I study the following paper about finite $p$-groups,
Finite groups in which the non-normal subgroups have nontrivial intersection, N. Blackburn, Journal of algebra, 3, 30-37 (1966).
In this paper the intersection of all non-normal subgroups of $G$ is denoted by $R(G)$. We say that a group $G$ is an $R$-group if $G$ is not a Dedekind group and $R(G)\neq 1$.
The main result of this paper is as follows,
Let $G$ be a finite $p$-group. If $G$ is an $R$-group, then $p=2$ and one of the following holds.\ (i) $G\cong Q_8\times C_4\times E$, where $E$ is elementary abelian.\ (ii) $G\cong Q_8\times Q_8\times E$, where $E$ is elementary abelian.\ (iii) $G$ is a $Q$-group, i.e. $G=\langle x,A\rangle$, where $A$ is an abelian subgroup which is not elementary abelian, $x^2\in A$ has order $2$, and $a^x=a^{-1}$ for all $a\in A$. In this case $R(G)=\langle x^2\rangle$. Also, if $A$ is cyclic then $G$ is a generalized quaternion group. In particular, if the centre of $G$ is cyclic, then $G$ is a generalized quaternion group.
I have a question about the third type ($Q$-groups) of this classification.
Let $c(G)$ denotes the nilpotency class of the group $G$ and $\displaystyle\bar{G}=\frac{G}{R(G)}$. Is it true that $c(G):=c(\bar{G})$?
Any commnt or answer will be appreciated!
The answer is no. If $G$ is a generalized quaternion group then $c(\overline{G}) = c(G)-1$.