# Semi-group In Algebra

In Algebra , semigroup is an algebraic system determined by the pair <G, *> , such that G is a non-empty set and * is a law of internal composition; therefore it fulfills the closing property, and is also associative.

In the case that <G, *> is a semigroup and the operation is commutative, it is said to be a commutative or abelian semigroup .

## Summary

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• 1
• 2
• 3 Other cases
• 4 Affirmations
• 5 References
• 6

## Definition.

Let be a set G and the binary operation * defined as * (x, y) = z , normally written as x * y = z that satisfies the axioms :

1. Closing: . * is closed.
2. Associativity: For all x , y , z of G , (x * y) * z = x * (y * z) .

It is said that G with the operation * is a semigroup .

Note that semigroups differ from groups in that these are semigroups that also have a neutral element and for each element in the set they have one and only one inverse element .

## Examples.

1. Every group <G, *>is a semigroup.
2. The following are semigroups represented in tabular form:
<{a, b}, @> <{a, b, c}, *>
 @ to b to to b b b to
 * to b c to to b c b b c to c c to b

## Other cases

• Let S be a non-empty set to a fixed element, let us define x * y = a, for any x, and elements of S. The closure of * is true.

The associative property: (x * y) * z = a * z = a = x * a = x * (z * t). Trivially, every set not empty with the law of internal composition *, is a semigroup. [one]

• The set N = {0,1,2 … n, …} of the naturals with addition and multiplication is a semigroup, since these two operations are laws of internal composition, that is, each pair of n . natural corresponds to a natural number that is its sum or product depending on the case. However, N is not a semigroup with the subtraction, since this is a partially defined operation, since for some ordered pairs of n. natural there is no difference; the same for the division of n. natural.
• In the case of integers, subtraction is a fully defined operation, it is a law of internal composition; however, the associative property is not fulfilled; for cases a- (bc) and (ab) -c in almost all cases. So the set Z of n. integers is not a semigroup with subtraction.

This same set of n. integers is a semigroup with the multiplication of even integers; since the product of two even numbers is equal and the associative property is fulfilled.

• Let T be the set of the powers of three, being T = {3 nn is an integer} the set T with the multiplication is a semigroup, since the product of two powers of 3 is a power of 3; In addition, the product in T inherits the associative property of the product of the integers, since it is a subset of Z.
• The set of odd integers with multiplication is a semigroup, since the product of odd is odd, it is also associative; but it is not semigroup with the addition, since the sum of two odd numbers is even, although the associativity is fulfilled.
• The set of prime integers is semigroup with neither addition nor multiplication; for the sum of two prime integers, in almost all cases, is not prime; and in the case of the product of two primes, the primality is broken, the product already has four divisors and therefore it is no longer prime.

## Affirmations

• Let t = {… [(b 12 ) b 3 ] a 4 …} a product of n elements b 1 , b 2 , … b n of a multiplicative semigroup, then any product t ‘formed by the factors b 1 , b 2 , … b n taken in that order is equal to t. This is the case of generalized associative property.
• In a semigroup it is possible to speak of the k-th powerof an element b , product of k elements b . It is denoted b k , carrying the rule

i b j = b i + j .

• In the case of an abelian-commutative semigroup, the product

t = b 1 b 2 … b m does not depend on the order of the factors.

• Every homomorphic image of a semigroup is a semigroup.
• In a semigroup S, center C is a stable (occasionally empty) part