general commutativity

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

\displaystyle\prod_{i=1}^{n}a_{\pi(i)}\;=\;\prod_{i=1}^{n}a_{i}.

(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.}\quad\pi(k)=m.

Then

\prod_{i=1}^{m}a_{\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 exactly the factors $a_{1},a_{2},\ldots,a_{m-1}$ in a certain , is also by the induction hypothesis equal to $\prod_{i=1}^{m-1}a_{i}$ . Thus we obtain

\prod_{i=1}^{m}a_{\pi(i)}\;=\;\prod_{i=1}^{m-1}a_{i}\cdot a_{m}\;=\;\prod_{i=1% }^{m}a_{i},

whence (1) is true for $n=m$ .

Note. There is mentionned in the Remark of the entry “http://planetmath.org/node/2148commutativity” a more general notion of commutativity.

Title	general commutativity
Canonical name	GeneralCommutativity
Date of creation	2014-05-10 21:59:41
Last modified on	2014-05-10 21:59:41
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Theorem
Classification	msc 20-00
Related topic	CommutativeLanguage
Related topic	GeneralAssociativity
Related topic	AbelianGroup2