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.
Definition of identity in a monoid
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group-theory
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definition
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0See also my [answer here](http://math.stackexchange.com/questions/6832/why-is-two-to-the-power-of-zero-equal-to-binary-one/6838#6838) on empty exponents, products and sums. – 2019-05-24
2 Answers
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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 $x_1x_0$.
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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.