisors. abstract algebra - In a finite ring every left-sided zero divisor is a right-sided zero divisor? - Mathematics Stack Exchange

Finite rings all have the division property for elements that are not zero divisors, and when multiplicative identity 1 is present, these elements form a group. So a finite ring with unity divides in two: a multiplicative-group part and a zero-divisors part. This is parallel to the decomposition of a finite monoid into a group and a semigroup. The zero divisors are the semigroup here.

Returning to Z4, the multiplicative group is 1, 3 and the zero divisors are 0, 2. They form an additive group.

Trying out Z6, the multiplicative group is 1, 5, and the zero divisors 0, 2, 3, 4. They do not form an additive group. In fact, Z6 = Z2*Z3 and 0 - (0,0), 1 - (1,1), 2 - (0,2), 3 - (1,0), 4 - (0,1), 5 - (1,2)

For Z2, the multiplicative group is 1 and the zero divisors 0, and for Z3, the multiplicative group is 1, 2, and the zero divisors 0 also.

If the zero divisors form an additive group, then the ring is called a "local ring". There is a theorem that says that every commutative ring with unity is a local ring or the product of local rings. Like Z4 or Z6 = Z2*Z3. The rings Z(p^m) are local.
 
Now I must mention something that I see a lot of: a kind of set called an ideal, a multimember version of a zero.

A groupoid G has left ideal J if G*J <= J, a right ideal J if J*G <= J, and a two-sided or plain ideal if an ideal is both left and right.

Some ideals always exist, and are thus trivial ones: the empty set and the groupoid's set.

A maximal ideal is an ideal J < G such that there is no ideal K such that J < K < G.

For integers under multiplication, (some number) * (integers) is an ideal of that algebra.

In the group + semigroup construction of a monoid, the semigroup part is an ideal -- the maximal ideal.

Groups, quasigroups, and loops only have the trivial ideals.

-

Turning to rings, their ideals are defined with the ring's multiplication, and they have the additional constraint that they must be groups over the ring's addition. There are two trivial ideals, all the ring's elements, and the zero ideal, (0).

Thus, Z4 has ideal (0,2) and Z6 ideals (0,2,4) and (0,3). Z12 has ideals (0,2,4,6,8,10), (0,3,6,9), (0,4,8), (0,6), and of these, the first two are maximal.

In general, ring Z(p^n) has ideals (p^m)*Z(p^(n-m)) where m = 1 is maximal, and ring Z has ideals a*Z, where the maximal ones are those where a is prime. Z(p^n) is thus local.

For a ring with identity, the ideals contain only zero divisors, and for a local ring, all the zero divisors are in a single maximal ideal.
 
Much like how a finite monoid can be divided up into a group and a semigroup, a unital ring (one with an identity) can be divided up into a group over the ring's multiplication and a set of zero divisors.
R = G + D

All the ring's nonself ideals are subsets of D. In a local ring, D is the maximal ideal of the ring.

Subrings R' thus have the constraint that G' <= G and D' <= D

Let us look at finite fields GF(p^n). Their D is the zero D, with only 0 in it. Their G is a group with order p^n-1, and it can be shown to be the cyclic group Z(p^n-1).

A subfield GF(p^m) would need some m where (p^n-1) is divisible by (p^m-1). That happens only when m divides n. Thus the subfields of GF(p^n) are the fields GF(p^(n/k)) where k evenly divides n.


An ideal J will have additive cosets: C(x) = x + J. These follow addition and multiplication laws C(x)+C(y) = C(x+y) and C(x)*C(y) = C(x*y), and one thus gets a ring of cosets, a quotient ring for J, just like a quotient group. The quotient ring's additive quotient group will be (ring's additive group) / (ideal's additive group).

As with groups, quotient rings are related to ring homomorphisms f, operators that preserve the ring properties. Here, f(J) = 0.

Let's consider ring Z6 again. One of its ideals is J = (0,2,4) with additive group Z3. That ideal's coset is K = (1,3,5). J + J = K + K = J, J + K = K + J = K, J*J = J*K = K*J = J, K*K = K. Thus, the quotient ring is Z2, with J -> 0 and K -> 1
 
I'll now consider zero divisors of rings more closely. They can be left zero divisors, a*x = 0 for nonzero x, or right zero divisors, y*x = 0.

If an element is not a left zero divisor, then a*x = a*y implies a*x - a*y = 0 (subtraction of right-hand side) and in turn, a*(x-y) = 0 But by left zero non-division, x-y = 0 and thus x = y. That makes a "left cancellative". For ring R, a*R <= R. But a*R and R have the same cardinality, and for a finite ring, a*R = (permutation of R). That need not happen for an infinite ring, however.

This gives a the "left division" property, that for every b, there is a unique x that satisfies a*x = b.

If a and b are left zero non-divisors, then a*b is also one. The left cancellation property also propagates in this fashion, as does the left division property.

All this is true for the right-side versions of these properties.


Let us consider a both-sides version. That makes the zero non-divisors a multiplicative subsemigroup of R, so let us consider a semigroup with division, equivalent to a quasigroup with associativity.

Consider e1*a = a*e2 = a for some a. Do e1 * both sides of e1*a = a. That means e1^2*a = e1*a, and by cancellation, e1^2 = e1. Now do * b by any element b: e1^2*b = e1*b and by cancellation on e1 *, that is e1*b = b for any b. Thus, e1 is a left identity. Likewise, e2 is a right identity. Since there is an identity on both sides, the two are equal and unique. Thus, one finds an identity element e.

Thus, this algebra is a monoid with division, or a loop with associativity. Using the division property again, for every a, there is a b such that a*b = e, and a c such that c*a = e. Now consider c*a*b -- by associativity, that is (c*a)*b = e*b = b, or else c*(a*b) = c*e = c. Giving b = c, meaning a unique inverse, and making this algebra a group.

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If a finite group has a left zero non-divisor but no a right one, let's see what happens. Some a such that a*x = 0 implies x = 0, but y*a = 0 is possible for some nonzero y.

Take w*a = 0 for nonzero w. Now do * b for all b in R. Then, w*(a*b) = 0, and since (a*b) covers the entire ring, we find that every ring element is a right zero divisor.

So if some ring elements are neither left nor right zero divisors, then this property is all or nothing, with no only-one-side zero divisors. That is,

R = (Mult group) + (Zero divisors on both sides)
 
So one finds a decomposition of rings much like a decomposition of monoids.

Monoid = group G + semigroup S
Ring = mult group G + zero divisors D
G*G = G
G * S/D = permutation on S/D
S/D * G = permutation on S/D
S/D * S/D <= S/D (no G)


I've gone rather far afield into the structures of these algebras, so I'll return to my original subject, counts of them. Many algebras can be reduced to direct or outer products of algebras:

for algebras 1 and 2, a = (a1,a2), b = (b1,b2), a*b = (a1*b1,a2*b2)
Number of elements / order / size = (order of algebra 1) * (order of algebra 2)

Thus, the total number in terms of irreducible number:
n1 -- n1
n2 -- n2
n3 -- n3
n4 -- n2(n2+1)/2 + n4
n5 -- n5
n6 -- n2*n3 + n6
...
The first term for n4 is not n2^2 because (algebra 1, algebra 2) is equivalent to (algebra 2, algebra 1).

More generally,
n(k) = sum over divisor sets of (product over divisors d of sympwr(n(d),m(d)))
The divisor sets are all divisors d with multiplicity m(d) that satisfy k = product over d of d^(m(d))

sympwr(n,m) = n(n+1)...(n+m-1)/m!
 
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