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

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Statement

It is possible to have a nilpotent group G and a 3-subnormal subgroup (?) H of G of order a prime number p such that the 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

Proof

Example of the dihedral group of order 16

Further information: dihedral group:D16, subgroup structure of dihedral group:D16

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.