Existence of a cyclic non-normal subgroup in a $p$-group

Question

Let $G$ be a finite non-abelian $p$-group, where $p$ is an odd prime, $N$ be a normal subgroup of $G$ of order $p$, where $\frac{G}{N}$ is non-abelian. Does there exist an element $g\in G$ such that $\langle g\rangle$ is NOT normal in $G$ and $N\langle g\rangle$ is normal in $G$?(Note that $G$ and $\frac{G}{N}$ are non-Dedekind groups and contain non-normal cyclic subgroups. By a Dedekind group I mean a group all of whose subgroups are normal, which are abelian groups or direct product of a quaternion group, an elemntary abelian group and an abelian group with all elements of odd order (Hamiltonian group)).

Thank you very much!

Added later(According to professor Holt's answer): Let $G$ be a finite non-abelian $p$-group and $N$ be a normal subgroup of $G$ of order $p$, where $\frac{G}{N}$ be also non-abelian ($p$ is an odd prime). Also let for each $g\in G$ $$\langle g\rangle\lhd G\Leftrightarrow N\langle g\rangle\lhd G.$$ Is it possible to classify such $G$?

Answer 1

Let $G = \langle x,y,z \rangle$ be a $3$-generated group of order $p^6$ of exponent $p$ and class $2$. So $Z(P) = [P,P]=\Phi(P)$ is elementary abelian of order $p^3$, and so is $G/Z(P)$.

Let $N = \langle [x,y] \rangle$. So $N \lhd G$ with $|N|=p$. Then $Z(G/N) = Z(G)/N$, so for $g \in G$, we have $$\langle g \rangle \lhd G \Leftrightarrow g \in Z(G) \Leftrightarrow \langle g, N \rangle < Z(G/N) \Leftrightarrow \langle g, N \rangle \lhd G/N.$$

Added later: An easier type of example is $P = Q \times N$ where $Q$ is any nonabelian $p$-group and $N$ is any abelian $p$-group.

Answer 2

Here's an easy example: take $G$ to be of class $3$ and exponent $p$ (take $p\gt 3$ just to be sure it is regular). Say $$K=\langle x,y\mid x^p=y^p=[y,x]^p=[y,x,x]^p=[y,x,y]^p=[y,x,x,x]=[y,x,x,y]=[y,x,y,x]=[y,x,y,y]=1\rangle.$$ Every element of $K$ can be written uniquely as $x^ay^b[y,x]^c[y,x,x]^d[y,x,y]^e$, with multiplication done using commutator collection.

Now mod out by one of the two commutators of weight $3$, say $[y,x,y]$, to get $G$. This is a group of class $3$. Let $N$ be the subgroup of $G$ generated by the other commutator of weight $3$, $[y,x,x]$. The quotient of $G$ by $N$ is the Heisenberg group of order $p^3$, hence nonabelian.

Finally, let $g=[y,x]$. The subgroup generated by $g$ is not normal in $G$, since it does not contain $[y,x]^x = [y,x][y,x,x]$. However, $gN$ is central in $G/N$, hence $\langle g,N\rangle$ is normal in $G$.