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!
One can prove the following proposition (see below):


  If $G$ is a finite group and $H$ is a normal subgroup then $\forall g\in G$, $o(gH)\mid o(g)$, where $o(g)$ denotes the order of $g$.


Suppose $o(g)=l$, $o(gH)=n.$ By using above proposition 

$$n\mid l\Rightarrow \exists r\ , \space l=nr.$$

Now let $b:=g^r$ and suppose $o(b)=k$. We have

$$b=g^r\Rightarrow b^n=g^{nr}=g^l=e$$

$$\Rightarrow k\mid n \qquad(*)$$

On the other hand 

$$b^k=e\Rightarrow g^{kr}=e\Rightarrow l\mid rk\Rightarrow rn\mid rk$$

$$\Rightarrow n\mid k \qquad(**)$$

By considering $(*)$ and $(**)$, $k=n$ is concluded.



Proof of the proposition:

Here's a slightly more general statement that might actually isolate the fact better.


  Let $f:G\rightarrow G^{\prime}$ be a homomorphism. Then the order of $f(g)$ divides the order of $g$.


In your case, you are just talking about the projection $f:G\rightarrow G/H$.

Suppose $o(g)=n$ and $o(f(g))=m$, so $g^n=e$ by cosidering that $f$ is homomorphism one can conclude:

$$f(g^n)=f(e)=e\Rightarrow (f(g))^n=e \Rightarrow m\mid n.$$

(Reminder: $g^k=e$ implies that the order of $g$ divides $k$.)
o(g)/o(gH).Hence o(g)/n. i.e. o(g)=nt, for some positive integer t.
Now, o(g^t)=o(g)/GCD[t,o(g)]=nt/GCD[t,nt]=nt/t=n.
Hence our required element in G is g^r.