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


*

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

*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. 
 A: 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 $nm$.
