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 fact about::3-subnormal subgroup $$H$$ of $$G$$ of order a prime number $$p$$ such that the fact about::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.

Related facts

 * Normal closure of 2-subnormal subgroup of prime order in nilpotent group is abelian
 * Normal closure of 2-subnormal subgroup of prime power order in nilpotent group has nilpotency class at most equal to prime-base logarithm of order

Example of the dihedral group of order 16
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.