Subgroup of finite field is invariant under Frobenius map

Question

Let $q$ be a prime power and $a\in\Bbb{F}_{q^2}$. Let $m$ be a positive integer dividing $q+1 $ and $H \subset\Bbb{F}_{q^2}^{\times}$ a subgroup of order $m(q-1)$.

If $a \in H \cup \{0\}$, why is $a^m$ invariant under the Frobenius map?

Answer 1

The group of units of a finite field of order $q^2$ is cyclic of order $q^2-1=(q-1)(q+1)$. In particular it has a unique subgroup $H$ of order $m(q-1)$. Then the $m$-th powers in $H$ form a subgroup of index $m$ in $H$, so this is the unique subgroup of order $q-1$ in the full unit group, which is of course $\Bbb{F}_q^{\times}$.