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I am reading Dummit and Foote's Abstract Algebra and in the section on basic axioms, they have left the generalized associative law to be proved by the reader but have given a proof outline as follows:


"For any $a_1,a_2,...,a_n$ $\in$ G the value of $a_1*a_2*...*a_n$ is independent of how the expression is bracketed(this is called the generalized associative law).

This is left as a good exercise using induction on n. First show the result is true for n = 1, 2, and 3. Next assume for any k $<$ n that any bracketing of a product of k elements $ b_1 * b_2*...*b_k$ can be reduced (without altering the value of the product) to an expression of the form $$b_1* (b_2 *(b_3* ( ... * b_k))...).$$ Now argue that any bracketing of the product $a_1 * a_2 * ... * a_n$, must break into 2 subproducts, say $(a_1 * a_2 * ... * a_k) * (a_{k+1} * a_{k+2} * ... * a_n)$, where each sub-product is bracketed in some fashion. Apply the induction assumption to each of these two sub-products and finally reduce the result to the form $a_1 * (a_2 * (a_3 * (... *a_n))...)$ to complete the induction."



I understand how to prove the base case and understand the induction hypothesis but I don't understand how to argue that any bracketing of the product $a_1 * a_2 * ... * a_n$ must break into 2 subproducts, and also how would I reduce the subproducts, after applying the induction hypothesis and getting the form $a_1 * (a_2 * (a_3 * (... *a_k))...)$ and $a_{k+1} * (a_{k+2} * (a_{k+3} * (... *a_n))...)$, into the form $a_1 * (a_2 * (a_3 * (... *a_n))...)$. Any help would be greatly appreciated.

Mohammad Areeb
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  • Does this help? If I vote it as a duplicate, it will close immediately, so I'm asking first. – Arturo Magidin Jul 30 '21 at 11:06
  • @ArturoMagidin thank you so much the post you linked is really clear – Mohammad Areeb Jul 30 '21 at 12:35
  • you can mark it as duplicate – Mohammad Areeb Jul 30 '21 at 12:35
  • @ArturoMagidin sorry for disturbing you but I have just one last question how can we apply the induction hypothesis to (n-1) factors, haven't we just assumed it to be true for k factors? – Mohammad Areeb Jul 30 '21 at 14:21
  • @MohammadAreeb: This is done using so-called "strong induction" or the second principle of induction: we assume it holds for all $k$ strictly smaller than $n$, and prove it for $n$. See, e.g., this question. – Arturo Magidin Jul 30 '21 at 14:24
  • @ArturoMagidin I don't get why n-1 is strictly less than or equal to k. (Sorry I am really dumb and trying to study things by myself) – Mohammad Areeb Jul 30 '21 at 14:29
  • @MohammadAreeb: Again: in strong induction, you assume the result holds for all number strictly smaller than the one you are at, and then prove it for that one. Don't get tangled up on whether you are using $n$, $k$, $n+1$, or $\aleph$. In that post, it is. necessary that $k$ be smaller than $n$ because of how it is defined: It cannot be equal to $n$ because of it definition. Look at the specific examples I give in the answer to see how $k$ is defined. It cannot be $n$, because that makes absolutely no sense. – Arturo Magidin Jul 30 '21 at 14:41
  • And, no, you don't need "$n-1$ strictly less than or equal to $k$". You need $k$ strictly less than $n$. – Arturo Magidin Jul 30 '21 at 14:44
  • @ArturoMagidin Right! Thank you so much, the notation tripped me up a bit but this statement really helped "in strong induction, you assume the result holds for all number strictly smaller than the one you are at". Thank you Thank you <3 – Mohammad Areeb Jul 30 '21 at 14:45
  • @ArturoMagidin Really nice proof btw – Mohammad Areeb Jul 30 '21 at 14:50

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