General Associativity Theorem
Theorem
If an operation is associative on $3$ entities, then it is associative on any number of them.
Formulation 1
Let $\struct {S, \circ}$ be a semigroup.
Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ be a sequence of elements of $S$.
Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a strictly increasing sequence of natural numbers such that $r_0 = p$ and $r_s = p+n$.
Suppose:
- $\ds \forall k \in \closedint 1 s: b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} {a_j}$
Then:
- $\ds \forall n \in \N_{>0}: \prod_{k \mathop = 1}^s {b_k} = \prod_{k \mathop = p \mathop + 1}^{p \mathop + n} {a_k}$
That is:
- $\ds \forall n \in \N_{>0}: \prod_{k \mathop = 1}^s \paren {a_{r_{k - 1} + 1} \circ a_{r_{k - 1} + 2} \circ \ldots \circ a_{r_k} } = a_{p + 1} \circ \ldots \circ a_{p + n}$
Formulation 2
Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
Let $\circ$ be an associative operation on $S$.
Let the set $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
- $\map {P_1} {a_1} = \set {a_1}$
- $\map {P_2} {a_1, a_2} = \set {a_1 \circ a_2}$
- $\map {P_n} {a_1, a_2, \ldots, a_n} = \set {x \circ y: x \in \map {P_r} {a_1, a_2, \ldots, a_r} \land y \in \map {P_s} {a_{r + 1}, a_{r + 2}, \ldots, a_{r + s} }, n = r + s}$
Then $\map {P_n} {a_1, a_2, \ldots, a_n}$ consists of a unique entity which we can denote $a_1 \circ a_2 \circ \ldots \circ a_n$.
Formulation 3
Let $\struct {S, \circ}$ be a semigroup.
Let $a_i$ denote elements of $S$.
Let $\circ$ be associative.
Let $n \in \Z$ be a positive integer such that $n \ge 3$.
Then all possible parenthesizations of the expression:
- $a_1 \circ a_2 \circ \cdots \circ a_n$
are equivalent.
Also known as
Also known as the general (or generalized) associative law.
Also see
Motivation
The General Associativity Theorem answers the following question:
It has been proved that, for example, union and intersection are associative in Union is Associative and Intersection is Associative.
That is:
- $R \cup \paren {S \cup T} = \paren {R \cup S} \cup T$
and the same with intersection.
However, are we sure that there is only one possible answer to $\ds \bigcup_{i \mathop = 1}^n S_i$ and $\ds \bigcap_{i \mathop = 1}^n S_i$?
That is, is it completely immaterial where we put the brackets in an expression containing an arbitrary number of multiple instances of one of these operations?
The question is a larger one than that: given any associative operation, is it completely associative?
This result shows that it is. Always.
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{C}$
- 1967: Michael Spivak: Calculus ... (previous) ... (next): Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers
- in the context of addition
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: $1.3$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 27 \beta, \gamma$
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Theorem $1.6$
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $2$: Examples of Groups and Homomorphisms: $2.1$ Definition
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 28$. Associativity and commutativity