A homomorphism $\Phi: G \to H$ where the image of $\Phi$ has an element of order n. Assume $\Phi: G \to H$ is a homomorphism and G is a finite group.  If the image of $\Phi$ has an element of order n, then G has an element of order n.
Proof: Assume $\Phi: G \to H$ is a homomorphism, and assume $\exists \Phi(g)\in Im \Phi$ such that $|\Phi(g)|=n$
$\implies \Phi(g)^n=e_{h}$ and by the homomorphism property we have $\Phi(g)^n=\Phi(g^n)=e_{h}$
Since $|G|<\infty$, it follows that $|g^n|\leq G$.
I am having trouble finishing this proof.
I had initially ended it by stating: 
We know $\Phi(e_g)=e_h \therefore e_g=(g^n)$
But I realized that this conclusion is incorrect. Any help on how to proceed is appreciated.

Let $h$ be an element of $\Phi(G)$ having order $n$, and choose $g\in G$ such that $\Phi(g)=h$. If $g$ has order $m$, then
$$ h^m=\Phi(g)^m=\Phi(g^m)=\Phi(e)=e $$
hence $n$ divides $m$, and $g^{\frac{m}{n}}$ has order $n$.
