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A group $G$ is metacyclic if it contains a cyclic normal subgroup $N$, such that $G/N$ is also cyclic. $N$ is called the kernel, and the order of $G/N$ is the index of the metacyclic group. There is an alternative definition (see Hempel, Comm. Algebra 28(8), 2000):

A group $G$ is metacyclic with a kernel of order $m$ and index $n$ if, and only if, it has a presentation of the form $$\langle a, b: a^n = b^l, b^m = 1, a^{-1}ba = b^r \rangle$$ where $n, l, m, r$ are positive integers, such that $r^n \equiv 1 (m)$ and $m|l(r-1)$. Now, replacing $b$ by a suitable power of itself, we may assume that $l|m$. In particular, I have been working with $l=m$. My question is:

What implies this restriction $l=m$? I.e. what subclass of metacyclic groups would that be?

Thank you very much in advance.

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    Your group has order $m\cdot n$, as $\langle b\rangle$ has order $m$ and index $n$. You also have that $\langle a\rangle\cap\langle b\rangle=1$, so your group is the semidirect product $\mathbb{Z}_n\rtimes\mathbb{Z}_m$, where the action is dependent on the $r$ and the $n$ (the other two bits of the semidirect product criteria are immediate).2012-03-09
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    These are called "split metacyclic" if I recall correctly. The term is used in many papers.2012-03-09
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    Thanks to both. Do you have an example of a metacyclic group that is not a semidirect product? An example of a non-split metacyclic group?2012-03-09
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    The quaternion group of order 8 is a non-split metacyclic group with kernel 4 and index 2.2012-03-10
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    @Derek Holt. Thanks to you too.2012-03-16
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    Do someone know which conditions must be imposed on $n$ and $m$ in order to have a non abelian split metacyclic group (i.e., $r>1$)?2015-04-21

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