I want to prove the following: If $G$ is a group under the operation $\star$, $\forall a_1,…,a_n\in G$, the value of $a_1\star a_2\star … \star a_n$ is independent of how the expression is bracketed.
My attemp at the proof:
We prove this with induction on $n$. Base case: $n=3$, $a_1\star (a_2\star a_3)=(a_1\star a_2)\star a_3$. This is true by the definition of a group operation. Because a group operation is a mapping $\star :G\times G\rightarrow G$, we can substitute any $a_i\star a_j$ with some $a_k\in G$. Therefore, $a_1\star a_2\star …\star a_n\star a_{n+1}$ is equivalent to $a_1\star a_2\star …\star a_k$, which is true by the inductive hypothesis.
I want to know if:
$1)$ There are any correction to be made, and
$2)$ Better wording for the proof.
Thank you all in advance.
Best Answer
I see what you're trying to do here, but the wording needs some work.
I actually proved this in an assignment once. It was a great work of passive agression against my lecturer who insisted vehemently that no steps be missed. So, I decided to try formally proving generalised associativity (and commutativity too). Here's a cleaned-up version of the proof I gave him.
All of that said, I do realise that the exposition at the end is a bit of a cop-out. In order to do things properly, you need to be able to encode the idea of placing brackets in order to form a well-formed expression. My suggestion would be to form a parse tree, and prune it inductively.
Anyway, I hope this helps in some way.