Reference : Algebra by Hungerford, Springer Verlag, GTM.
Let G be a semigroup. Given , with latex a_{1}a_{2}\ldots a_{n}$ so as to yield a “meaningful” product in G of these n elements in this order. Furthermore, it is plausible that any two such products can be proved equal by repeated use of the associative law. A necessary prerequisite for further study of groups and rings is a precise statement and proof of these conjectures and related ones.
Given any sequence of elements of a semigroup G, define inductively a meaningful product of (in this order) as follows: If , the only meaningful product is . If , then a meaningful product is defined to be any product of the form where and and are meaningful products of m and elements respectively. (to show that this statement is well-defined requires a version of the Recursion Theorem). Note that for each there may be meaningful products of . For each we single out a particular meaningful product by defining inductively the standard n product of as follows:
, and for ,
The fact that this definition defines for each a unique element of G (which is clearly a meaningful product) is a consequence of the Recursion Theorem.
Theorem: Generalized Associative Law:
If G is a semigroup and , then any two meaningful products in in this order are equal.
Proof:
We use induction to show that for every n any meaningful product is equal to the standard n product . This is certainly true for . For , by definition, for some . Therefore by induction and associativity:
QED.
Corollary: Generalized Commutative Law:
If G is a commutative semigroup and , then for any permutation of 1, 2, …,n
Proof: Homework.
Definition:
Let G be a semigroup with and . The element is defined to be the standard n product with for . If G is a monoid, is defined to be the identity element e. If G is a group, then for each , is defined to be .
It can be shown that this exponentiation is well-defined. By definition, then , , …and so on. Note that it is possible that even if , we may have .
Regards.
Nalin Pithwa