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

Statement
Suppose $$G$$ is a nilpotent group and $$H$$ is a fact about::2-subnormal subgroup of $$G$$ of order $$p^r$$ for some prime number $$p$$. Then, the fact about::normal closure $$H^G$$ of $$H$$ in $$G$$ is a nilpotent group and its nilpotency class is at most equal to $$r$$.

Note that since prime power order implies nilpotent, the result applies in particular whenever $$G$$ itself is a finite p-group.

Similar facts

 * 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

Facts used

 * 1) uses::Normal of prime power order implies contained in upper central series member corresponding to prime-base logarithm of order in nilpotent
 * 2) uses::Upper central series members are characteristic
 * 3) uses::Characteristic of normal implies normal
 * 4) uses::Nilpotency is subgroup-closed (more specifically, nilpotency of fixed class is subgroup-closed).

Proof
Given: A nilpotent group $$G$$, a 2-subnormal subgroup $$H$$ of $$G$$. $$H$$ has order $$p^r$$, with $$p$$ a prime number.

To prove: The normal closure $$H^G$$ has nilpotency at most $$r$$

Proof: