If $G$ is a finite group, $H$ is a subgroup of $G$, and there is an element of $G/H$ of order $n$, then there is an element of $G$ of order $n$ 
$G$ is a finite group, $H$ is a normal subgroup. There is an element in $G/H$ of order $n$. Prove that there is an element in $G$ of order $n$.

Note: I have proved that there exists an element in $G$, say $b$, such that $b^n=e$. I'm stuck in proving that $n$ is the smallest element, i.e. $n$ is the order of $b$.
Please provide a new proof if possible!

Let $a\in G/H, |a|=n$ and $c\in G$ is a preimage of $a$. Then $[\langle c\rangle H:H]=n$. Since $\langle c\rangle H/H \cong \langle c\rangle/\langle c\rangle\cap H$ we have $|\langle c\rangle| =n|\langle c\rangle\cap H|$. Therefore the cyclic group $\langle c\rangle$ contains an element of order $n$.
