Properties of Abelian Groups I'm trying to prove that if $G$ is an Abelian group under $\cdot$, $\forall a,b \in G. \forall z \in \mathbb{Z}. (a \cdot b)^n = a^n \cdot b^n.$ I was originally considering doing this problem using an AFSOC, but I realized that originally assuming that $(a \cdot b)^n \neq a^n \cdot b^n$ would be rather difficult to contradict with the given properties of an Abelian group. Thus, I considered inducting in two ways on $n$, first going through all the positives and then going through all the negatives in the integers. However, I'm worried about what I do with the case of $n = 0.$ I understand that in algebraic terms $a^0$ is an abbrevation for ``$a \cdot a \cdot ... \cdot a$ with 0 many $a$'s'', but I am confused as to what this represents in the Abelian group $G.$ Perhaps I need to prove some properties of $a^0$ for all $a \in G$?
 A: $G$ is an abelian group, so let $a,b\in G$ be given, and fix $n$. Then
$$ (ab)^n=(ab)(ab)(ab)\cdots(ab).$$
Because $G$ is abelian
$$ (ab)(ab)(ab)\cdots(ab)=(aa\cdots a)(bb\cdots b)=a^nb^n.$$
This follows naturally from the definition of commutativity.
A: $a^n$ is a notation device.  As the binary operation is associative $a^na^m = a^{n+m}$ and $(a^m)^n = a^{mn}$ follows implicitly and doesn't need proving.  Declaring $a^0 = e$ as a definition is acceptable.
But none of this is needed.  $(ab)^n = ababab.....ababab = aaa....aabbbb....b = a^nb^n$ and that is all there is to it.
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Oh... for all integers.  Okay...
Before we can show it for 0 and negative integers we have to define what $a^0$ and $a^{-m}; m > 0$ mean.
$a^0:= e$ is simply notation.  $(ab)^0 = e = e*e = a^0b^0$ is obvious.
For $m > 0$, $a^m(a^{-1})^m = (a(a(a.....(aa^{-1})...a^{-1})a^{-1})a^{-1}) = e$.  So $(a^m)^{-1} = (a^{-1})^m$.  So $a^{-m} := (a^{-1})^m = (a^m)^{-1}$ is well defined.
Note: $(ab)(b^{-1}a^{-b}) = e$  So $(ab)^{-1}=b^{-1}a^{-1}$.
So $(ab)^{-m} = ((ab)^{-1})^m = (b^{-1}a^{-1})^m = (a^{-1}b^{-1})^m = (a^{-1})^m(b^{-1})^m = a^{-m}b^{-m}$.
That's it.
