Associativity and Commutativity Properties

Theorem

Let $\circ$ be a binary operation on a set $S$.

Let $\circ$ be associative.

Let $x, y, z \in S$.

• If $x$ commutes with both $y$ and $z$, then $x$ commutes with $y \circ z$.

• If $x$ and $y$ both commute with $z$, then $x \circ y$ commutes with $z$.

General Theorem

Let $\circ$ be a binary operation on a set $S$.

Let $\circ$ be associative.

Let $\left \langle {a_k} \right \rangle_{1 \le k \le n}$ be a sequence of terms of $S$.

Let $b \in S$.

If $b$ commutes with $a_k$ for each $k \in \left[{1 \,.\,.\, n}\right]$, then $b$ commutes with $a_1 \circ \cdots \circ a_n$.

Proof

The following are demonstrated by associativity of $\circ$ and the defined commutativity relations.

• If $x$ commutes with both $y$ and $z$, then $x$ commutes with $y \circ z$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \circ \left({y \circ z}\right)\)	\(=\)	\(\displaystyle \left({x \circ y}\right) \circ z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({y \circ x}\right) \circ z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle y \circ \left({x \circ z}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle y \circ \left({z \circ x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({y \circ z}\right) \circ x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

• If $x$ and $y$ both commute with $z$, then $x \circ y$ commutes with $z$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \circ y}\right) \circ z\)	\(=\)	\(\displaystyle x \circ \left({y \circ z}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \circ \left({z \circ y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \circ z}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({z \circ x}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z \circ \left({x \circ y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

The truth of the general theorem can be proved by induction.

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $2.16$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 4$: Theorem $4.5: \ 1^\circ$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 18$: Theorem $18.6$