Any answer will be greatly appreciated!

In the proof of Theorem 1 of the paper,

The number of conjugacy classes of non-normal cyclic subgroups in nilpotent groups of odd order, Journal of Group Theory. Volume 1, Issue 2, Pages 165–171,

pages 167-169 $G$ is a finite nilpotent group and $\nu^*(G)$ and $\Phi(G)$ denote the number of conjugacy classes of non-normal cyclic subgroups of $G$ and the Frattini subgroup of $G$, respectively. I have the following questions in step (d) in this proof:

  1. Why do we have $U=\langle u,w\rangle$?

2.Why is $\langle w,\Phi(U)\rangle$ a maximal subgroup of $G$?

3.Why $x\notin\langle w,\Phi(U)\rangle$?

Many thanks.


$\langle u \rangle$ is normal in $G$, so $\langle x^p,u \rangle$ has index $p$ in $\langle x,y \rangle =U$, so $\langle x^p,u \rangle = \langle u,\Phi(U) \rangle$. So if $w \not\in \langle x^p,u \rangle$, then (since $U$ is a $2$-generator group), $\langle u,\Phi(U),w\rangle = U$ and hence $U=\langle w,u \rangle$.

It says that $\langle w,\Phi(U) \rangle$ is a maximal subgroup of $U$, not of $G$, and this follows from the fact that $U$ is a $2$-generator group.

It can't contain $x$, because it's normal in $G$, but $U$ is the normal closure of $\langle x \rangle$.

  • $\begingroup$ Dear Professor Holt, thank you very much for your nice answer! $\endgroup$ – sebastian Feb 4 '17 at 18:43

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