# 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 2-subnormal subgroup (?) of $G$ of order $p^r$ for some prime number $p$. Then, the 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.

## Facts used

1. Normal of prime power order implies contained in upper central series member corresponding to prime-base logarithm of order in nilpotent
2. Upper central series members are characteristic
3. Characteristic of normal implies normal
4. 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:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There exists a normal subgroup $K$ of $G$ such that $H$ is a normal subgroup of $K$. Definition of 2-subnormal $H$ is 2-subnormal in $G$.
2 $K$ is nilpotent. Fact (4) $G$ is nilpotent. Step (1)
3 $H$ is contained in $Z^r(K)$, i.e., the $r^{th}$ member of the upper central series of $K$, which is a characteristic subgroup of $K$ of nilpotency class at most $r$. Facts (1), (2) $H$ has order $p^r$ Steps (1), (2) Fact+Given+Step direct.
4 $Z^r(K)$ is a normal subgroup of $G$ of nilpotency class at most $r$. Fact (3) Steps (1), (3) [SHOW MORE]
5 The normal closure $H^G$ is contained in $Z^r(K)$. Steps (3), (4) [SHOW MORE]
6 $H^G$ has nilpotency class at most $r$. Steps (4), (5) [SHOW MORE]