The second axiom says: when combining three elements, it doesn’t matter where you put the parentheses.

$$(a \cdot b) \cdot c = a \cdot (b \cdot c)$$

This means we can write $a \cdot b \cdot c$ without ambiguity.

Why it matters

Without associativity, the expression $a \cdot b \cdot c$ would be ambiguous — you’d need to specify which pair to combine first, and different choices might give different results.

With associativity, you can chain as many operations as you like — $a \cdot b \cdot c \cdot d \cdot e$ — and evaluate them in any order, always getting the same result. This is what makes groups computable: you never need to track grouping.

Where associativity holds

Addition: $(2 + 3) + 4 = 5 + 4 = 9$ and $2 + (3 + 4) = 2 + 7 = 9$. Always.

Multiplication: $(2 \times 3) \times 4 = 24$ and $2 \times (3 \times 4) = 24$.

Permutation composition: composing rearrangements is associative — doing $\sigma$ then $\tau$ then $\rho$ gives the same result regardless of grouping.

Matrix multiplication: $(AB)C = A(BC)$ always.

Where associativity fails

Subtraction: $(7 - 3) - 2 = 2$ but $7 - (3 - 2) = 6$. Different!

Division: $(12 \div 4) \div 3 = 1$ but $12 \div (4 \div 3) = 9$.

Exponentiation: $(2^3)^2 = 64$ but $2^{(3^2)} = 512$.

Cross product of vectors: $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \neq \mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ in general.

Commutativity — the axiom groups DON’T require

Note that associativity ($a \cdot b \cdot c$ grouping) is different from commutativity ($a \cdot b = b \cdot a$ ordering). Groups require associativity but NOT commutativity. In $S_3$, composing permutations $(12)$ then $(13)$ gives a different result from $(13)$ then $(12)$.

Groups that are also commutative are called abelian (after Niels Henrik Abel — whom we’ll meet again when we discuss the quintic).