# Definition of identity in a monoid

I'm having trouble understanding the way the identity element is defined in Lang's Algebra. Below is the relevant information. Suppose we have a monoid G with elements $x_{1},...,x_{n}$. We can define their products like so, $$\Pi_{v=1}^{n} = (x_{1}...x_{n-1})x_{n}$$ We define $$e=\Pi_{v=1}^{0}x_{n}$$ where $e$ is the identity element of the monoid. I do not understand why this definition of $e$ makes sense. It seems to me that this is saying $e=x_{1}x_{0}$, so in particular, $x_{0}$ is the inverse of $x_{1}$. Thank you in advance.

-
In this definition $v$ is supposed to run through the set $\{1,2,\cdots,n\}$. So if $n=0$ this set is empty, and so you have the empty product which is defined to be the neutral element of the monoid -- this is the only reasonable way to define an empty product. So Lang isn't defining the neutral element there, he's just giving a convention for empty products. –  Ryan Budney Jun 11 '11 at 6:16
See also my answer here on empty exponents, products and sums. –  Bill Dubuque Jun 11 '11 at 18:23

This is just an empty product - the notation $\prod_{v=i}^j$ is shorthand for $\prod_{i\leq v\leq j}$, and there is no $v$ such that $1\leq v\leq 0$, so $\prod_{v=1}^0 x_v$ is a product of no elements, not $x_1x_0$.
This expression is the empty product and not the product of $x_0$ with $x_1$. As it is mentioned in p.4 below this expression, it is a matter of convention to denote the identity element as the empty product.