Join of two 3-subnormal subgroups may be proper and contranormal

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Statement

It is possible to have a group G and two 3-subnormal subgroup (?)s H and K of G such that the join \langle H, K \rangle is not a Subnormal subgroup (?) of G. In fact, it can happen that the join is a proper Contranormal subgroup (?).

Related facts

Similar facts

Opposite facts

Proof

Construction of the counterexample

The construction involves the following steps:

  • Let S be the set of all subsets X of \mathbb{Z} such that there exists integers l(X) \le L(X) such that X contains all integers less than l(X) and no integer greater than L(X).
  • Let A be an elementary abelian 2-group with basis (as a vector space over the field of two elements) given by a_X, where X ranges over S.
  • Let B be an elementary abelian 2-group with basis (as a vector space over the field of two elements) given by b_X, where X ranges over S. (Note that A and B are isomorphic).
  • Let M be the direct product of A and B.
  • For every n \in \mathbb{Z}, define a_{X*n} = a_{X \cup \{ n \}} if n \notin X, and 0 if n \in X. Analogously, define b_{X * n}. Now define, for n \in \mathbb{Z}:
    • Automorphisms u_n:M \to M given on the basis by u_n(a_X,b_Y) = (a_X, b_{X*n} + b_Y).
    • Automorphisms v_n:M \to M given on the basis by v_n(a_X,b_Y) = (a_{Y*n} + a_X, b_Y).
  • Let H be the subgroup of \operatorname{Aut}(M) generated by the u_n and K be the subgroup of \operatorname{Aut}(M) generated by the v_n.
  • Define J = \langle H, K \rangle, again as a subgroup of \operatorname{Aut}(M).
  • Define G as the external semidirect product M \rtimes J, with the action of J the usual action by automorphisms.

Then, both H and K are 3-subnormal subgroups of G, but J = \langle H, K \rangle is not a subnormal subgroup of G.

Preliminary computations

Claim: [H,A] = B and [K,B] = A.

Proof: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

Proof that H and K are both 3-subnormal

We prove that H is 3-subnormal in three steps:

  • The normal closure of H in G is H^KM: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
  • The normal closure of H in H^KM is HB: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
  • H is normal in \langle H, B \rangle: In fact, \langle H, B \rangle is an internal direct product of H and B.

Proof that J is proper and contranormal

The normal closure of J in G contains both H^A = H[H,A] = HB and K^B = K[K,B] = KA. Thus, the normal closure of J in G contains H,K,A,B, and hence must be the whole group G.

That J is proper follows because it is the non-normal part of a semidirect product with a nontrivial group M.

References

Textbook references