According to Burnside's classification in his book "Theory of groups of finite order", one of the types of non-abelian groups of order $pq^2$ ($p$ and $q$ are distinct primes), has the presentation $\langle a, b \vert a^p=b^{q^2}=1, aba^{-1}=b^i, {\rm Ord}_{q^2}(i)=p\rangle$.

  1. Is it true that all maximal subgroups of such group are cyclic?

  2. Is it possible to classify all finite non-abelian groups $G$ of order $pq^m$ with all cyclic Sylow subgroups, $m\geq 2$ and $p<q$ such that all maximal subgroups of $G$ are cyclic?

(We know that such groups are supersolvable, because all Sylow subgroups of them are cyclic. Thus all maximal subgroups of such groups are of prime index).

Thank you very much!

  • $\begingroup$ The answer is to 1 is no. The maximal subgroup $\langle a,b^q \rangle$ of index $q$ is nonabelian. $\endgroup$ – Derek Holt Sep 26 '14 at 14:27
  • $\begingroup$ Dear prof. Holt, Is it possible to give me more details of your comment? How could we show that $\langle a, b^q\rangle$ is not abelian? Thanks! $\endgroup$ – sebastian Sep 26 '14 at 14:42
  • $\begingroup$ The subgroup of the automorphism group of a cyclic subgroup of order $q^2$ that centralizes the subgroup of order $q$ has order $q$. So it contains no element of order $p$. Hence conjugation by $a$ cannot fix $b^q$; in other words, $a$ does not centralize $b^q$. $\endgroup$ – Derek Holt Sep 26 '14 at 15:45

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