Q:

Finding multiplicative order

Assume that $a$ has multiplicative order $l$ $\pmod m$. Let $k$ be given, find the multiplicative order of $a^k$ $\pmod m$.
So we have $a^l\equiv 1\pmod m$ and we're looking for the smallest $h$ such that $[a^k]^h\equiv 1\pmod m$.
The given proof is as follows:
Let $gcd(k,l)=d$ such that $k=k'd$ and $l=l'd$ and $gcd(k',l')=1$
$a^{kh}\equiv 1\pmod m$ $\iff$ $kh\equiv 0\pmod l$ $\iff dk'h\equiv 0\pmod {l'd}$ $\iff k'h\equiv 0\pmod {l'}$ and since $gcd(k',l')=1$, then $k'$ is invertible $\pmod {l'}$ which implies $h\equiv 0\pmod {l'}$
How can I justify the first and third equivalences? And does this mean that $h=\frac{ql}{d}$ for $q\in \mathbb{Z}$?

A:

The first equivalence just results from the definition of the order of $l$: it's the positive generator of $\bigl\{n\in\mathbf Z\mid a^n\equiv 1\pmod m\bigr\}$.
For the 3rd equivalence, $dk'h$ is divisible by $l'd$ if and only if $k'h$ is divisible by $l'$, this should be clear.
Last question : yes, since $\dfrac ld=l'$.

