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

Statement
It is possible to have a group $$G$$ and two fact about::3-subnormal subgroups $$H$$ and $$K$$ of $$G$$ such that the join $$\langle H, K \rangle$$ is not a fact about::subnormal subgroup of $$G$$. In fact, it can happen that the join is a proper fact about::contranormal subgroup.

Similar facts

 * Subnormality is not finite-join-closed is an easy corollary.
 * 3-subnormality is not finite-join-closed is another easy corollary.
 * 3-subnormal not implies finite-automorph-join-closed subnormal: This is not a corollary of the statement, but the same example works.
 * 4-subnormal not implies finite-conjugate-join-closed subnormal

Opposite facts

 * Join of normal and subnormal implies subnormal of same depth
 * 2-subnormal implies join-transitively subnormal
 * 3-subnormal implies finite-conjugate-join-closed subnormal

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:

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 $$H^KM$$:
 * The normal closure of $$H$$ in $$H^KM$$ is $$HB$$:
 * $$H$$ 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 $$M$$.