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Theorem. If the binary operation ``$\cdot$ '' on the set $S$ is commutative, then for each $a_1,\,a_2,\,\ldots,\,a_n$ in $S$ and for each permutation $\pi$ on $\{1,\,2,\,\ldots,\,n\}$ , one has
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(1) |
Proof. If $n = 1$ , we have nothing to prove. Make the induction hypothesis, that (1) is true for $n = m\!-\!1$ . Denote $$\pi^{-1}(m) \;=\; k, \quad \mbox{i.e.}\;\; \pi(k) = m.$$ Then $$\prod_{i=1}^ma_{\pi(i)} \;=\; \prod_{i=1}^{k-1}a_{\pi(i)}\cdot a_{\pi(k)}\cdot\prod_{i=1}^{m-k}a_{\pi(k+i)} \;=\; \left(\prod_{i=1}^{k-1}a_{\pi(i)}\cdot\prod_{i=1}^{m-k}a_{\pi(k+i)}\right)\cdot a_m,$$ where $a_m$ has been moved to the end by the induction hypothesis. But the product in the parenthesis, which contains exactly the factors $a_1,\,a_2,\,\ldots,\,a_{m-1}$ in a certain order, is also by the induction hypothesis equal to $\prod_{i=1}^{m-1}a_i$ . Thus we obtain $$\prod_{i=1}^ma_{\pi(i)} \;=\; \prod_{i=1}^{m-1}a_i\cdot a_m \;=\; \prod_{i=1}^ma_i,$$ whence (1) is true for $n = m$ .
Note. There is mentionned in the Remark of the entry ``commutativity'' a more general notion of commutativity.
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