Take ${G^n}$ to be {${g^n:} $ g in G}, ${G_n}$ to be {g in G: ${g^n = 1}$}. Show that ${G^n}$ is isomorphic to ${G/}$${G_n}$.
I'm not sure how to go about proving this. My initial thought was to use one of the isomorphism theorems, particularly the first one. If I can find a homomorphism from G to H s.t. the image of the homomorphism is ${G^n}$ and the kernel is ${G_n}$ I could use the first isomorphism theorem, but I'm not sure if this is a good idea.
Intuitively, ${G_n}$ seems to be all elements of order n in G. Why do the cosets of ${G}$ isomorphic to ${G^n}$?
For this to work, $G$ has to be an abelian group (else $G^n$ might not be a subgroup!)
The homomorphism idea discussed in the comments is good. You are right, the obvious homomorphism is $\phi: G \to G^n$ with $\phi(g) = g^n$. (You should check this is a homomorphism and its image is $G$; both of these are easy.)
What is the kernel of $\phi$? Well it is exactly those elements $g$ such that $g^n = 1$...