# 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

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

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 |