Normalizer of 2-subnormal subgroup may have arbitrarily large subnormal depth

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This is a statement of the form: there exist subnormal subgroups of arbitrarily large subnormal depth satisfying certain conditions.
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Statement

Suppose k is a positive integer. Then, we can find a 2-subnormal subgroup of a finite group whose normalizer is (k+1)-subnormal but not k-subnormal.

Related facts

Facts used

  1. There exist subgroups of arbitrarily large subnormal depth
  2. Subnormality satisfies image condition: The image of a k-subnormal subgroup under a surjective homomorphism is k-subnormal in the image.
  3. Subnormality satisfies inverse image condition: The inverse image of a k-subnormal subgroup for any homomorphism is k-subnormal.

Proof

The construction is as follows:

  1. (uses: Fact (1)): Find a finite group G and a subgroup H of G such that H is (k + 1)-subnormal in G but is not k-subnormal in G.
  2. Let V be the additive group of the group ring of G over any finite field, and consider the semidirect product K := V \rtimes G where G acts by left multiplication. Then, V is an Abelian normal subgroup of G, so any subgroup of V is 2-subnormal in G.
  3. Let W be the subspace of V spanned by the basis elements corresponding to H. W is 2-subnormal in G, and its normalizer in K = V \rtimes G is V \rtimes H. Thus, we have a 2-subnormal subgroup whose normalizer in the whole group is the subgroup V \rtimes H.
  4. Consider the projection map from K = V \rtimes G to G with kernel V.
    • By fact (2), we observe that if V \rtimes H is k-subnormal in K, then H is k-subnormal in G. Thus, V \rtimes H is not k-subnormal in G.
    • By fact (3), we find that since H is (k+1)-subnormal in G, its full inverse image V \rtimes H is (k+1)-subnormal in K.