# 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 (?).

## 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 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 $M$.