Given an element in a quotient group $G/H$, find an element with the same order in $G$ Problem

A finite abelian group $G$ has an element $g$ of order $n$, which generates the subgroup $H = \langle g \rangle$.  Let $\gamma$ be an element of order $m$ in $G/H$.  Show that there exists an element $x \in G$ such that $\lvert x \rvert = \lvert \gamma \rvert = m$.

I copied my instructor's solution, but I missed the last part, and I can't proceed from what's on my notebook.

Suppose that $x_0 \in G$ such that $\overline{x_0} = \gamma$.  Since $\lvert \gamma \rvert = m$, $x_0^m \in H = \langle g \rangle$, and we denote $x_0^m = g^k$.  Divide $k$ by $m$ to get $k = mq + r$ with $r \in \{0,\dots,m-1\}$.
  \begin{align}
x_0^m &= g^k = g^{mq + r} \\
g^r &= x_0^m g^{-mq} = \left(x_0 g^{-q} \right)^m
\end{align}
  We want to show $r=0$.
  Let $x' = x_0 g^{-q}$.  Observe that $\overline{\mathstrut x'}=\overline{\mathstrut x_0}=\gamma$ and $m \mid \lvert x' \rvert$ because $$e = x'^{\mathstrut \lvert x' \rvert} = x_0^{\mathstrut \lvert x' \rvert} \underbrace{g^{\mathstrut -q \lvert x' \rvert}}_{\in H}.$$
  $$\therefore \lvert (x')^m \rvert = \frac{\lvert x' \rvert}{\gcd(m,\lvert x' \rvert)} = \frac{\lvert x' \rvert}{m}$$

I don't know how to continue with this.
 A: This argument seems overly complicated (and the conditions overly strict).
Claim. Let $G$ be a finite group and $H\lhd G$ a normal subgroup. Let $\gamma\in H$ be an element of order $m$. Then $G$ has an element $x$ of order $m$.
Proof. Let $\pi\colon G\to G/H$ denote the canonical projection. Pick $x_0\in G$ with $\pi({x_0})=\gamma$. Then $\langle x_0\rangle$ is a cyclic group with $\pi(\langle x_0\rangle)=\langle \gamma\rangle$, hence is a cyclic group of order divisible by $m$. We know that such a cyclic group contains an element of order $m$. $\square$
A: Suppose $[x_0]=\gamma$.
It means that $$x_0^m=g^k$$
for some $k$ becuse $x_0^m$ is the unit in the quotient. Now just elevate to the smallest number $t$ making $n|kt$. We get $$(x_0^t)^m = 1$$ thus $ord (x_0^t) \leq m$. Now it can't be strictly less than $m$ because you would find that $g$ has order less than $n$. We were looking for $x:=x_0^t$.
A: Is this true? Consider $\mathbb Z_4$ and the subgroup $\{0,2\}$. Notice that the odd coset has order $2$ but no element in it has order $2$, they all have order $4$. It is true when $(n,m)=1$ however.
