Normal closure of 2-subnormal subgroup of prime order is abelian

Statement

Suppose $G$ is a group and $H$ is a 2-subnormal subgroup (?) of $G$ of order $p$ for some prime number $p$. Then, the Normal closure (?) of $H$ in $G$ is an abelian group.

Related facts

• 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
• Normal closure of 2-subnormal subgroup of prime order in nilpotent group is abelian

Opposite facts

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