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Let $(G, * )$ be a group . For any $a_1,...,a_n \in G$, the value of $a_1 * ... * a_n$ is independent of how the expression is bracketed.

Attempt:

hint is to use inducition . Notice that for $n=3$ we have

$$ (a_1 * a_2 ) * a_3 = a_1 * (a_2 * a_3) $$

and there is not other way to place the parenthesis but is true by definition. Now, $n=4$ let try see all possible ways to place the parentesis:

$$ a_1a_2a_3a_4 = ((a_1a_2)a_3)a_4 = (a_1 (a_2a_3)) a_4 = a_1 ((a_2a_3)a_4) = a_1(a_2 (a_3a_4)) = (a_1a_2)(a_3a_4)$$

Let us assume it is possible to do so for some $n$: that is

Let $S_n$ be the statement that: value of $a_1...a_n$ is independent of how we place parenthesis. We prove $S_{n+1}$ is true.Let $A = a_1...a_n$. Now express this as

$$ A = (...(a_1...)))a_{n-1}) a_n $$

Now, consider $A a_{n+1}$. That is,

$$ A * a_{n+1} = (...(a_1...)))a_{n-1}) a_n * a_{n+1} $$

but here I get stuck. Im not quite sure whether the LHS actually can be done... any help would be greatly appreciated.

Community
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James
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1 Answers1

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So we want to prove that every expressions $a_1*\cdots *a_n$, regardless of bracketing is equal to $a_1*(a_2*(a_3*(\cdots *a_n)\ldots )$.

By induction.

The case for $n=1$ is trivially true.

So assume that $n>1$ is given and that our hypothesis holds for any $1\leq m <n$. An arbitrary bracketing on the expression $a_1*\cdots *a_n$ can be written as $x*y$ where $x=a_1*\cdots*a_k$ and $y=a_{k+1}*\cdots * a_n$ are subbracketings for $1\leq k<n$. Since both subsequences are of length $<n$ then we can apply the induction hypothesis to them and we obtain

$$\big(a_1*(a_2*(a_3*(\cdots *a_k)\ldots)\big)*\big(a_{k+1}*(a_{k+2}*(a_{k+3}*(\cdots * a_n)\ldots)\big)$$

Next we apply associativity in order to move $a_1$ to the top level:

$$a_1*\bigg(\big(a_2*(a_3*(\cdots *a_k)\ldots )\big)*\big(a_{k+1}*(a_{k+2}*(a_{k+3}*(\cdots * a_n)\ldots)\big)\bigg)$$

Now the top sub-bracketing (the one starting from $a_2$) is again of length $<n$ and so the induction hypothesis applies to it. We obtain

$$a_1*(a_2*(a_3*(\cdots *a_n)\ldots )$$

which finishes the proof. $\Box$

John
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freakish
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  • Thanks for your reply! I am still unsure about the arbitrary positionin of the brackets. When you write the 9th line, is this bracketing of $x$ and $y$ arbitrary? – James Jun 05 '20 at 07:45
  • @James bracketing of $x$ and $y$ depends on the original $a_1\cdots a_n$ bracketing of course. For example if we have the following bracketing $(a_1a_2)(a_3a_4)$ then $x=a_1a_2$ and $y=a_3a_4$. Or with $((a_1a_2)a_3)a_4$ we get $x=(a_1a_2)a_3$ and $y=a_4$. There's always top level "$$" which divides entire sequence into two pieces $xy$. By definition: bracketing is nothing else then the order on the composition of "$*$" treated as a function. – freakish Jun 05 '20 at 09:07