External Direct Product Associativity/General Result

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

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

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * If $\circ_1, \ldots, \circ_n$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$.

Basis for the Induction
$\map P 1$ is true, as this just says:
 * $\circ_1$ is associative.

$\map P 2$ is the case:
 * If $\circ_1$ and $\circ_2$ are both associative, then so is the external direct product $\circ$ of $\circ_1$ and $\circ_2$.

This has been proved in External Direct Product Associativity.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * If $\circ_1, \ldots, \circ_k$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_k$.

Then we need to show:
 * If $\circ_1, \ldots, \circ_{k + 1}$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_{k + 1}$.

Induction Step
This is our induction step:

Let $a, b, c \in S^{k + 1}$:
 * $a = \tuple {a_1, a_2, \ldots, a_k, a_{k + 1} }$
 * $b = \tuple {b_1, b_2, \ldots, b_k, b_{k + 1} }$
 * $c = \tuple {c_1, c_2, \ldots, c_k, c_{k + 1} }$

Note that in the below, by abuse of notation, $\circ$ is to be used for two separate operations:
 * $\tuple {a_1, a_2, \ldots, a_k, a_{k + 1} } \circ \tuple {b_1, b_2, \ldots, b_k, b_{k + 1} }$

and:
 * $\tuple {a_1, a_2, \ldots, a_k} \circ \tuple {b_1, b_2, \ldots, b_k}$

Thus:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore: For all $n \in \N_{> 0}$:
 * If $\circ_1, \ldots, \circ_n$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$.

Also see

 * External Direct Product Commutativity
 * External Direct Product Identity
 * External Direct Product Inverses