# Associativity on Four Elements

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## Contents

## Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $a, b, c, d \in S$.

Then:

- $a \circ b \circ c \circ d$

gives a unique answer no matter how the elements are associated.

## Proof

As $\struct {S, \circ}$ is a semigroup:

- it is closed
- $\circ$ is associative

From Parenthesization of Word of $4$ Elements, there are exactly $5$ different ways of inserting brackets in the expression $a \circ b \circ c \circ d$.

As $\circ$ is associative, we have that:

- $\forall s_1, s_2, s_3 \in S: \paren {s_1 \circ s_2} \circ s_3 = s_1 \circ \paren {s_2 \circ s_3}$

As $\struct {S, \circ}$ is closed, we know that all products of elements from $\set {a, b, c, d}$ are in $S$, and are likewise bound by the associativity of $S$.

So:

\(\displaystyle \paren {\paren {a \circ b} \circ c} \circ d\) | \(=\) | \(\displaystyle \paren {a \circ \paren {b \circ c} } \circ d\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \circ \paren {\paren {b \circ c} \circ d}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \circ \paren {b \circ \paren {c \circ d} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \circ b} \circ \paren {c \circ d}\) |

$\blacksquare$

## Also see

- General Associativity Theorem, which demonstrates this rule for any number of elements.

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $2.1: \ 11$ - 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.2$. Commutative and associative operations - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 27 \alpha$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 28$. Associativity and commutativity - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions