# External Direct Product Associativity

## Theorem

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.

Let $\circ_1$ and $\circ_2$ be associative.

Then $\circ$ is also associative.

### General Result

Let $\displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$.

If $\circ_1, \ldots, \circ_n$ are all associative, then so is $\circ$.

## Proof

 $\displaystyle \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3}$ $=$ $\displaystyle \tuple {\paren {s_1 \circ_1 s_2} \circ_1 s_3, \paren {t_1 \circ_2 t_2} \circ_2 t_3}$ $\displaystyle$ $=$ $\displaystyle \tuple {s_1 \circ_1 s_2 \circ_1 s_3, t_1 \circ_2 t_2 \circ_2 t_3}$ Definition of Associative Operation $\displaystyle$ $=$ $\displaystyle \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} }$ $\displaystyle$ $=$ $\displaystyle \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} }$

Thus $\circ$ is associative.

$\blacksquare$