Ex. 8 of the first chapter. The statement is as follows.
This exercise is higly recommended if you want or need to practice induction proofs. We define $a^2$ as $a*a$, $a^3$ as $a*a*a$, and for $n $ a positive integer $a^n=a*a*a*...*a$ ($n$ copies of a).
a) If $(G,*)$ is an abelian group, then show that $(a*b)^n=a^n*b^n$ for all positive integers $n$.
[This I proved it easily with mathematical induction using the commutativity and associativity properties of an abelian group.]
b) If n is positive integer show $(a^n)^{-1}=(a^{-1})^n $.
Here I am puzzled how to proceed. I did not have any success with induction.
Thanks for your help.
$(a^{n})^{-1} *a^{n}=e(identity)$
$(a^{-1})^{n}*a^{n}=(a^{-1}*a^{-1}*...*a^{-1})*(a*a*...*a)=e(identity) $ using associativity.
So
$(a^{n})^{-1}=(a^{-1})^{n}$ by uniqueness of inverse.