I've been starting to play around with some properties of groups, and I wanted to prove this claim to make things simpler for me in the future. If $G$ is closed under and operation $\cdot$ that is also associative, I want to show that $\forall a_1, a_2, ..., a_n \in G$, no matter how I choose to bracket $a_1 \cdot a_2 \cdot ... \cdot a_n$ while retaining the order, I will end up with the same element in $G.$ My friend told me that induction on the number of terms would probably be my best bet, and I see how the $n = 1 \wedge n = 2$ cases are trivial with the $n = 3$ case implied by associativity. However I am having some problems Figuring out how to work out the induction step. It seems to me that my induction hypothesis is not helping me when proving for brackets that include the $n+1$th element together with the $n$th element.
1 Answers
It's pretty messy and most books just say "it's obvious". If you do want do do it, probably the easiest way is to be specific, with something like this.
Theorem. A product of terms $a_1,\ldots,a_n$, in that order, with any order of bracketing, is equal to $$(\cdots((a_1*a_2)*a_3)*\cdots*a_{n-1})*a_n\ .$$
Sketch of proof. In a given product, identify the "last" $*$ to be performed. Then the product looks like $$\eqalign{ (\cdots)*(\cdots*a_n) &=(\cdots)*((\cdots)*a_n)\qquad\hbox{[using inductive assumption for the}\cr &\qquad\qquad\qquad\qquad\qquad\hbox{number of terms in the second bracket]}\cr &=((\cdots)*(\cdots))*a_n\qquad\hbox{[associative law for three terms]}\cr &=(\cdots(a_1*a_2)*\cdots)*a_n\qquad\hbox{[inductive assumption for $n-1$ terms]}\cr}$$
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