# Element of a group, its order and inferring that it belongs to a subgroup generated by another element

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 $(h.g^{-x_{0}})\in<g^{q}>$?

• No${}{}{}{}{}{}$. – Angina Seng Nov 19 '17 at 22:36
• I am pretty sure I can, since this is part of the Pohlig-Hellman algorithm proof. I just don't know how. – popololvic Nov 19 '17 at 22:38
• Then you should say that $G$ is a finite abelian group since that information changes everything. – Levent Nov 19 '17 at 22:39
• @jkbestami You need a lot more then $G$ just being a group to use the P-H algorithm. – Angina Seng Nov 19 '17 at 22:40
• You are totally right Levent, and I made the edit. I know that you need more for the P-H algorithm but I think I have everything else except for the justification of this claim. – popololvic Nov 19 '17 at 22:41

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 $h\cdot g^{-x_0}\in \langle g^{e+1-d}\rangle\subseteq\langle g^q\rangle$.