Central graphs and intersections of good subnormal subgroups

Problem

A subnormal subgroup $H$ of $G$ is good if its join with every subnormal subgroup is subnormal. Is the intersection of two good subgroups good?

\begin{aligned} \operatorname{Good}_G(H)&\iff H\mathrel{\mathrm{sn}}G\ \land\ \forall S\mathrel{\mathrm{sn}}G,\ \langle H,S\rangle\mathrel{\mathrm{sn}}G,\\ \operatorname{Good}_G(H)\land\operatorname{Good}_G(K) &\stackrel{?}{\Longrightarrow}\operatorname{Good}_G(H\cap K). \end{aligned}

Setup and result

Write $H\mathrel{\trianglelefteq\!\trianglelefteq}G$ when $H$ is subnormal in $G$ : there is a finite chain

H=H_0\trianglelefteq H_1\trianglelefteq\cdots\trianglelefteq H_d=G.

Following Thompson’s terminology in Kourovka Problem 8.74 (1), a subgroup $H\mathrel{\trianglelefteq\!\trianglelefteq}G$ is good if

S\mathrel{\trianglelefteq\!\trianglelefteq}G\quad\Longrightarrow\quad \langle H,S\rangle\mathrel{\trianglelefteq\!\trianglelefteq}G.

The question asks whether the intersection of two good subgroups is good.

Theorem 1. There is a countable group $G$ with good subgroups $H,K$ and a subnormal subgroup $J$ such that

H\cap K\mathrel{\trianglelefteq\!\trianglelefteq}G, \qquad \langle H\cap K,J\rangle\not\mathrel{\trianglelefteq\!\trianglelefteq}G.

In particular, good subnormal subgroups need not be closed under intersections.

Groups with two subnormal subgroups whose join is not subnormal are classical. We use a version of the transvection construction presented in Robinson (2, §13.1, Theorem 13.1.11); related constructions occur in Smith (3, §5). The additional step concerns goodness. A subgroup normal in a normal subgroup is good. If such a subgroup $U$ admits a homomorphism $f$ to an abelian group, then two graphs associated to $f$ are good in a central direct product. Their intersection identifies $\ker f$ . This permits a non-subnormal join involving $\ker f$ to obstruct goodness of the intersection.

All groups below are abstract. We use the commutator convention $[x,y]=xyx^{-1}y^{-1}$ .

Good subgroups and central graphs

We first record the elementary subnormality arguments used in the construction. Images and inverse images of finite subnormal chains are finite subnormal chains. Intersecting a chain with an intermediate subgroup gives the corresponding restriction of subnormality.

Lemma 2. If $U\trianglelefteq N\trianglelefteq E$ , then $U$ is good in $E$ .

Proof. Let $S\mathrel{\trianglelefteq\!\trianglelefteq}E$ , and put $M=\langle N,S\rangle$ . The image of $S$ in $E/N$ is subnormal, so $M\mathrel{\trianglelefteq\!\trianglelefteq}E$ . Let $R$ be the normal closure of $U$ in $M$ .

Since $M=SN$ , each $g\in M$ has the form $g=sn$ , with $s\in S$ and $n\in N$ . For $u\in U$ ,

gug^{-1}=s(nun^{-1})s^{-1}\in\langle U,S\rangle.

Thus $U\le R\le\langle U,S\rangle$ , and

\langle U,S\rangle=\langle R,S\rangle.

Now $S\mathrel{\trianglelefteq\!\trianglelefteq}M$ and $R\trianglelefteq M$ . Passing to $M/R$ and taking the inverse image shows that $\langle R,S\rangle\mathrel{\trianglelefteq\!\trianglelefteq}M$ . Hence $\langle U,S\rangle\mathrel{\trianglelefteq\!\trianglelefteq}E$ . ◻

Lemma 3. Let $Z\le Z(E)$ . If $XZ$ is good in $E$ , then $X$ is good in $E$ .

Proof. Since $X\trianglelefteq XZ$ , the subgroup $X$ is subnormal in $E$ . If $S\mathrel{\trianglelefteq\!\trianglelefteq}E$ and $L=\langle X,S\rangle$ , then

L\trianglelefteq LZ=\langle XZ,S\rangle\mathrel{\trianglelefteq\!\trianglelefteq}E.

Therefore $L\mathrel{\trianglelefteq\!\trianglelefteq}E$ . ◻

Proposition 4. Suppose that $U\trianglelefteq N\trianglelefteq E$ , that $C$ is abelian, and that $f:U\to C$ is a homomorphism. In $E\times C$ , the subgroups

H=U\times\{1\}, \qquad K=\{(u,f(u)):u\in U\}

are good. If $Q\mathrel{\trianglelefteq\!\trianglelefteq}E$ and $\langle\ker f,Q\rangle\not\mathrel{\trianglelefteq\!\trianglelefteq}E$ , then $H\cap K$ is not good.

Proof. Set $Z=\{1\}\times C$ . Then $Z\le Z(E\times C)$ and

HZ=KZ=U\times C\trianglelefteq N\times C\trianglelefteq E\times C.

Lemmas 2 and 3 prove that $H$ and $K$ are good. Moreover,

H\cap K=(\ker f)\times\{1\}.

The subgroup $J=Q\times\{1\}$ is subnormal in $E\times C$ . Projection onto $E$ maps $\langle H\cap K,J\rangle$ onto $\langle\ker f,Q\rangle$ . The former subgroup therefore cannot be subnormal. ◻

Square-free transvections

Let $\mathcal F$ be the set of finite subsets of $\mathbb N$ , and let

A=\bigoplus_{S\in\mathcal F}\mathbb F_2 e_S.

For $i\in\mathbb N$ , define an endomorphism $d_i$ of $A$ by

\tag{11} d_i e_S= \begin{cases} 0,&i\in S,\\ e_{S\cup\{i\}},&i\notin S. \end{cases}

Checking the basis vectors gives

\tag{12} d_i^2=0,\qquad d_i d_j=d_jd_i.

Put $v_0=e_\varnothing$ and $v_{n+1}=d_n v_n$ . Then

\tag{13} v_n=e_{\{0,\ldots,n-1\}}\ne0 \qquad(n\ge0),

where the set indexing $v_0$ is empty.

On $V=A\oplus A$ , define

\tag{14} a_i(x,y)=(x+d_i y,y),\qquad b_i(x,y)=(x,y+d_i x).

These maps are involutory linear automorphisms. Let

P=\langle a_i:i\in\mathbb N\rangle,\quad Q=\langle b_i:i\in\mathbb N\rangle,\quad T=\langle P,Q\rangle, \quad Z=\langle[a_i,b_j]:i,j\in\mathbb N\rangle.

Lemma 5. The subgroup $Z$ is central in $T$ , preserves both coordinate subspaces of $V$ , and satisfies

PZ\trianglelefteq T,\qquad QZ\trianglelefteq T.

Every element of $P$ fixes $A\oplus0$ pointwise and induces the identity on $V/(A\oplus0)$ . The analogous statements hold for $Q$ and $0\oplus A$ .

Proof. Using (12) and characteristic two in the product of the four transvections gives

\tag{17} [a_i,b_j](x,y)=(x+d_i d_jx,y+d_i d_jy).

The endomorphism $d_i d_j$ commutes with every $d_k$ . Thus (17) commutes with every $a_k$ and $b_k$ , and $Z\le Z(T)$ . Each map in (17), and hence each element of $Z$ , preserves both coordinate subspaces.

Conjugation of $a_i$ by $b_j$ changes $a_i$ by an element of $Z$ . Since $Z$ is central, every generator of $T$ normalizes $PZ$ . This proves $PZ\trianglelefteq T$ ; the other assertion is symmetric.

Products and inverses of upper transvections have the form $(x,y)\mapsto(x+d y,y)$ for a linear endomorphism $d$ of $A$ . Such a map fixes $A\oplus0$ and preserves the second coordinate. Lower transvections give the corresponding assertions for $Q$ . ◻

Form $E=V\rtimes T$ , using the multiplication

(v,t)(w,s)=(v+t w,ts).

We identify $V$ and $T$ with their canonical subgroups of $E$ , and similarly regard $P,Q$ as subgroups of $E$ . Put

U=(A\oplus0)\rtimes P,\qquad W=(0\oplus A)\rtimes Q,

and

N_P=V\rtimes PZ,\qquad N_Q=V\rtimes QZ.

Lemma 6. The following are normal chains in $E$ :

P\trianglelefteq U\trianglelefteq N_P\trianglelefteq E, \qquad Q\trianglelefteq W\trianglelefteq N_Q\trianglelefteq E.

The map $f:U\to V$ given by $f(u,p)=u$ is a homomorphism with kernel $P$ .

Proof. Lemma 5 gives $N_P\trianglelefteq E$ . Also $P\trianglelefteq PZ$ because $Z$ centralizes $P$ . For $(v,t)\in N_P$ and $(u,p)\in U$ ,

\tag{22} (v,t)(u,p)(v,t)^{-1} =(v+t u-(tpt^{-1})v,\ tpt^{-1}).

Here $tpt^{-1}\in P$ , $t u\in A\oplus0$ , and $v-(tpt^{-1})v\in A\oplus0$ by Lemma 5. Hence $U\trianglelefteq N_P$ .

Since $P$ fixes $A\oplus0$ pointwise, multiplication in $U$ satisfies

(u,p)(u',p')=(u+u',pp').

Consequently $f$ is a homomorphism with kernel $P$ , so $P\trianglelefteq U$ . Interchanging the coordinates proves the second chain. ◻

The non-subnormal join

Lemma 7. If $L\mathrel{\trianglelefteq\!\trianglelefteq}E$ , there is an integer $d\ge0$ such that, for every $c_0\in E$ and every sequence $t_0,t_1,\ldots$ in $L$ , the recursion

c_{n+1}=[t_n,c_n]

satisfies $c_d\in L$ .

Proof. Induct along a finite normal chain from $L$ to $E$ . The case $L=E$ has $d=0$ . Suppose $L\trianglelefteq L_1\mathrel{\trianglelefteq\!\trianglelefteq}E$ and that $d$ works for $L_1$ . For a sequence in $L$ , the induction hypothesis gives $c_d\in L_1$ . Therefore

c_{d+1}=t_d(c_d t_d^{-1}c_d^{-1})\in L.

◻

Proposition 8. The subgroup $T=\langle P,Q\rangle$ is not subnormal in $E$ .

Proof. Take $c_0=(0,v_0)\in V$ and put $t_n=a_n$ for even $n$ and $t_n=b_n$ for odd $n$ . Under the identification of $V$ with its translation subgroup,

[t,v]=t v-v.

Equations (14) and (13) now give, by induction,

c_n= \begin{cases} (0,v_n),&n\text{ even},\\ (v_n,0),&n\text{ odd}. \end{cases}

Every $c_n$ is nonzero. Since $V\cap T=1$ in $E$ , no $c_n$ belongs to $T$ . This contradicts Lemma 7 if $T\mathrel{\trianglelefteq\!\trianglelefteq}E$ . ◻

Proof of Theorem 1. Apply Proposition 4 to the chain $U\trianglelefteq N_P\trianglelefteq E$ , the abelian group $C=V$ , and the homomorphism $f$ in Lemma 6. In $G=E\times V$ , take

H=U\times\{0\},\qquad K=\{(u,f(u)):u\in U\},\qquad J=Q\times\{0\}.

Both $H$ and $K$ are good, while $J\mathrel{\trianglelefteq\!\trianglelefteq}G$ . Since $\ker f=P$ and $\langle P,Q\rangle=T\not\mathrel{\trianglelefteq\!\trianglelefteq}E$ , the join $\langle H\cap K,J\rangle$ is not subnormal. Intersecting a subnormal chain for $H$ with $K$ , then appending a subnormal chain for $K$ , proves $H\cap K\mathrel{\trianglelefteq\!\trianglelefteq}G$ .

Finally, $\mathcal F$ and $A$ are countable. The group $T$ is generated by the countable family $\{a_i,b_i:i\in\mathbb N\}$ and is therefore a quotient of a free group on a countable set. Thus $V$ , $E$ and $G$ are countable. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved Problems in Group Theory, 21st ed., September 2026 update, Problem 8.74, arXiv:1401.0300.

D. J. S. Robinson, A Course in the Theory of Groups, 2nd ed., Graduate Texts in Mathematics 80, Springer, New York, 1996.

H. Smith, Groups with the subnormal join property, Canad. J. Math. 37 (1985), 1–16.