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 $n$

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.

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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 $k$.)

If $gH$ has order $n$ we get $g^n\in H$. Let $t$ be the order of $g^n$. Then the order of $g$ is $nt$ (why?). In the cyclic subgroup generated by $g$ there exists an element of order $n$ (why?).

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 $n$.