PlanetMath: general associativity

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general associativity

(Theorem)

If an associative binary operation of a set is denoted by “ $\cdot$ ”, the associative law in is usually expressed as

$\displaystyle (a\!\cdot\!b)\!\cdot\!c = a\!\cdot\!(b\!\cdot\!c),$

or leaving out the dots,

. Thus the common value of both sides may be denoted as

. With four elements of

we can calculate, using only the associativity, as follows:

$\displaystyle (ab)(cd) = a(b(cd)) = a((bc)d)= (a(bc))d = ((ab)c)d$

So we may denote the common value of those five expressions as

Theorem 1 The expression formed of elements

, ...,

represents always the same element of

independently on how one has joined them together with the associative operation and parentheses, if only the order of the elements is every time the same. The common value is denoted by $a_1a_2\ldots a_n$ .

Note. The elements can be joined, without changing their order, in $\frac{(2n-2)!}{n!(n-1)!}$ ways (see the Catalan numbers).

The theorem is proved by induction on . The cases and have been stated right above.

Let $n \in \mathbb{Z}_+$ . The expression $aa \ldots a$ with equal “factors” may be denoted by and called a power of . If the associative operation is denoted “additively”, then the “sum” $a\!+\!a\!+\cdots+\!a$ of equal elements is denoted by and called a multiple of ; hence in every ring one may consider powers and multiples. According to whether is an even or an odd number, one may speak of even powers, odd powers, even multiples, odd multiples.

The following two laws can be proved by induction:

$\displaystyle a^m\cdot a^n = a^{m+n}$

$\displaystyle (a^m)^n = a^{mn}$

In additive notation:

$\displaystyle ma\!+\!na = (m\!+\!n)a,$

$\displaystyle n(ma) = (mn)a$

Note. If the set together with its operation is a group, then the notion of multiple resp. power can be extended for negative integer and zero values of by means of the inverse and identity elements. The above laws remain in force.