Associativity on Four Elements

Theorem

Let $\left({S, \circ}\right)$ 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 $\left({S, \circ}\right)$ is a semigroup:

• it is closed
• $\circ$ is associative

It can be shown that 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:

$\forall s_1, s_2, s_3 \in S: \left({s_1 \circ s_2}\right) \circ s_3 = s_1 \circ \left({s_2 \circ s_3}\right)$

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

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\left({a \circ b}\right) \circ c}\right) \circ d\)	\(=\)	\(\displaystyle \left({a \circ \left({b \circ c}\right)}\right) \circ d\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \circ \left({\left({b \circ c}\right) \circ d}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \circ \left({b \circ \left({c \circ d}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a \circ b}\right) \circ \left({c \circ d}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Comment

This entry is clearly superseded by the General Associativity Theorem, which demonstrates this rule for any number of elements.

The entry is included, however, for its instructive and illustrative nature.

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 4.2$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 27 \alpha$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 28$