I have a couple of questions that I a unsure on. The first part have done but hopefully you can confirm I have done this right the rest I am not sure on. Can you point me in the right direction.
Suppose G is a group with elements a and b or orders p and q resectively and p and q are distinct primes
a) Show that (ab)^k=a(ab)^(k-1) b
This is what I have. The only way I could figure to do this was to start from the RHS and prove that it equals the LHS.
a(ab)^(k-1) b=a*a^(k-1) *b(k-1) *b
=a*a^k *a^-1 *b^k*b^-1 *b
=a*a^k *a^-1 *b^k e (as b^-1 b =e)
=a*a^k *a^-1 *b^k
Now since multiplication of integers is associative, aa^ka^-1=aa^-1a^k. Thus
=e*a^k *b^k (as aa^-1=e)
=(ab)^k as required
Have I done this correctly?
b)Show that (ab)^k=e if and only if a^k=b^k=e.
Now I know that I need to do mathematical induction here but I always have trouble with this. Can you give me a starting point or a similar example so I can follow what you are saying as you do it.
c)Show that the order of ab is pq.
I think I must be having a stupid day as to me if a and b have orders p and q respectively then obviously ab=pq and I am not sure how to use mathematical induction to get the answer.