I am interested in the following related questions in metacyclic groups of the form $\mathbb{Z}\_n \ltimes\_r \mathbb{Z}\_m$, where $r^n \equiv 1 \pmod{m}$: 


1. The order of an arbitrary element $g = (\alpha, 0)\*(0, \beta)$ - or some upper bound on the order - where \* is the group operation.
2. The exponent of the group


I know that the first question reduces to finding the smallest integer $k$ such that: 


$k \alpha \equiv 0\pmod{n}$, and 


$\beta \frac{r^{k \alpha} - 1}{r^\alpha - 1} \equiv 0 \pmod{m}$, 


but that's about it. Thank you very much in advance. 



Exponent of metacyclic groups
Here is an answer which is probably far from optimal (I am no expert). Let
$$ t:=\mathrm{ord}\_m r, \qquad k:=\mathrm{lcm}\left(\frac{n}{\gcd(n,\alpha)},\frac{mt}{\gcd(t,\alpha)}\right), $$
then clearly $k\alpha\equiv 0\pmod{n}$, and I claim that $\frac{r^{k\alpha}-1}{r^\alpha-1}\equiv 0\pmod{m}$. For the latter observe that
$$ \alpha\ \Big|\frac{t\alpha}{\gcd(t,\alpha)}\quad\text{and}\quad 
\frac{mt\alpha}{\gcd(t,\alpha)}\ \Big|\ k\alpha, $$
so that
$$ \frac{r^\frac{mt\alpha}{\gcd(t,\alpha)}-1}{r^\frac{t\alpha}{\gcd(t,\alpha)}-1}\ \Big|\ \frac{r^{k\alpha}-1}{r^\alpha-1}. $$


So it suffices to show that the left hand side is divisible by $m$. The fraction equals
$$ \sum\_{j=0}^{m-1} r^\frac{jt\alpha}{\gcd(t,\alpha)}. $$
Here each exponent is divisible by $t$, hence each term in the sum is $\equiv 1\pmod{m}$. There are $m$ terms, hence the sum is divisible by $m$ as claimed.


It also follows that the exponent of the group divides $\mathrm{lcm}(n,mt)$. Note that the last quantity is in between $\mathrm{lcm}(n,m)$ and $nm$.


