$\newcommand{\ord}{\operatorname{ord}}$
If $\alpha,\beta \in \mathbb{GF}(q)$ and $\ord(\alpha)=n_1$ and $\ord(\beta)=n_2$, then what is $\ord(\alpha\beta)$ ?
Edit:
if $k=\ord(\alpha\beta)$ then
$$(\alpha\beta)^k=1 \Longrightarrow \alpha^k=\beta^{(-k)} \Longrightarrow$$ $$\ord(\alpha^k)= \ord(\beta^{(-k)})= \ord(\beta^k) \Longrightarrow$$ $$\frac{n_1}{\gcd(n_1,k)}=\frac{n_2}{\gcd(n_2,k)}$$
Also it is clear that $k \mid \rm lcm (n_1,n_2)$.
There is insufficient information to find the order of $\alpha\beta$ exactly. Consider the following case. Assume that $n_1=n_2=p>2$ is a prime. Then it is possible that $\alpha=\beta$, in which case $\alpha\beta=\alpha^2$ is of order $p$. But it is also possible that $\alpha=\beta^{-1}$, in which case $\alpha\beta=1$ has order $1$.
Let $\nu_p(m)$ be the multiplicity of prime $p$ as a factor of $m$. The above example generalizes as follows. If $\nu_p(n_1)=\nu_p(n_2)=\ell$, then we have no way of figuring out $\nu_p(k)$. All we can say in this case is that $0\le \nu_p(k)\le\ell$. Think about this in the additive group $\Bbb{Z}_{p^\ell}$ to see that all possibilities occur.
On the other hand, if $\nu_p(n_1)\neq\nu_p(n_2)$, then we know that $$ \nu_p(k)=\max\{\nu_p(n_1),\nu_p(n_2)\}=:\ell. $$ Again, this is easy to see in the additive group $\Bbb{Z}_{p^\ell}$, and the structure theorem of finitely generated abelian groups (here $GF(q)^*$) reduces the general case to this.