Show that $G^n$ is a subgroup of $G$ if $G$ is abelian Let $G$ be a group. For $n\in\Bbb N$, let $G^n=\{x^n : x\in G\}$
Show that $G^n$ is a subgroup of $G$ if $G$ is abelian
I know that $x\in G^n$ therefore $G^n$ is a non empty subset of G. If we let $x^n, y^n\in G^n$, then as G is abelian, we know $x^n.y^n$ = $(xy)^n$ by commuting the $x$'s and $y$'s; therefore, $G^n$ is closed under multiplication
I'm trying to show that if $x^n\in G^n$ then $(x^n)^{-1}\in G^n$ but $(x^n)^{-1}=x^{-n}$ and the original condition was that $n\in\Bbb N$ so $-n$ would not be a natural number and therefore the inverse is not an element of $G^n$, which is obviously contradicting what I'm trying to show.
Where have I gone wrong here?
 A: If $G$ is abelian, then $xy = yx$ for all $x,y\in G$. Then, note that, if $$x,y \in G \Rightarrow x,y,y^{-1}\in G \Rightarrow x^n, y^n, y^{-n} \in G^n \Rightarrow x^ny^{-n} = (xy^{-1})^n \in G^n$$
The last step can be expanded as follows:
$$
 (xy^{-1})^n = (xy^{-1})(xy^{-1})\dots(xy^{-1}) = (xy^{-1}y^{-1}x)\dots(xy^{-1}) = (x^2y^{-2})\dots(xy^{-1}) = \cdots
$$
Then you are done because the condition you need to check for a subgroup is $s, t \in S \Rightarrow st^{-1} \in S$, and this has been verified above for $s = x^n, t = y^n$.
A: First of all, there is a mistake in the beginning of your proof. The statement that $x\in G^n$ does not make sense since $x$ is not defined. Every element in $G^n$ must be able to be written in the form of $x^n$ for some $x\in G$. In fact, you can prove that $G^n$ is nonempty by saying that $e=e^n\in G^n$ where $e$ is the identity element of $G$.  
To prove that $G^n$ is a subgroup of $G$, let $a,b\in G^n$, you need to show that $ab,a^{-1}\in G^n$. Since $a,b\in G^n$, by definition, $a=x^n,b=y^n$ where $x,y\in G$. You did the part that $ab\in G^n$ correctly. So now we want to show that $a^{-1}\in G^n$. Notice $a^{-1}=(x^n)^{-1}=(x^{-1})^n=z^n$ where $z=x^{-1}\in G$. This means that $a^{-1}\in G^n$ and you can conclude that $G^n$ is a subgroup of $G$.  
The property that you missed is that for any group $G$(not necessarily abelian) and $x\in G$, $n\in \Bbb{Z}$, $$(x^{n})^{-1}=(x^{-1})^n$$
It is a good exercise for you to verify this by induction.
