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Simple semigroup problem
Let the semigroup g be: _ | ABC...n _ _ _

a | a 1 1 a 12 a 13...a 1n

b|a2 1 a22 a23...a2n

c|a3 1 a32 a33...a3n

. | ..................

. | ..................

N|an 1 an2 an3 ...

|

Where aiji ∈ g, the order of g is n, and g has at least one idempotent a.

In order to understand semigroups by definition, I introduce the concept of "weakening" semigroups myself. The main properties of semigroups are closure and satisfying the associative law, so it is semigroups that satisfy these two properties. To this end, we will convert the above g into h:

__|a b c ... noun _ _ _ _

Answer | answer ... a

a22 a...a

a33...a

. | ..................

. | ..................

One by one ... Ann

|

We can prove that H is a semigroup as follows:

When n = 1, obviously; When n = 2, h is:

__|a b _____

Answer | answer

a22

Obviously closed, for the combination: (aa)a=a(aa) (aa)b=a(ab)

a = a(ba)(ba)a = b(aa)(bb)a = b(ba)(ba)b = b(ab)

b=a(bb) (bb)b=b(bb)

Therefore, whether a22 is B or A, H is a semigroup.

Assuming n=k, the conclusion holds, that is, the following is a semigroup:

__|a b c...k _____

Answer | answer ... a

a22 a...a

a33...a

. | ..................

. | ..................

k|a a a...akk

|

When n=k+ 1, h is:

__|a b c...k k+ 1_____

Answer | answer ... answer

A22 a ... Answer

A33 ... Answer

. | ...................

. | ...................

k|a a a...akk a

K+ 1 | aa aa ... an ak+ 1, k+1

|

In H, there are only operations between ak+ 1, k+ 1 and A and B, ..., ak+ 1, k+ 1.

Therefore, we only discuss whether the following laws are satisfied:

*(k+ 1)(k+ 1),(k+ 1)*(k+ 1),(k+ 1)(k+ 1)*,**(k+ 1),

(k+ 1)**,*(k+ 1)*,(k+ 1)(k+ 1)(k+ 1); Where * refers to k numbers A, B, ..., K. It is easy to verify that all the above meet the combination.

Therefore, H: _ | ABC...N _ _ _

Answer | answer ... a

a22 a...a

a33...a

. | ..................

. | ..................

One by one ... Ann

|

It is a semigroup.

The above process is my weakened idea: g->; H

In order to simplify the complexity of operation in a semigroup and ensure that it is still a semigroup.

In addition, we can prove that the following changes are still semigroups:

__|a b c...n-k n-k+ 1 ... noun _ _ _ _ _

Answer | answer ... answer ... a

A22 a ... answer ... a

A33 ... Answer ... a

. |a ...........................

. |a............n-k a...a

. |....................a...a

. | .............................

One by one ... ahhh ... a

|

To save time, I won't be here. Same method as before!

With this weakening, we will eventually get:

__|a b c...n ___

Answer | answer ... a

a a a a...a

c|a a...a

. | ................

. | ................

One by one ... a

|

This is the final form of our weakening, that is, any finite semigroup with at least one idempotent element can be weakened to the above form, which is not only the simplest overall form of this semigroup, but also the simplest operation formula.

Suppose there is a finite semigroup without any idempotents. According to the above ideas, it must be weakened into the following form:

__|a b c...n ___

A | A ...

a a a a...a

c|a a...a

. | ................

. | ................

One by one ... a

|

Where the position of c can be other elements in the semigroup except a;

It can be verified that (aa)c=a, and a(ac)=c, obviously the associative law is not established, which is related to semigroup G->; The conclusion of semigroup H is contradictory! (—& gt; In order to weaken the sign of semigroups, there is no finite semigroup without idempotents.

For infinite semigroups, we can also construct as follows:

__|a b c...n......___

Answer | answer ... a ......

b|a a...a ......

c|a a...a ......

. | .......................

. | .......................

One by one ... a ......

. | .......................

. | .......................

|

This infinite semigroup obviously contains idempotent element aa=a (this is only in the case of infinite countability)

For infinite semigroups without idempotent elements, I haven't found an example yet, which needs all of us to find!

————— All the above certificates belong to wtf888.

I hope metallicanko can give some advice, thank you.