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

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 $G$?

• The answer is that it depends on $G$. Are you looking for an example in which there is no such $g \in G$? – Derek Holt Aug 13 '17 at 11:59
• There are so many different examples both of groups that satisfy and of groups that do not satisfy this condition, that it seems very unlikely that you could classify them in any meaningful way. – Derek Holt Aug 13 '17 at 15:48

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.

• I think you mean $|N|=p$, though of course it also works if $p=3$. But note that he wants $\langle g\rangle$ not normal but $N\langle g\rangle$ normal, hence he wants $\langle g\rangle\not\triangleleft G$ and $\langle g,N\rangle\triangleleft G/N$, no? – Arturo Magidin Aug 13 '17 at 12:20
• Yes of course I meant $|N|=p$. I was unsure what he wanted.It is unclear whether he means do some groups satisfy the condition or do all groups satisfy the condition. I thought he might be looking for an example in which there exists no $g \in G$ that satisfies the condition. – Derek Holt Aug 13 '17 at 12:39
• Good point! And I gave him an example where such a $g$ can exist, so I guess we've covered the bases. – Arturo Magidin Aug 13 '17 at 12:56
• Thank you very much for your counterexample and example for my question. Dear professor Holt is it possible to classify all finite $p$-groups that do not satisfy the mentioned property(All counterexamples for this question)? – sebastian Aug 13 '17 at 13:01
• @sebastian:It's unclear to me, at least, what "all counterexamples for this question" may mean (it was unclear what the question was to begin with, on reflection). – Arturo Magidin Aug 13 '17 at 13:20

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 $G$.

• Perhaps an easier example is $G = P \times Q$ where $P$ is nonabelian of order $p^3$, $Q$ is any nonabelian $p$-group, $N = Z(P)$, and $g \in P \setminus Z(P)$. – Derek Holt Aug 13 '17 at 13:23