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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.

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It seems like a homework. Voted to close. –  Mark Sapir Mar 1 '12 at 15:41
    
It is not homework. If it looks so easy to you, could you please give me some hint? –  Hebert Mar 1 '12 at 16:30
    
This question was posted two days ago at MathStackExchange, and it hasn't got any answer so far. That's why I have posted it here. It is a legitimate research question. –  Hebert Mar 1 '12 at 16:57
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See the paper C. E. Hempel, Metacyclic groups, Comm. Algebra 28 (2000), no. 8, 3865--3897. In particular, Lemma 2.1 gives the answer to your question. –  Primoz Mar 1 '12 at 19:34
    
I still haven't got access to the paper, but thank you very much anyway. I'm looking forward to see it. –  Hebert Mar 1 '12 at 21:23
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1 Answer 1

up vote 1 down vote accepted

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 $nm$.

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It is indeed helpful; thank you very much. –  Hebert Mar 1 '12 at 20:53
    
Actually in the definition of $k$ one can lower $m$ to $m/\gcd(m,\beta)$. –  GH from MO Mar 1 '12 at 21:05
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Well, it turns out that the exponent of the group is exactly $\mbox{lcm}(n,m)$. –  Hebert Mar 2 '12 at 21:56
    
@Hebert: Thank you. Can you provide an explanation, e.g. as a response to your own question? From Hempel's Lemma 2.1 the answer is not obvious to me, e.g. I don't know the values $k,l,m,n$ in your situation. An elementary number theoretic argument would be even better. –  GH from MO Mar 3 '12 at 10:08
    
@GH: In our case, $k=n$, $m=m$, $l=m$, and $n=r$ (the left-hand sides are Hempel's variables, and the right-hand sides are mine). So, the value of $\lcm(n,m)$ follows from Lemma 2.1 of Hempel's paper. I would still like to obtain that same value by lowering $k$ in your argument. –  Hebert Mar 4 '12 at 14:52
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