Let $G$ be a commutative monoid, $x_1,\dots,x_n\in G$. Let $\psi\in S_n$. Then $\prod_{\nu = 1}^nx_{\psi(\nu)}=\prod_{\nu=1}^nx_\nu$

Question

I started learning from Algebra by Serge Lang. In page 5, he presented an equation

Let $G$ be a commutative monoid, and $x_1,\ldots,x_n$ elements of $G$. Let $\psi$ be a bijection of the set of integers $(1,\ldots,n)$ onto itself. Then $$\prod_{\nu = 1}^n x_{\psi(\nu)} = \prod_{\nu=1}^n x_\nu$$

In this equation, the mapping $\psi(\nu) = \nu $, resulting the same value, I don't understand why $x_{\psi(\nu)}$ bothers to index with a mapping, rather than a number.

Answer 1

Here $\psi(v)$ is a different ordering of the actions of multiplication. The $\Pi x_{\psi(v)}$ refers to a different order of the terms in the product than $\Pi{x_v}$. In words, what that means is "Regardless of the order in which the operation is carried out, the result is the same". Contrast with a non-commutative form of multiplication such as matrix multiplication.

Answer 2

$\psi(\nu)$ is not a mapping, but the value of the map $\psi $ at $\nu$. In other words, the $n$-uple $(x_{\psi(1)},x_{\psi(2)},\dots,x_{\psi(n)})\:$ is the $n$-uple $(x_1, x_2,\dots,x_n)$ enumerated in another order.

For instance, suppose $n=5$ and $\psi$ is the cycle $(1\,3\,4\,2\,5)$, then the product $$x_1\, x_2\, x_3\, x_4\, x_5 \enspace\text{becomes the product}\enspace x_3\,x_5\,x_4\, x_2\, x_1. $$

Answer 3

This is an easy exercise if you use induction on $n$. Here $\psi$ represents a different order of the elements of $\{1,\dots, n\}$ and hence a different order of the $x_\nu$ (assuming $\psi$ is not the identity map, although the result still holds in that case).

Answer 4

Suggestion: consider the special case when $n = 3$ and write $a, b, c$ for $x_1, x_2, x_3$ respectively. What Lang is saying in that case is that we have all of the following equalities: $$ abc = bac = cab = acb = bca = cba $$ I.e., the product $x_1x_2x_3$ is independent of the way we order the factors $x_1$, $x_2$ and $x_3$. Now try to figure out how to generalise this to arbitrary $n$: what you will need is something to represent how the factors have been reordered: that is Lang's bijection $\psi$ on the index set $\{1, \ldots, n\}$.