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

[lpetrich]

For inverse semigroups, the inverse of the inverse of an element is that element itself -- inversion is an involution.

The inverse of a product is the reverse of the inverses of each one: (a*b)^-1 = b^-1*a^-1

(a*b)*(b^-1*a^-1)*(a*b) = a*(b*b^-1)*(a^-1*a)*b = a*(a^-1*a)*(b*b^-1)*b = a*b

(b^-1*a^-1)*(a*b)*(b^-1*a^-1) = b^-1*(a^-1*a)*(b*b^-1)*a^-1 = b^-1*(b*b^-1)*(a^-1*a)*a^-1 = b^-1*a^-1

If e is an idempotent, then a*e*a^-1 is also an idempotent. a*e*a^-1*a*e*a^-1 = a*a^-1*a*e*e*a^-1 = a*e*a^-1

For every idempotent e and element a, there is an idempotent f satisfying a*e = f*a, and one, f', satisfying e*a = a*f'. Proof of the first one, with the second one similar:

a*e = a*a^-1*a*e = a*e*a^-1*a = (a*e*a^-1)*a = f*a

 

[lpetrich]

The paper then does an excursion into homomorphisms: F(S) = T. If S is an inverse semigroup, then T is also, and F(a^-1) = (F(a))^-1. Also, if F(a) is an idempotent, then there is some idempotent e in S such that F(e) = F(a). Easy to prove by construction:

F(a) = b -- F(a*a^-1) = F(a)*(F(a))^-1 = b*b^-1 = b*b = b = F(a)


There is a natural ordering of inverse semigroups, and not just semilattices. For semilattices, a <= b when a = a*b. For inverse semigroups more generally,

a <= b when a = b*a^-1*a = b*e = f*b = a*a^-1*b for some idempotents e and f.

I can't find a proof of transitivity.

This ordering has properties:

a <= b -> a^-1 <= b^-1

a1 <= b1 and a2 <= b2 -> a1*a2 <= b1*b2

If a and b are idempotents, then a <= b for a = a*b

If b is an idempotent, then a must also be an idempotent.

Proof: in a = b*(a^-1*a) both factors on the rhs are idempotents, and the rhs is thus an idempotent, making a an idempotent.

If these orderings are all equalities, then the algebra is a group, and vice versa. Proof: for two idempotents e, f: e <= f and f <= e, meaning that e*f = e = f -- only one idempotent, thus a group. But if a group: for a <= b, a = b*a^-1*a = b*e = b (e = identity). Thus, b <= a also.

 

[lpetrich]

Returning to near-rings, like rings but with a noncommutative additive group, I must note that the product elements, R*R for near-ring R, need not form a group under addition, though their sums are abelian.

From left and right distributivity for (a+c)*(d+b), a*b + c*d = c*d + a*b. If the near-ring has a multiplicative identity, then the additive group is abelian, making the near-ring a proper sort of ring.

As I'd derived earlier, R*a for each a is a homomorphism of R onto some subgroup of R, and that is also true of a*R for each a. So R must have some normal-subgroup quotient group that is isomorphic to some subgroup of R. The trivial one is the identity group, {0}, so what nontrivial ones exist?

Each group R*a and a*R is abelian over addition, meaning that if nontrivial, the additive group must have a commutator subgroup that is smaller than it. This is because the commutator subgroup's quotient group is the maximal abelian quotient group, meaning that all other abelian quotient groups are subgroups of it.

The abelianness of R*R has an additional consequence. For conjugacy classes A and B, A*B = c where c is one element. Also, for a*b = c, order(c) evenly divides gcd(order(a),order(b)), where order(a) is the smallest positive integer of a's that make a+a+a+...+a = 0.

None of the proofs of these results use another ring property, associativity of multiplication -- they only use multiplication of two variables.

 

[lpetrich]

Or for a*b, order(c) evenly divides both order(a) and order(b).

Having done D3, I'll do D4. Dihedral group D(n) has elements of form a^k or a^k*b where b² = e, aⁿ = e, and a*b = b*a^-1.

For D3, r1 = a, r2 = a², s1 = b, s2 = a*b, and s3 = a²*b

For D4, there are three subgroups with quotient group Z2. With their cosets, {e, a, a², a³; b, a*b, a²*b, a³*b} and {e, a², b, a²*b; a, a³, a*b, a³*b} and {e, a², a*b, a³*b; a, a³, b, a²*b} the third one being the second one with b -> a*b.

There is one with an order-4 quotient group: Z2*Z2. With its cosets, {e, a²; a, a³; b, a²*b; a*b, a³*b}

I'll take on the Z2 case, with the subgroup H and its coset J. H*H = H*J = J*H = e and J*J = e or a² or a^k*b -- H or J.

The associative property is satisfied.

For the Z2*Z2 case, the subgroup is H and its cosets are, in order, J, K1, K2. The products map onto e or a² or a^k*b, where for each X, X*Y for each y maps onto all H's or two H's and two of the others. Also, H*X = X*H = e. Since e is in H, that makes H a zero.

For associativity, I use semigroup results to proceed further. I have a list of all order-4 semigroups to within relabeling and to within operation-order reversal. There are 126 of them, but I can trim down that list by using constraints. Having a zero gives 67. Using the homomorphism constraint gives 3.

I find {K1,K2}*{K1,K2} or else {J,K2}*{J,K2} or else {J,K1}*{J,K1} = e or a² or a^k*b -- H or K1 or K2 -- with all the others e.

So I'll leave off here unless I can get some further insight.