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

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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.

Related facts

Similar facts

Opposite facts

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]