counter example $\frac{G}{G^{n}}$ isomorphic with $G_{n}$

Question

Let's say that $G,.$ is a commutative group. We define: $G_{n}$={$g \in G| g^{n} =e$} and $G^{n}$={$g^{n}|g \in G$}.

First I prove that $\frac{G}{G_{n}}$ is isomorphic with $G^{n}$. Let's define a map $\phi: G\mapsto G^{n}$:$g \mapsto g^{n}$. I prove that this is a morphism. Take $x,y\in G$ random then $\phi(x.y)$=$(x.y)^{n}$=$x^{n}.y^{n}$=$\phi(x).\phi(y)$ so $\phi$ is a morphism. Now I know that $ker(\phi)$={$f\in G|\phi(f)=e$}={$f\in G| f^{n}=e$}=$G_{n}$

So now it follows that $\phi'$:$\frac{G}{ker(\phi)} \mapsto G^{n}$:$gG_{n}\mapsto \phi(g)$ is an injective group morphism. So I proved it.

But now I'm stuck because I'm looking for an counter example that $\frac{G}{G^{n}}$ isn't isomorphic to $G_{n}$. Can someone help me with this.

Answer 1

You can take $G=\mathbb Z, +$ and $n=3$. Then $G_3=\{0\}$ and $G^3= 3\mathbb Z$, but $G/G^3=\mathbb Z_3$.