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

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

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

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 $l(X)\leq L(X)$ 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 $n\in \mathbb {Z}$ $n\in \mathbb {Z}$ , define $a_{X*n}=a_{X\cup \{n\}}$ $a_{X*n}=a_{X\cup \{n\}}$ if $n\notin X$ $n\notin X$ , and $0$ $0$ if $n\in X$ $n\in X$ . Analogously, define $b_{X*n}$ $b_{X*n}$ . Now define, for $n\in \mathbb {Z}$ $n\in \mathbb {Z}$ :
- 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 $H^{K}M$ : PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
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The normal closure of $H$ in $H^{K}M$ is $HB$ : PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
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$H$ 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

A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, Page 389, ^{More info}