This is Artin Proposition 1.4 (from his book Algebra)
Suppose there is a associative law of composition given on a set S. There is a unique way to define, for every integer n, a product of n elements $a_1 , . . . a_n$ of S with the following properties:
i) the product of one element is the element itself.
ii) the product $a_1a_2$ is given by the law of composition.
iii) for any natural n, $[a_1a_2…a_n]=[a_1…a_i][a_{i+1}…a_n]$
Proof:
The product is defined by (i) and (ii) for $n≤2$ and satisfies (iii) for $n=2$. Suppose that we have defined the product of $r$ elements when $r≤n-1$ and that it is the unique product satisfying (iii). We then define the product of $n$ elements by the rule$$[a_1…a_n]=[a_1…a_{n-1}][a_n]$$
where the terms on the right side are those already defined. If a product satisfying (iii) exists, then this formula gives the product because it is (iii) when $i =n-1$.
So if it exists, the product is unique.We must now check (iii) for $i< n-1$:
\begin{align} [a_1…a_n]&=[a_1…a_{n-1}][a_n] &&\text{(our definition)}\\
&=([a_1…a_i][a_{i+1}…a_{n-1}])[a_n] &&\text{(induction hypothesis)}\\
&=[a_1…a_i]([a_{i+1}…a_{n-1}][a_n]) &&\text{(associative law)} \\
&=[a_1\cdots a_i][a_{i+1}\cdots a_n] &&\text{(induction hypothesis)}
\end{align}
This completes the proof.
why is it necessary to prove the case for $i< n-1$, since normally in inductive proofs we only prove the case for $i=n$?
Is it because we want to end with the formula in that particular format $(a_1 . . . a_i)(a_{i+1}. . .a_n)$
or is it because it is a inductive proof where we assumed the product exists for all $i< n-1$ as or inductive step (instead of just existing for n-1)?
Best Answer
Let us reformulate statement and proof. This will make clear what is going on.
Suppose there is a associative law of composition given on a set $S$. There is a unique way to define, for every integer $n$, a product of n elements $a_1 , \dots, a_n$ of $S$, denoted by $[a_1\dots a_n]$, with the following properties:
(1) $[a_1] = a_1$.
(2) For any $n > 1$ the following property is satisfied:
$P(n) \quad$ $[a_1a_2...a_n]=[a_1...a_i]\cdot[a_{i+1}...a_n]$ for all $i$ with $0 < i < n$.
Note that the orginal (ii) is nothing else than $P(2)$ since in this case $i=1$ is the only possible choice.
Proof:
The product is defined by (1) for $n = 1$.
For $n > 1$ the product is defined by induction.
Suppose that we have defined the product of $r$ elements for $r =1,\dots,n-1$ and that it is the unique product satisfying $P(r)$ for all these $r$. If a product of $n$ elements satisfying $P(n)$ exists, then it must satisfy the equation $$D(n) \quad [a_1...a_n]=[a_1...a_{n-1}] \cdot [a_n]$$ because this is $P(n)$ when $i =n-1$. Hence it will be unique because $[a_1...a_{n-1}]$ is unique.
Now let us define $ [a_1...a_n]$ by equation $D(n)$. We must check that $P(n)$ holds true. For $i = n-1$ this is nothing else than the definition. For $0 < i < n-1$ we have
\begin{align} [a_1...a_n]&=[a_1...a_{n-1}][a_n] &&\text{(our definition)}\\ &=([a_1...a_i][a_{i+1}...a_{n-1}])[a_n] &&\text{(induction hypothesis)}\\ &=[a_1...a_i]([a_{i+1}...a_{n-1}][a_n]) &&\text{(associative law)} \\ &=[a_1\cdots a_i][a_{i+1}\cdots a_n] &&\text{(induction hypothesis)} \end{align} This completes the proof.