product of multiple factors

Question

If $a_1,a_2,\dots,a_n$ are elements of a group $G$, $n\geq 2$, then we define the product $a_1a_2\cdots a_n$ with complete induction to $n$ with \begin{align} a_1a_2&=\text{product of $a_1$ and $a_2$ in the group}& (n=2),\\ a_1a_2\cdots a_n&=(a_1a_2\cdots a_{n-1})\cdot a_n& (n>2). \end{align} For example: $abcde=(((ab)c)d)e)$. With complete induction to $n$ we can deduce easily from the associate law that \begin{align} (a_1a_2\cdots a_k)\cdot(a_{k+1}\cdots a_n)=a_1a_2\cdots a_n&&(1\leq k\leq n-1). \end{align}

What was the induction hypothesis here? Was it for some $n\in\mathbb N_{>2}$, it holds that for all $k\leq n$, $$a_1a_2\cdots a_k=(a_1a_2\cdots a_{k-1})a_k.$$

Do I have to say the following;

Consider $a_1a_2\cdots a_k$ and $a_{k+1}$. If we multiply these two, we can write $(a_1a_2\cdots a_k)a_{k+1}$. Apply the induction hypothesis, and we're there.

I don't really get what they're doing; are they defining where to put the brackets? But a group obeys the associativity law, so why would they do that?

EDIT

This is the second induction proof:

Base case: $(a_1a_2)a_3=a_1(a_2a_3)$, by associativity law. Assume that for some $n>3$, it holds that $a_1\cdots a_n=(a_1\cdots a_k)\cdot(a_{k+1}\cdots a_n)$, for $1\leq k\leq n-1$. Consider the product with $n+1$ factors: $a_1\cdots a_{n+1}$. We can write this as $(a_1\cdots a_n)a_{n+1}$. The IH gives that this equals $((a_1\cdots a_k)\cdot(a_{k+1}\cdots a_n))a_{n+1}$. We can use the associativity law to set this equal to $(a_1\cdots a_k)\cdot((a_{k+1}\cdots a_n)a_{n+1})=(a_1\cdots a_k)\cdot (a_{k+1}\cdots a_{n+1})$. Because we know that $a_1\cdots a_{n+1}=(a_1\cdots a_n)a_{n+1}$, it is also true for $k=n$, so we've show that it holds for $1\leq k \leq n$.

Answer 1

First of all, there are two places where they mention "complete induction":

1. in the definition of $a_1a_2\cdot a_n$;
2. and in regards to the claim that $(a_1a_2\cdots a_k)\cdot(a_{k+1}\cdots a_n)=a_1a_2\cdots a_n$ for all $1\leq k\leq n-1$, saying that a proof can be done by induction.

Are you asking about the meaning of complete induction in both of them or only in one of them?

But basically, yes: they are defining where to put brackets, and then proving that the result is independent of wherever we put those brackets. (Well, there's no proof here, only a hint that such a proof can be done by induction.)

Why do we need to go thru all this trouble? First of all, there's only one operation in a group, and it's a binary one. It defines what $ab$ means, but NOT what "$abc$" means. And since we can't view "$abc$" as a single ternary operation, we have to see it as two multiplications, and then we have to specify which one goes first. In other words, taken literally, the expression "$abc$" is meaningless. Meaningful expressions are either $(ab)c$ or $a(bc)$. But since a group has the associative law, $(ab)c=a(bc)$, and that's what allows us to use an abbreviated notation "$abc$": it's actually ambiguous, but we can take it to mean either $(ab)c$ or $a(bc)$, because thankfully they are equal.

But that's as far as a reference to associative law, without any further work, goes. Note that the law says only that $(ab)c=a(bc)$; it does NOT say anything about products by four or more elements. We're so much accustomed to saying that by associativity $abcde$ is unambiguous in a group and to taking that for granted. But it's NOT the associativity axiom! This more general associativity property is a theorem. And that's what they do here: they state it and hint at how it can be proven.