Question

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.

Answer 1

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 $nm$.