If $(G,*)$ is an abelian group, then show that $(a * b)^n = a^n * b^n$.

Question

Define $a^2$ as $a * a, a^3 = a * a * a$, and for a positive integer $n, a^n= a * a * \cdots * a$ ($n$ copies of $a$).

(a) If $(G,*)$ is an abelian group, then show that $(a * b)^n = a^n * b^n$ for all positive integers $n$.
(b) If $n$ is a positive integer, then show that $(a^n)^{-1} = (a^{-1})^n$.

By being an abelian group, commutative property w.r.t. $*$ is there.
(a) Need show that $(a * b)^n = a^n * b^n,\ \forall n\in\mathbb{N}$.

Stuck ...

(b) If $n$ is a positive integer, then show that $(a^n)^{-1} = (a^{-1})^n$.
Unable to think of any approach.

Answer 1

(a) Your answer makes no sense, since there is no operation $+$ here. You can do it by induction:

• $(a*b)^1=a*b=a^1*b^1$;
• if $(a*b)^n=a^n*b^n$, then\begin{align}(a*b)^{n+1}&=(a*b)^n*a*b\\&=a^n*b^n*a*b\\&=a^n*a*b^n*b\\&=a^{n+1}*b^{n+1}.\end{align}

(b) $a^n*(a^{-1})^n=(a*a^{-1})^n=e^n=e$

Answer 2

Use induction: For $n=1$ we must show that $ab=ab$ which holds.

Suppose $(ab)^n = a^n b^n$ then $(ab)^{n+1} = (ab)^n (ab) = a^n b^n ab = a^n a b^n b$ by Abelianness and so this equals $a^{n+1}b^{n+1}$, as required.

(b) follows from $1= aa^{-1} = (aa^{-1})^n$ which equals by (a) $a^n (a^{-1})^n$ So by the uniqueness of the inverse $(a^n)^{-1}= (a^{1-})^n$