# Questions about a finite $p$-group in a paper

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 steps (e) and (f) in this proof:

1. In part (e), the author says that "$\frac{G}{\langle u^p\rangle}$ satisfies the assumption of the theorem." Why do we have $\nu^*(\frac{G}{\langle u^p\rangle})=1$? (According to Lemma 1 of this paper, I know that $\nu^*(\frac{G}{\langle u^p\rangle})\leq \nu^*(G)=1$. Now why do we have $\nu^*(\frac{G}{\langle u^p\rangle})\neq 0$?)

1. By induction hypothesis we have $\frac{G}{\langle u^p\rangle}\cong M(p^n)$. Why do we have $x^p\in \langle u^p\rangle$?

2. In part (f), the author has concluded that $\nu^*(G)=\nu^*(H)=1$. How could we conclude it from this fact that $U$ is an elementary abelian $p$-group of order $p^2$?

Thank you so much for your thought on my questions! Here is the proof of the Theorem

• For you first question: each group has a neutral element. This element forms its own conjugacy class, so a group always has at least 1 conjugacy class: $\{e}\$, where $e$ denotes the neutral element. – Student Feb 11 '17 at 7:41
• The complete reference is Li, Shirong. The number of conjugacy classes of non-normal cyclic subgroups in nilpotent groups of odd order. J. Group Theory 1 (1998), no. 2, 165-171. – Andreas Caranti Feb 11 '17 at 10:57