Simpler proof of $g\,h\,g^{-1} = h^a \Rightarrow g^n\,h\,g^{-n} = h^{a^n}$

Question

In a rather easy online lecture on group theory (which included many obvious statements such as "the only divisors of a prime number $p$ are $1$ and $p$"), the professor began a proof by assuming that $g\,h\,g^{-1} = h^a$, where $g$ and $h$ are elements of some group $G$ having prime orders $p$ and $q$, respectively, and from this he immediately asserted, without proof (i.e. as if it were obvious):

$$a^p \equiv 1 \mod{q}.$$

This blew me away. The assertion does not look obvious to me at all. Is it?

It's easier to see that the assertion above must be true if one has already established the more general implication (valid for any two elements $g$ and $h$ in any multiplicative group)

$$g\,h\,g^{-1} = h^a \;\; \Rightarrow \;\; g^n\,h\,g^{-n} = h^{a^n}, \forall \, n \in \mathbb{N},$$

...but he had not established this, and it is not something that jumps out at me as obvious either (even though I can prove it¹).

Am I somehow missing some immedate proof of either of the two assertions?

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<sub>¹ My proof is by induction. Assuming that $g^k\,h\,g^{-k} = h^{a^k}$ holds for all $k\in\{1\dots n-1\}$, then $$g^n\,h\,g^{-n} = g (g^{n-1}\,h\,g^{-(n-1)})g^{-1} = g\,h^{a^{n-1}}\,g^{-1} = (g\,h\,g^{-1})^{a^{n-1}} = (h^a)^{a^{n-1}} = h^{a^n},$$ ...where the second and fourth equalities follow from the induction hypothesis for $k = n - 1$ and $k = 1$, respectively.

EDIT: I thought of a slight improvement on my original proof: $g\,h\,g^{-1} = h^a$ implies that, for any positive integer $k$, $g\,h^k\,g^{-1} = (g\,h\,g^{-1})^k = (h^a)^k = h^{ka}$. It also holds for any negative integer $k$, because, by taking inverses of both sides of the preceding equality, $(g\,h^k\,g^{-1})^{-1} = g\,h^{-k}\,g^{-1} = h^{-ka}.$ Finally, the equality also holds (trivially) if $k = 0$. The assertion $g\,h^k\,g^{-1} = (g\,h\,g^{-1})^k = h^{ka} = (h^k)^a, \forall\,k \in \mathbb{Z}$ says that conjugation by $g$ of any element $h^k$ of the cyclic group $\langle h \rangle$ is the same as raising the element to the $a$-th power. Hence, $n$ successive conjugations by $g$ (or equivalently, conjugation by $g^n$) is the same as raising the conjugated element to the $a^n$-th power.

Granted, this revised proof is significantly longer than my original one, but I like it better because it clarifies for me the scope of validity (namely, all of $\langle h \rangle$) of identifying the operation $h\mapsto g\,h\,g^{-1}$ with the operation $h\mapsto h^a$. The assertion I proved originally then follows trivially: each side of that equality represents $n$ successive applications one of these two equivalent operations.</sub>

Answer 1

Identify $\langle h\rangle$ by $\mathbb{Z}/q$. If $ghg^{-1} = h^a$, then action of $\langle g \rangle$ on $\langle h\rangle$ by conjugation is $g \mapsto$ multiplication by $a$.

Answer 2

$gh g^{-1} = h^a$ says that $h$ is raised to its $a$th power when conjugated with $g$. Thus, when we conjugate it $n$ times, we have to raise it to the $a^n$th power. This is a complete proof of $g^n h g^{-n} = h^{a^n}$. If it doesn't convince you, here is a slightly more formal formulation of this argument.

Recall that $c_g : h \mapsto g h g^{-1}$ is a group automorphism. It satisfies $c_g(h)=h^a$. Induction shows $(c_g)^n(h)=h^{a^n}$ (for example $c_g(c_g(h))=c_g(h^a)=c_g(h)^a = (h^a)^a = h^{a^2}$, etc.). Then $c_{g^n}(h)=(c_g)^n(h)=h^{a^n}$.