Doubt in a proof of dropping parentheses with associativity

Question

This question is from Artin Algebra:

Artin gives this proof of why we can get rid of parentheses in composition if associativity is assumed:

What I can't understand is what these sentences have to say:'If a product satisfying (iii) exists, then this formula gives the product because it is (iii) when $i=n-1$. So, if the product of $n$ elements exists, it is unique.' Also, do we get to know whether a product satisfying (iii) exists till the end of the proof?

Answer 1

We are looking for a product that has certain properties. Property (iii) consists, in fact, of $n-1$ properties:

• $[a_1\ldots a_n]=[a_1][a_2a_3\ldots a_n]$;
• $[a_1\ldots a_n]=[a_1a_2][a_3\ldots a_n]$;
• $\cdots$
• $[a_1\ldots a_n]=[a_1\ldots a_{n-1}][a_n]$.

So, Artin uses the last of these properties to define $[a_1\ldots a_n]$ and then he proves that it satisfies all the others.

Besides, the product, defined this way, must be unique, because he is assuming (induction hypothesis) that $[a_1\ldots a_{n-1}]$ is unique.

Answer 2

Answering your last question first: yes, it is not until the end that we know that there is a product satisfying (iii). We define a product recursively via $[a_1\cdots a_n]=[a_1\cdots a_{n-1}][a_n]$, but until the end of the proof, we don't know that this satisfies (iii).

As for the part you quoted, the idea is the definition just given is equal to any other definition of the product that satisfies (iii). Let's say we had another product $\langle a_1\cdots a_n\rangle$ assumed to satisfy (iii). Then the $i=n-1$ case of (iii) states, $\langle a_1\cdots a_n\rangle=\langle a_1\cdots a_{n-1}\rangle\langle a_n\rangle$. By the uniqueness part of induction hypothesis, $\langle a_1\cdots a_{n-1}\rangle = [a_1\cdots a_{n-1}]$ and $\langle a_n \rangle = [a_n]$, so $$\langle a_1\cdots a_n\rangle=\langle a_1\cdots a_{n-1}\rangle\langle a_n\rangle = [a_1\cdots a_{n-1}][a_n] = [a_1\cdots a_n]$$ and thus the products coincide.