As part of a larger proof, I want to make the following statement:
Let G be finite abelian group.
Let $g\in G$ such that the order of $g^{q}$ is $q^{e}$ where $e\geq1$ and q is prime.
Let $h\in G$ such that $h$ is a power of $g$.
Moreover, let $x_{0}$ be a solution to the Discrete Logarithm Problem applied to $g^{q^{e}}$ and $h^{q^{e}}$.
That is, $(g^{q^{e}})^{x_{0}}=h^{q^{e}}$.
Therefore, $(g^{q^{e}})^{x_{0}}=h^{q^{e}} \iff h^{q^{e}}.(g^{q^{e}})^{-x_{0}}=1\iff(h.g^{-x_{0}})^{q^{e}}=1$.
Now, can I say that $(h.g^{-x_{0}})\in<g^{q}>$?
For a cyclic group $H$ of order $n$, it is a theorem that there exists a unique subgroup $K\leq H$ of order $d$ for every divisor $d$ of $n$. Notice that the subgroup generated by $g$ is cyclic of order $q^{e+1}$. Now $h\cdot g^{-x_0}$ has order at most $q^e$. Say $h\cdot g^{-x_0}$ has order $q^d$. Then this element lies in the unique subgroup of order $q^d$ which is necessarily generated by $g^{e+1-d}$. Hence $h\cdot g^{-x_0}\in \langle g^{e+1-d}\rangle\subseteq\langle g^q\rangle$.