Equivalence of definitions of weakly abnormal subgroup

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This article gives a proof/explanation of the equivalence of multiple definitions for the term weakly abnormal subgroup
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

Here are two equivalent definitions of weakly abnormal for a subgroup H of a group G:

  1. For any g \in G, let H^{\langle g \rangle} be the closure in G of H under the action by conjugation of the cyclic group generated by g. Then, g \in H^{\langle g \rangle}.
  2. If H \le K \le G, then K is a Self-normalizing subgroup (?) of G.
  3. If H \le L \le G, then H is a Contranormal subgroup (?) of L.

Proof

(1) implies (2)

Given: H \le G such that g \in H^{\langle g \rangle} for all g \in G. H \le K \le G.

To prove: K = N_G(K).

Proof: Suppose g \in N_G(K). Then, K^{\langle g \rangle} = K, so H^{\langle g \rangle} \le K^{\langle g \rangle} \le K. In particular, g \in H^{\langle g \rangle} \le K, so g \in K. Thus, N_G(K) \le K. K \le N_G(K) is tautological, so N_G(K) = K.

(2) implies (1)

Given: H \le G such that, for any intermediate subgroup K, K = N_G(K).

To prove: g \in H^{\langle g \rangle} for any g \in G.

Proof: Let K = H^{\langle g \rangle}. Then, by definition of H^{\langle g \rangle}, both g and g^{-1} map K to within itself, so conjugation by g gives an automorphism of K. Thus, g \in N_G(K). By assumption, K = N_G(K), so g \in K, yielding g \in H^{\langle g \rangle}, as desired.

(2) implies (3)

Given: A subgroup H of G with the property that N_G(K) = K for every subgroup K of G containing H.

To prove: If H \le L \le G, H is contranormal in L.

Proof: Let K be the normal closure of H in L. Since K is normal in L, L \le N_G(K). Since H \le K, N_G(K) = K, so L \le K, forcing K = L. Thus, H is contranormal in L.

(3) implies (2)

Given: A subgroup H of G with the property that if H \le L \le G, H is contranormal in L.

To prove: N_G(K) = K for any subgroup K of G containing H.

Proof: Let L = N_G(K). Then H \le K \le L \le G. By assumption, H is contranormal in L, and K is normal in L, forcing K = L. Thus, we get K = N_G(K).