Prob. 19, Sec. 2.3, in Herstein's TOPICS IN ALGEBRA, 2nd ed: Bracketing any $n$-tuple of elements in a set with an associative binary operation

Question

Here is Prob. 19, Sec. 2.3, in the book Topics in Algebra by I.N. Herstein, 2nd edition:

If $S$ is a set closed under an associative operation, prove that no matter how you bracket $a_1a_2 \ldots a_n$, retaining the order of the elements, you get the same element in $S$ (e.g., $\left( a_1 \cdot a_2 \right) \codt \left( a_3 \cdot a_4 \right) = a_1 \cdot \left( a_2 \cdot \left( a_3 \cdot a_4 \right) \right)$; use induction on $n$).

My Attempt:

Of course the assertion holds for $n = 1, 2, 3$. Suppose it holds for every $k = 1, 2, 3, \ldots, n$ for some $n \in \mathbb{N}$ such that $n \geq 3$.

Let $a_1, a_2, a_3, \ldots, a_n, a_{n+1}$ be any elements of set $S$.

How to proceed from here? How to prove rigorously that, no matter how we bracket $a_1a_2a_3\ldots a_n a_{n+1}$, retaining the order of the elements, we obtain the same element of $S$.

Answer 1

Let's consider a string which is simply a finite sequence $w$.

Define an association of the string $w$ to be any string that is built by inserting characters "$($" and "$)$" into $w$ according to following rules. Let $N=\text{Length}(w)$.

• If $N \le 2$ then $w$ is the only association of $w$.
• If $N \ge 3$, first choose $n$ with $1 \le n \le N-1$, then decompose $w$ as a concatenation $w=uv$ where $\text{Length}(u)=n$ and $\text{Length}(v)=N-n$, and then:
  • If $n=1$, choose $v'$ to be any association of $v$, and then $u(v')$ is an association of $w$.
  • If $n=N-1$, choose $u'$ to be any association of $u$, and then $(u')v$ is an association of $w$.
  • If $2 \le n \le N-2$, choose $u'$ to be an association of $u$ and choose $v'$ to be an association of $v$, and then $(u')(v')$ is an association of $w$.

Now, as suggested, you should proceed by induction on the length of $w$, to prove that the outcome of multiplication is independent of the choice of association of $w$.