# Normal closure of 3-subnormal subgroup of prime order in nilpotent group need not be abelian

## Statement

It is possible to have a nilpotent group $G$ and a 3-subnormal subgroup (?) $H$ of $G$ of order a prime number $p$ such that the Normal closure (?) $H^G$ of $H$ in $G$ is not an abelian group.

In fact, we can take $G$ itself to be a finite p-group.

## Proof

### Example of the dihedral group of order 16

Further information: dihedral group:D16, subgroup structure of dihedral group:D16

Let $G$ be the dihedral group of order 16 (degree 8), given explicitly by the presentation:

$G = \langle a,x \mid a^8 = x^2 = e, xax^{-1} = a^{-1} \rangle$

Suppose $H$ is the subgroup $\langle x \rangle$ of $G$. Then:

• $H$ is a subgroup of order 2 in $G$.
• $H$ is 3-subnormal in $G$.
• The normal closure of $H$ in $G$ is D8 in D16, the subgroup $\langle a^2,x \rangle$, which is isomorphic to dihedral group:D8. This is not an abelian group.