1
$\begingroup$

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 $n+1$?

$\endgroup$
  • 7
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.