I am just curious. We have a group $G$ and its subgroup $H$ with a generator element $h \in H$. How difficult is it to check for $x \in G$ that $x \in \langle h \rangle$? Is there a better way than testing $x \equiv h^1,\ x \equiv h^2,\ x \equiv h^3,\ ...$?
If the group $G$ is finite and cyclic (and hence Abelian), it has exactly one subgroup of order $s$, for each $s$ which divides the order of $G$.
Since any subgroup of a cyclic group is also cyclic and hence Abelian, you might easily implement the operation in $H$ that is the equivalent of modular exponentiation in $Z_n^*$.
If you also already know the order $s$ of $h$, you know that, because of associativity and commutativity, that $(h^s)^t = (h^t)^s = 1$ for any $t$. Hence, checking that $x \in H$ is just a question of checking that $x^s = 1$.