General Commutativity Theorem

Theorem
Let $\left({S, \circ}\right)$ be a semigroup.

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

Suppose that:
 * $\forall i, j \in \left[{1 \,.\,.\, n}\right]: a_i \circ a_j = a_j \circ a_i$

Then for every permutation $\sigma: \N^*_n \to \N^*_n$:


 * $a_{\sigma \left({1}\right)} \circ \cdots \circ a_{\sigma \left({n}\right)} = a_1 \circ \cdots \circ a_n$

where $\N^*_n$ denotes the initial segment of $\N_{>0}$:
 * $\N^*_n = \left\{{1, 2, \ldots, n}\right\}$

Proof
The proof will proceed by the Principle of Finite Induction on $\N_{>0}$.

Let $T$ be the set of all $n \in \N_{>0}$ such that:
 * $a_{\sigma \left({1}\right)} \circ \cdots \circ a_{\sigma \left({n}\right)} = a_1 \circ \cdots \circ a_n$

holds for all sequences $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ of $n$ elements of $S$ which satisfy:
 * $\forall i, j \in \left[{1 \,.\,.\, n}\right]: a_i \circ a_j = a_j \circ a_i$

for every permutation $\sigma: \N^*_n \to \N^*_n$.

Basis for the Induction
We have that when $n = 1$:


 * $\sigma \left({1}\right) = 1$

and so:


 * $a_{\sigma \left({1}\right)} = a_1$

for all $\sigma: \left\{{1}\right\} \to \left\{{1}\right\}$.

Thus $1 \in T$.

This is our basis for the induction.

Induction Hypothesis
It is to be shown that, if $m \in T$ where $m \ge 1$, then it follows that $m + 1 \in T$.

This is the induction hypothesis:


 * $a_{\sigma \left({1}\right)} \circ \cdots \circ a_{\sigma \left({m}\right)} = a_1 \circ \cdots \circ a_m$

holds for all sequences $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le m}$ of $m$ elements of $S$ which satisfy:
 * $\forall i, j \in \left[{1 \,.\,.\, m}\right]: a_i \circ a_j = a_j \circ a_i$

for every permutation $\sigma: \N^*_m \to \N^*_m$.

It is to be demonstrated that it follows that:


 * $a_{\sigma \left({1}\right)} \circ \cdots \circ a_{\sigma \left({m + 1}\right)} = a_1 \circ \cdots \circ a_{m + 1}$

holds for all sequences $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le m + 1}$ of $m + 1$ elements of $S$ which satisfy:
 * $\forall i, j \in \left[{1 \,.\,.\, m + 1}\right]: a_i \circ a_j = a_j \circ a_i$

for every permutation $\sigma: \N^*_{m + 1} \to \N^*_{m + 1}$.

Induction Step
This is our induction step:

Suppose $m \in T$.

Let $\left \langle {a_k} \right \rangle_{1 \le k \le m+1}$ be a sequence of $m+1$ elements of $S$ which satisfy:
 * $\forall i, j \in \left[{1 \,.\,.\, n+1}\right]: a_i \circ a_j = a_j \circ a_i$

Let $\sigma: \N^*_{m+1} \to \N^*_{m+1}$ be a permutation of $\left[{1 \,.\,.\, m+1}\right]$.

There are three cases to consider:


 * $(1): \quad \sigma \left({m+1}\right) = m + 1$
 * $(2): \quad \sigma \left({1}\right)= m + 1$
 * $(3): \quad \sigma \left({r}\right) = m + 1$ for some $r \in \left[{2 \,.\,.\, m}\right]$

$(1): \quad$ Suppose $\sigma \left({m+1}\right) = m + 1$.

Then the restriction of $\sigma$ to $\N^*_m$ is then a permutation of $\N^*_m$.

From the induction hypothesis:
 * $m \in T$:

Thus:
 * $a_{\sigma \left({1}\right)} \circ \cdots \circ a_{\sigma \left({m}\right)} = a_1 \circ \cdots \circ a_m$

from which:

$(2): \quad$ Suppose $\sigma \left({1}\right) = m + 1$.

Let $\tau: \N^*_m \to \N^*_m$ be the mapping defined as:


 * $\forall k \in \left[{1 \,.\,.\, m}\right]: \tau \left({k}\right) = \sigma \left({k + 1}\right)$

From Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor:
 * $\left[{1 \,.\,.\, m + 1}\right] = \left[{1 \,.\,.\, m}\right] \cup \left\{{m + 1}\right\}$

Thus $\tau$ is clearly a permutation on $\left[{1 \,.\,.\, m}\right]$.

Thus, as $m \in T$:


 * $a_{\tau \left({1}\right)} \circ \cdots \circ a_{\tau \left({m}\right)} = a_1 \circ \cdots \circ a_m$

So:

$(3): \quad$ Suppose $\sigma \left({r}\right) = m + 1$ for some $r \in \left[{2 \,.\,.\, m}\right]$.

Let $\tau: \N^*_{m + 1} \to \N^*_{m + 1}$ be a mapping defined by:


 * $\forall k \in \N^*_{m + 1}: \tau \left({k}\right) = \begin{cases}

\sigma \left({k}\right) & : k \in \left[{1 \,.\,.\, r - 1}\right] \\ \sigma \left({k + 1}\right) & : k \in \left[{r \,.\,.\, m}\right] \\ m + 1 & : k = m + 1\end{cases}$

Clearly $\tau$ is a permutation of $\N^*_{m+1}$.

So, by the first case:


 * $a_{\tau \left({1}\right)} \circ \cdots \circ a_{\tau \left({m + 1}\right)} = a_1 \circ \cdots \circ a_{m + 1}$

Thus:

So in all cases, $m + 1 \in T$.

Thus by the Principle of Finite Induction:
 * $T = \N_{>0}$

The result follows.