general commutativity

Theorem.  If the binary operation “$\cdot$” on the set $S$ is commutative, then for each $a_1,a_2,\mathrm{\dots },a_n$ in $S$ and for each permutation $\pi$ on  $\{1,\mathrm{\hspace{0.17em}2},\mathrm{\dots },n\}$,  one has

${\displaystyle \prod _{i=1}^n}{a}_{\pi (i)}={\displaystyle \prod _{i=1}^n}{a}_i.$(1)

Proof.  If  $n=1$,  we have nothing to prove. Make the induction hypothesis, that (1) is true for $n=m-1$.  Denote

$${\pi }^{-1}(m)=k,\text{i.e.}\mathit{\hspace{1em}}\pi (k)=m.$$

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 inductionhypothesis.  But the product in the parenthesis, which exactly the factors$a_1,a_2,\mathrm{\dots },a_{m-1}$ 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.