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:
\(\ds a \circ \paren {b \circ c}\) | \(=\) | \(\ds \tuple {a_1, a_2, \ldots, a_k, a_{k + 1} } \circ \paren {\tuple {b_1, b_2, \ldots, b_k, b_{k + 1} } \circ \tuple {c_1, c_2, \ldots, c_k, c_{k + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\tuple {a_1, a_2, \ldots, a_k}, a_{k + 1} } \circ \paren {\tuple {\tuple {b_1, b_2, \ldots, b_k}, b_{k + 1} } \circ \tuple {\tuple {c_1, c_2, \ldots, c_k}, c_{k + 1} } }\) | Definition of Ordered Tuple | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\tuple {a_1, a_2, \ldots, a_k} \circ \paren {\tuple {b_1, b_2, \ldots, b_k} \circ \tuple {c_1, c_2, \ldots, c_k} }, a_{k + 1} \circ_{k + 1} \paren {b_{k + 1} \circ_{k + 1} c_{k + 1} } }\) | Definition of Operation Induced by Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\paren {\tuple {a_1, a_2, \ldots, a_k} \circ \tuple {b_1, b_2, \ldots, b_k} } \circ \tuple {c_1, c_2, \ldots, c_k}, a_{k + 1} \circ_{k + 1} \paren {b_{k + 1} \circ_{k + 1} c_{k + 1} } }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\paren {\tuple {a_1, a_2, \ldots, a_k} \circ \tuple {b_1, b_2, \ldots, b_k} } \circ \tuple {c_1, c_2, \ldots, c_k}, \paren {a_{k + 1} \circ_{k + 1} b_{k + 1} } \circ_{k + 1} c_{k + 1} }\) | Basis for the Induction | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\tuple {\tuple {a_1, a_2, \ldots, a_k}, a_{k + 1} } \circ \tuple {\tuple {b_1, b_2, \ldots, b_k}, b_{k + 1} } } \circ \tuple {\tuple {c_1, c_2, \ldots, c_k}, c_{k + 1} }\) | Definition of Operation Induced by Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\tuple {a_1, a_2, \ldots, a_k, a_{k + 1} } \circ \tuple {b_1, b_2, \ldots, b_k, b_{k + 1} } } \circ \tuple {c_1, c_2, \ldots, c_k, c_{k + 1} }\) | Definition of Ordered Tuple | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ b} \circ c\) | By Hypothesis |
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 $\circ_1, \ldots, \circ_n$.
$\blacksquare$
Also see
- External Direct Product Commutativity
- External Direct Product Identity
- External Direct Product Inverses
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.10: \ 1^\circ$