Generalized associative law, gen comm law etc.

Reference : Algebra by Hungerford, Springer Verlag, GTM.

Let G be a semigroup. Given a_{1}, a_{2}, a_{3}, \ldots, a_{n} \in G, with n \geq 3, it is intuitively plausible that there are many ways of inserting parentheses in the expression 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, (a_{1}a_{2}\ldots a_{n}) define inductively a meaningful product of a_{1}, a_{2}, \ldots, a_{n} (in this order) as follows: If n=1, the only meaningful product is a_{1}. If n>1, then a meaningful product is defined to be any product of the form (a_{1}\ldots a_{m})(a_{m+1}\ldots a_{n}) where m<n and (a_{1}\ldots a_{m}) and (a_{m+1} \ldots a_{n}) are meaningful products of m and n-m elements respectively. (to show that this statement is well-defined requires a version of the Recursion Theorem). Note that for each n \geq 3 there may be meaningful products of a_{1}, a_{2}, \ldots, a_{n}. For each n \in \mathcal{N}^{*} we single out a particular meaningful product by defining inductively the standard n product \prod_{i=1}^{n}a_{i} of a_{1}, \ldots, a_{n} as follows:

\prod_{i=1}^{1}a_{i}=a_{1}, and for n>1, \prod_{i=1}^{n}a_{i}=(\prod_{i=1}^{n-1})a_{n}

The fact that this definition defines for each n \in \mathcal{N}^{*} 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 a_{1}, a_{2}, \ldots, a_{n} \in G, then any two meaningful products in a_{1}a_{2}\ldots a_{n} in this order are equal.

Proof:

We use induction to show that for every n any meaningful product a_{1}a_{2} \ldots a_{n} is equal to the standard n product \prod_{i=1}^{n}a_{i}. This is certainly true for n=1, 2. For n>2, by definition, a_{1}a_{2} \ldots a_{n} = (a_{1}a_{2}\ldots a_{m})(a_{m+1} \ldots a_{n}) for some m<n. Therefore by induction and associativity:

(a_{1}a_{2} \ldots a_{n}) = (a_{1}a_{2} \ldots a_{m})(a_{m} \ldots a_{n}) = (\prod_{i=1}^{m}a_{i})(\prod_{i=1}^{n-m})a_{m+i}

= (\prod_{i=1}^{m}a_{i}) ((\prod_{i=1}^{n-m-1}a_{m+i})a_{n} ) = ((\prod_{i=1}^{m})(\prod_{i=1}^{n-m-1}a_{m+i}))a_{n} = (\prod_{i=1}^{n-1}a_{i})a_{n} = \prod_{i=1}^{n}a_{i}

QED.

Corollary: Generalized Commutative Law:

If G is a commutative semigroup and a_{1}, \ldots, a_{n}, then for any permutation i_{1}, \ldots, i_{n} of 1, 2, …,n a_{1}a_{2}\ldots a_{n} = a_{i_{1}}a_{i_{2}}\ldots a_{i_{n}}

Proof: Homework.

Definition:

Let G be a semigroup with a \in G and n \in \mathcal{N}^{*}. The element a^{n} \in G is defined to be the standard n product \prod_{i=1}^{n}a_{i} with a_{i}=a for 1 \leq i \leq n. If G is a monoid, a^{0} is defined to be the identity element e. If G is a group, then for each n \in \mathcal{N}^{*}, a^{-n} is defined to be (a^{-1})^{n} \in G.

It can be shown that this exponentiation is well-defined. By definition, then a^{1}=a, a^{2}=aa, a^{3}=(aa)a=aaa, \ldots, a^{n}=a^{n-1}a…and so on. Note that it is possible that even if n \neq m, we may have a^{n} = a^{m}.

Regards.

Nalin Pithwa

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