General Commutativity Theorem

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

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

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

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


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

Proof

 * Let $$T$$ be the set of all $$n \in \N^*$$ 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 $$\left \langle {a_k} \right \rangle_{1 \le k \le n}$$ of $$n$$ terms 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 $$\sigma: \N^*_n \to \N^*_n$$.


 * It is clear that $$1 \in T$$.


 * Suppose $$n \in T$$.

Let $$\left \langle {a_k} \right \rangle_{1 \le k \le n+1}$$ be a sequence of $$n+1$$ terms in $$S$$ which satisfy:


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

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

There are three cases to consider:


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


 * Suppose $$\sigma \left({n+1}\right) = n + 1$$:

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

Thus, as $$n \in T$$:


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

from which:

$$ $$ $$


 * Suppose $$\sigma \left({1}\right)= n + 1$$:

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


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

From Closed Interval of Successor, $$\left[{1 \,. \, . \, n+1}\right] = \left[{1 \,. \, . \, n}\right] \cup \left\{{n+1}\right\}$$.

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

Thus, as $$n \in T$$:


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

So:

$$ $$ $$ $$ $$


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

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



\forall k \in \N^*_{n+1}: \tau \left({k}\right) = \begin{cases} \sigma \left({k}\right): & k \in \left[{1 \,. \, . \, m-1}\right] \\ \sigma \left({k+1}\right): & k \in \left[{m \,. \, . \, n}\right] \\ n + 1: & k = n + 1 \end{cases} $$

Clearly $$\tau$$ is a permutation of $$\N^*_{n+1}$$. So, by the first case:


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

Thus:

$$ $$ $$ $$ $$

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

Thus by induction $$T = \N^*$$, and the result follows.