Basically I need to prove that a group $(G,*)$ is associative in the general case. To do this I know I have to use induction to show that no matter where I insert parentheses into the equation $a_1*a_2*a_3...*a_n$ I will always get the same result.

The first step is to show that $(G,*)$ is associative in the smallest possible case, where $n=3$. Clearly if $n=3$ we get $$a*(b*c)=(a*b)*c$$ as $(G,*)$ is a group and we have exhausted all the possibilities as to where we can put the parentheses.

Next we assume $(G,*)$ is associative for $n$ elements and show that it is still associative for $n+1$ elements. We have $$a_1*(a_2*(a_3*(...*(a_{n-1}*(a_n*a_{n+1})...))))$$ This is where I'm stuck, how do I show that if it works for $n$ elements it must work for $n+1$?

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    $\begingroup$ Hint: Use strong induction. Show that if the result is true for all $k\lt n$, it is true for $n$. $\endgroup$ – André Nicolas Jul 4 '14 at 20:33

The outermost bracketing is $$(a_1a_2 \cdots a_k)(a_{k+1}a_{k+1}\cdots a_{n+1})$$ for some $k$ with $1 \le k \le n$. By induction, the inner bracketing of the two bracketed terms makes no difference to the result.

The result will follow if we can show that the product for any value of $k$ is the same as it is for $k>1$. So suppose that $k>1$. Then the product is equal to

$$(a_1(a_2 \cdots a_k))(a_{k+1}a_{k+1}\cdots a_{n+1}),$$

which, by the associativity group axiom is equal to

$$a_1((a_2 \cdots a_k))(a_{k+1}a_{k+1}\cdots a_{n+1})),$$

which is the case $k==1$, and we are done.

  • $\begingroup$ I'm confused as to why $a_{k+1}$ appears twice in the second term. Also, I'm not sure if I'm not understanding or just nitpicking but is there an extra parenthesis in the last equation? $\endgroup$ – leibnewtz Jul 6 '14 at 0:51

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