A question about associativity in monoids. Lang's "Algebra" (on pg. 4) says the following:

Let $G$ be a monoid. Then $\Pi{x_i}$ is defined as $(x_2x_2\dots)x_n$. 

This probably means $\Pi x_i=(((x_1x_2)x_3)\dots)x_n$.
He then says 

We then have the following rule $\Pi_a^bx_i.\Pi_{b+1}^c x_j=\Pi_a^cx_k$. This ensures that parenthesis can be placed in any way. 

I don't understand this. A monoid anyway has associative property. Shouldn't the fact that parentheses can be placed in any way have followed from the first statement itself?
Thanks in advance!
 A: Lang's point is that these intuitive  normalizations using the assoicative law require formal proof. Below is one way to do so from Bergman's superb Companion to Lang's Algebra, from pp. 5-7 of the Introduction, and Notes to Chapter I. I highly recommend that anyone reading Lang's Algebra have Bergman's notes close at hand. I suggest that you attempt to prove the Lemma's below before consulting their proofs in Bergman's notes.



A: I am just getting into this book, and came up with the following proof, which I thought was simple enough. I am curious what you all would think....
Let $G$ be a monoid. Fix an integer $m \ge 1$. Suppose $n=1$.
Then we have
$$(\Pi_{p=1}^{m} x_p) * (\Pi_{q=1}^{n} x_{q+m})$$
$$ = (x_1 * ... * x_m) * x_{m+1}$$
$$ = \Pi_{p=1}^{m+1} x_p.$$
Suppose $n=k$ for some positive integer $k$. Assume this inductive hypothesis:
$$(\Pi_{p=1}^{m} x_p) * (\Pi_{q=1}^{k} x_{q+m}) = (\Pi_{p=1}^{m+k} x_p)$$
Then we have
$$(\Pi_{p=1}^{m} x_p) * (\Pi_{q=1}^{k+1} x_{q+m})$$
$= (\Pi_{p=1}^{m} x_p) * \bigl((x_{1+m} * ... * x_{k+m}) * x_{k+1+m}\bigr)$ by definition of product
$= \bigl((\Pi_{p=1}^{m} x_p) * (x_{1+m} * ... * x_{k+m})\bigr) * x_{k+1+m}$ by associativity of the law of composition of $G$.
$= \bigl((\Pi_{p=1}^{m} x_p) * (\Pi_{q=1}^{k} x_{q+m})\bigr) * x_{k+1+m}$ by definition of product
$= (\Pi_{p=1}^{m+k} x_p) * x_{k+1+m}$ by our inductive hypothesis
$= \Pi_{p=1}^{k+1+m} x_p$ by our definition of product.
By the principle of mathematical induction, that proves the point for all $n$.
