Should the order of $a^k$ be $h/k$ as opposed to $h/(h,k)$?

Question

Previously shown:

Let $m\in\mathbb{N}$ and $a\in\mathbb{Z}$ s.t. $(a,m)=1$. Then the order $d$ of $a$ modulo $m$ exists and $d\mid\phi(m)$.

Moreover, whenever $a^k\equiv 1\pmod{m}$, one has $d\mid k$

Trying to show: Suppose that $a$ has order $h$ modulo $m$. Then $a^k$ has order $h/(h,k)$ modulo $m$

Proof: From above, one has $(a^k)^j\equiv 1\pmod{m}$ iff $h\mid kj$.

But, $h\mid kj\Leftrightarrow h/(h,k)\mid (k/(h,k))j\Leftrightarrow h/(h,k)\mid j$

Thus the least positive inter $j$ s.t. $(a^k)^j\equiv 1\pmod{m}$ is $j=h/(h,k)$

Point of contention: Surely here we have $(h,k)=k$, so we have shown that the order of $a^k$ has order $h/k$ as opposed to $h/(h,k)$

Answer 1

As requested by OP, in the comments:

$h/(h,k)$ and $k/(h,k)$ are relatively prime. If $r$ and $s$ are relatively prime, and $r$ divides $st$, then $r$ divides $t$. Therefore, if $h/(h,k)$ divides $(k/(h,k))j$, then it divides $j$.

Answer 2

One can also prove it without Euclid's Lemma, instead using the gcd Distributive Law

$$ h\mid kj\iff h\mid kj,hj\iff h\mid \underbrace{(kj,hj)}_{\Large (k,h)\,j}\iff h/(k,h)\mid j\qquad $$