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 $⟨H,K⟩$ 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

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

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 $\mathrm{A}\mathrm{u}\mathrm{t}(M)$ generated by the $u_n$ and $K$ be the subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(M)$ generated by the $v_n$.
• Define $J=⟨H,K⟩$, again as a subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(M)$.
• Define $G$ as the external semidirect product $M⋊J$, with the action of $J$ the usual action by automorphisms.

Then, both $H$ and $K$ are $3$-subnormal subgroups of $G$, but $J=⟨H,K⟩$ 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]

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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 $⟨H,B⟩$: In fact, $⟨H,B⟩$ 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}