Indicator is Well-Defined
Let $G$ be a finite group, and $a \in G$.
Let $H$ be a subgroup of $G$.
Then the indicator of $a$ in $H$ is well defined.
If $a \in H$, then for $n = 1$, $a^n \in H$.
Moreover, by Element of Finite Group is of Finite Order, there is $n \in \N$ such that $a^n = e \in H$.
Therefore, for any $g \in G$, there is some $n \in \N$ such that $a^n \in H$.