Cardinal bounds for cyclic ascent and separation of overnilpotent radicals in 2-groups

Problem

For infinite cardinals $\mu_1<\mu_2$ , does there exist a nontrivial $\mu_2$ -overnilpotent group with trivial $\mu_1$ -overnilpotent radical?

\forall\mu_1<\mu_2,\quad \exists G\ne1:\quad G\in\mathcal N_{\mu_2},\quad R_{\mu_1}(G)=1.

A group is $\mu$ -overnilpotent when every cyclic subgroup admits a continuous ascending normal series of ordinal length having cardinality less than $\mu$ . The cardinals need not be regular.

Ascending series and the separation theorem

Let $\mu$ be an infinite cardinal. A subgroup $H\le G$ is $\mu$ -ascendant if there are an ordinal $\gamma$ and subgroups $(H_\alpha)_{\alpha\le\gamma}$ such that

\begin{align*} H_0&=H,& H_\gamma&=G,& \lvert \gamma\rvert&<\mu,\tag{1a}\\ H_\alpha&\trianglelefteq H_{\alpha+1}&& (\alpha<\gamma),\tag{1b}\\ H_\lambda&=\bigcup_{\alpha<\lambda}H_\alpha &&(0<\lambda\le\gamma\text{ a limit ordinal}).\tag{1c} \end{align*}

The series is increasing. Only successive normality is required in (1b). The group $G$ is $\mu$ -overnilpotent if every cyclic subgroup of $G$ is $\mu$ -ascendant. We use the radical

\tag{2} \mathrm R_\mu(G)=\bigvee\{N\trianglelefteq G: N\text{ is }\mu\text{-overnilpotent}\}.

Here $\bigvee$ denotes the subgroup generated by the indicated family.

Write $\mathcal N_\mu$ for the class of $\mu$ -overnilpotent groups. Vovsi (1) proved

\mu_1<\mu_2\quad\Longrightarrow\quad \mathcal N_{\mu_2}\setminus\mathcal N_{\mu_1}\ne\varnothing.

Plotkin’s Problem 3.43 asks for the stronger separation

\exists G\in\mathcal N_{\mu_2}\quad \mathrm R_{\mu_1}(G)=1.

We obtain it by iterating extensions based on Vovsi’s group (2, Lemma 1), so that every nonidentity element obstructs a short cyclic ascent.

Theorem 1. For every infinite cardinal $\kappa$ there is a nontrivial $2$ -group $G_\kappa$ with the following properties.

Every cyclic subgroup of $G_\kappa$ has a continuous ascending normal series to $G_\kappa$ whose terminal ordinal has cardinality at most $\kappa$ .
If $g\in G_\kappa$ and $\langle g\rangle$ is $\kappa$ -ascendant in $G_\kappa$ , then $g=1$ .

In particular, for every $\mu>\kappa$ ,

\tag{5} \mathrm R_\kappa(G_\kappa)=1, \qquad \mathrm R_\mu(G_\kappa)=G_\kappa.

Neither cardinal is required to be regular.

A $2$ -group here means a group in which every element has finite order a power of $2$ . We write $[a,b]=aba^{-1}b^{-1}$ and $A\vee B=\langle A,B\rangle$ . In a regular wreath product we use

\tag{6} A\wr Q=A^Q\rtimes Q, \qquad {}^qd(r)=d(q^{-1}r).

For $a\in A$ let $\delta_r(a)$ denote the function with value $a$ at $r$ and value $1$ elsewhere. Then

\tag{7} q\delta_r(a)q^{-1}=\delta_{qr}(a).

Operations on ascending series

Images, inverse images and intersections preserve successive normality and continuity. At limits this follows from

f\Bigl(\bigcup_{\alpha<\lambda}H_\alpha\Bigr) =\bigcup_{\alpha<\lambda}f(H_\alpha), \qquad f^{-1}\Bigl(\bigcup_{\alpha<\lambda}K_\alpha\Bigr) =\bigcup_{\alpha<\lambda}f^{-1}(K_\alpha).

Concatenation adds ordinal lengths; constant extension pads a series to any larger length.

Lemma 2. Let $(C_\alpha)_{\alpha\le\eta}$ be a continuous increasing chain. Suppose that, for every $\alpha<\eta$ , there is a continuous ascending normal series from $C_\alpha$ to $C_{\alpha+1}$ of length at most a fixed nonzero limit ordinal $\rho$ . Then there is such a series from $C_0$ to $C_\eta$ of length $\rho\eta$ .

Proof. Extend each series constantly to length $\rho$ . If its terms are $B_{\alpha,\beta}$ , set

D_{\rho\alpha+\beta}=B_{\alpha,\beta} \qquad(\alpha<\eta,\ \beta<\rho), \qquad D_{\rho\eta}=C_\eta.

Successive normality holds inside each block. At the end of a block, continuity of the extended series gives $C_{\alpha+1}$ . At a limit of blocks, continuity follows from that of $C$ . The uniqueness of division by $\rho$ makes the definition consistent. ◻

Lemma 3. Let $C_0\le C_1\le\cdots$ have union $K$ . Suppose $\kappa$ is infinite and each inclusion $C_n\le C_{n+1}$ has a continuous ascending normal series of length $\gamma_n$ with $\lvert \gamma_n\rvert\le\kappa$ . Then $C_0\le K$ has such a series of cardinal length at most $\kappa$ .

Proof. Put

\rho=\sup_{n<\omega}\gamma_n+\omega.

This is a nonzero limit ordinal, $\gamma_n\le\rho$ , and

\lvert \rho\rvert\le \sum_{n<\omega}\lvert \gamma_n\rvert+\aleph_0 \le\kappa\aleph_0=\kappa.

Apply Lemma 2 with $\eta=\omega$ and $C_\omega=K$ . The resulting length $\rho\omega$ has cardinality at most $\kappa$ . This argument uses no cofinality assumption on $\kappa$ . ◻

We shall also use the following join operation. If $N\trianglelefteq G$ and $H$ ascends to $K$ in $G$ , then $H\vee N$ ascends to $K\vee N$ with the same length: take the image series in $G/N$ and then its inverse image in $G$ .

Elementary abelian normal layers

Lemma 4. Let $V$ be an elementary abelian $2$ -group. If an action of $H$ on $V$ factors through a finite $2$ -group $P$ , then the complement $H$ is subnormal in $V\rtimes H$ in finitely many steps.

Proof. First consider $V\rtimes P$ . For the given action $\phi$ , the map

(v,p)\longmapsto\bigl((q\longmapsto\phi(q^{-1})(v)),p\bigr)

embeds $V\rtimes P$ into $V\wr P$ . The homomorphisms $V\to C_2$ separate points, since $V$ is a vector space over $\mathbf F_2$ . Applying each homomorphism to every base coordinate therefore embeds

V\wr P\hookrightarrow \prod_{\chi\in\operatorname{Hom}(V,C_2)}(C_2\wr P).

The group $C_2\wr P$ is a finite $2$ -group, hence nilpotent. Every factor in the displayed product has the same finite nilpotency class, so the product and $V\rtimes P$ are nilpotent.

In a nilpotent group $L$ of class $c$ , every subgroup $U$ has the finite ascending normal series

U=UZ_0(L)\trianglelefteq UZ_1(L)\trianglelefteq\cdots \trianglelefteq UZ_c(L)=L.

For a factorisation $H\to P$ of the action, pull the series for the complement $P\le V\rtimes P$ back along $V\rtimes H\to V\rtimes P$ . Its first term is exactly $H$ . ◻

An elementary normal series in $G$ is a continuous increasing series

\tag{15} 1=N_0\le N_1\le\cdots\le N_\gamma=G, \qquad N_\alpha\trianglelefteq G,

whose successive sections $N_{\alpha+1}/N_\alpha$ are elementary abelian $2$ -groups. Thus

\tag{16} [N_{\alpha+1},N_{\alpha+1}]\le N_\alpha, \qquad x^2\in N_\alpha\quad(x\in N_{\alpha+1}).

Lemma 5. If $G$ has an elementary normal series of length $\gamma$ , every finite $2$ -subgroup $U\le G$ has a continuous ascending normal series to $G$ of length $\omega\gamma$ .

Proof. Consider $C_\alpha=U\vee N_\alpha=UN_\alpha$ . This chain is continuous. In $G/N_\alpha$ , let $\overline U$ be the image of $U$ and $V$ the image of $N_{\alpha+1}$ . The subgroup $V$ is elementary abelian and normal, and $\overline U$ is a finite $2$ -group. The multiplication map

V\rtimes\overline U\longrightarrow V\overline U

is surjective and sends the complement to $\overline U$ . Lemma 4, followed by image and inverse image, gives a finite ascending normal series from $UN_\alpha$ to $UN_{\alpha+1}$ . Pad each such series to length $\omega$ and apply Lemma 2. ◻

Lemma 6. Let $(E_j)_{j\in J}$ be a family of normal elementary abelian $2$ -subgroups of $G$ .

The subgroup $N=\bigvee_{j\in J}E_j$ has an elementary normal series, with all terms normal in $G$ , of length $\theta$ for some ordinal $\theta$ with $\lvert \theta\rvert=\lvert J\rvert$ .
If $H\le G$ is subnormal in $H\vee E_j$ in finitely many steps for every $j$ , then $H$ has an ascending normal series to $H\vee N$ of length $\omega\theta$ .

Proof. Choose a well-order $(j_\alpha)_{\alpha<\theta}$ of $J$ and put

N_\alpha=\bigvee_{\beta<\alpha}E_{j_\beta}.

All terms are normal in $G$ . At a limit, every element of the join is a finite product and therefore lies in an earlier term. The section

N_{\alpha+1}/N_\alpha =(N_\alpha E_{j_\alpha})/N_\alpha

is an image of $E_{j_\alpha}$ , which proves the first assertion. For the second, join a finite series from $H$ to $H\vee E_{j_\alpha}$ with $N_\alpha$ . This gives a finite series from $H\vee N_\alpha$ to $H\vee N_{\alpha+1}$ . Apply Lemma 2 with blocks of length $\omega$ . ◻

Vovsi’s group and the cardinal obstruction

Fix a set $I$ of infinite cardinality $\kappa$ . Let

\tag{20} A=A(I)=\left\langle x_i\ (i\in I)\ \middle|\ x_i^2=1,\ [x_i,w^{-1}x_iw]=1\ (i\in I,\ w\in F(I))\right\rangle,

where $F(I)$ is the free group on $I$ . This is the case $p=2$ of Vovsi’s presentation (2, Lemma 1). Write $N_i=\langle x_i\rangle^{A}$ for the normal closure of $x_i$ .

Lemma 7. Each $N_i$ is elementary abelian. For every finite subset $S\subseteq I$ , killing all $x_i$ with $i\notin S$ defines a retraction

r_S:A(I)\longrightarrow A(S)

onto a finite $2$ -group. These retractions separate points. The group $A(I)$ is a $2$ -group.

Proof. Any two conjugates of $x_i$ commute by (20), and each has square $1$ . Hence $N_i$ is elementary abelian. Every map on generators that sends each generator either to a generator or to $1$ preserves the defining relations. The inclusions $A(S)\to A(I)$ are therefore split by $r_S$ .

Finiteness of $A(S)$ follows by induction on $\lvert S\rvert$ . Removing one generator gives the quotient by its elementary abelian normal closure. The quotient is a finite $2$ -group by induction. The kernel is finitely generated, by Schreier’s theorem, since it has finite index in the finitely generated group $A(S)$ , and an elementary abelian $2$ -group generated by finitely many elements is finite. Thus $A(S)$ is a finite extension of finite $2$ -groups.

Every element of $A(I)$ is represented by a word using finitely many generators. For such a set $S$ , the element belongs to the embedded copy of $A(S)$ and is fixed by the corresponding retraction. Thus the retractions separate points and every element has $2$ -power order. ◻

A disjoint-label commutator is a fully parenthesised commutator in which each leaf is one of the generators $x_i$ and no generator label occurs twice. A single generator is allowed. Its support is the finite set of its labels.

Lemma 8. Every disjoint-label commutator in $A(I)$ is nonidentity.

Proof. List its $m$ leaves in their left-to-right order. Map the corresponding generators to

t_j=1+E_{j,j+1}\in\operatorname{UT}_{m+1}(\mathbf F_2) \qquad(0\le j<m)

and all other generators to $1$ . This respects (20). Indeed, each $t_j$ has order $2$ , and the nonzero entries of the off-diagonal part of any conjugate lie in rows at most $j$ and columns at least $j+1$ . The product of any two such off-diagonal parts is zero, so $t_j$ commutes with all its conjugates.

For $a<b<c$ , direct matrix multiplication gives

[1+E_{a,b},1+E_{b,c}]=1+E_{a,c}.

Induction on the parenthesisation shows that the specified commutator maps to $1+E_{0,m}\ne1$ . ◻

Lemma 9. There is no continuous ascending series $(K_\alpha)_{\alpha\le\gamma}$ from $1$ to $A(I)$ satisfying

\tag{24} \lvert \gamma\rvert<\kappa, \qquad [K_{\alpha+1},K_{\alpha+1}]\le K_\alpha \quad(\alpha<\gamma).

Proof. Fix an infinite regular cardinal $\rho$ with $\lvert \gamma\rvert<\rho\le\lvert I\rvert$ . For $K\le A(I)$ , write $\mathsf L_\rho(K)$ if there are disjoint-label commutators $(c_\xi)_{\xi<\rho}$ such that

c_\xi\in K,\qquad \operatorname{supp}(c_\xi)\cap\operatorname{supp}(c_\zeta)=\varnothing \quad(\xi\ne\zeta).

Call such a subgroup large. The generators witness $\mathsf L_\rho(A(I))$ , whereas Lemma 8 gives $\neg\mathsf L_\rho(1)$ .

If $K_{\alpha+1}$ is large, pair off its family and take the commutator of each pair. The resulting words are still disjoint-label commutators, still have pairwise disjoint supports, and lie in $K_\alpha$ by (24). Thus $K_\alpha$ is large. If a limit term $K_\lambda$ is large, continuity assigns each word in a large family to an earlier term. Since

\lvert \lambda\rvert\le\lvert \gamma\rvert<\rho=\operatorname{cf}(\rho),

one earlier term contains a subfamily of cardinality $\rho$ and is large. Transfinite induction now shows that no term is large, contradicting $K_\gamma=A(I)$ .

For arbitrary infinite $\kappa$ , take $\rho=\aleph_0$ when $\gamma$ is finite and $\rho=(\lvert \gamma\rvert)^+$ otherwise. Then $\rho$ is regular and

\lvert \gamma\rvert<\rho\le\kappa=\lvert I\rvert.\quad\Box

◻

Proposition 10. Let $Q$ be any group and $1\ne t\in Q$ . The pure top cyclic subgroup $\langle t\rangle$ of $A(I)\wr Q$ is not $\kappa$ -ascendant.

Proof. Suppose $(H_\alpha)_{\alpha\le\gamma}$ is a contrary series and set

K_\alpha=\{a\in A(I):\delta_1(a)\in H_\alpha\}.

Inverse image along $\delta_1$ gives a continuous increasing series with $K_0=1$ and $K_\gamma=A(I)$ . Fix $\alpha<\gamma$ and $a,b\in K_{\alpha+1}$ . Every term $H_\alpha$ contains $t$ . In $H_{\alpha+1}/H_\alpha$ , equation (7) therefore identifies $\delta_1(a)$ with $\delta_t(a)$ . Since $t\ne1$ , the latter commutes with $\delta_1(b)$ . Hence

[\delta_1(a),\delta_1(b)]\in H_\alpha, \qquad [a,b]\in K_\alpha.

This contradicts Lemma 9. ◻

The extension step

Let $H$ be a $2$ -group with a separating family of surjections onto finite $2$ -groups,

\tag{30} q_j:H\twoheadrightarrow P_j\quad(j\in J), \qquad \bigcap_{j\in J}\ker q_j=1, \qquad \lvert J\rvert\le\kappa.

Define the restricted product

\begin{align*} B&=\bigoplus_{j\in J} A(I)^{P_j},\tag{31a}\\ H^+&=B\rtimes H, &({}^h b)_j(p)&=b_j(q_j(h)^{-1}p).\tag{31b} \end{align*}

Thus only finitely many orbit coordinates $b_j$ of a base element are nonidentity. We regard $H$ as the top complement of $H^+$ .

Lemma 11. The group $H^+$ is a $2$ -group with a separating family of finite $2$ -quotients indexed by $J\times\{S\subseteq I:S\text{ finite}\}$ . Every nonidentity $h\in H$ has an image under a surjection

H^+\twoheadrightarrow A(I)\wr P_j

which is a nonidentity pure top element.

Proof. Each finite power $A(I)^{P_j}$ is a $2$ -group. A base element has finite support, so a common $2$ -power kills all its coordinates. Thus $B$ is a $2$ -group. For $x\in H^+$ , some $2$ -power of $x$ belongs to $B$ , and a further $2$ -power is $1$ .

Projection onto orbit $j$ , together with $q_j$ on the top, gives a surjective homomorphism

\pi_j:H^+\longrightarrow A(I)\wr P_j.

For a finite $S\subseteq I$ , follow $\pi_j$ by the basewise retraction $r_S$ to obtain

\tag{34} \pi_{j,S}:H^+\twoheadrightarrow A(S)\wr P_j.

The target is a finite $2$ -group. If $x=(b,h)$ has $h\ne1$ , choose $j$ with $q_j(h)\ne1$ ; every $\pi_{j,S}(x)$ has nonidentity top coordinate. If $h=1$ and $b\ne1$ , choose $j,p$ with $b_j(p)\ne1$ and then a finite $S$ with $r_S(b_j(p))\ne1$ . This proves separation. Finally, for $1\ne h\in H$ , the first choice of $j$ gives the asserted image under $\pi_j$ . ◻

For $(j,i)\in J\times I$ , let $E_{j,i}$ be the subgroup supported on orbit $j$ with values in $N_i$ :

\tag{35} E_{j,i}=N_i^{P_j}\le B.

Lemma 12. Assume (30). Then:

$H$ has a continuous ascending normal series to $H^+$ of cardinal length at most $\kappa$ ;
if $H$ has an elementary normal series of cardinal length at most $\kappa$ , so does $H^+$ .

Proof. The groups $E_{j,i}$ are elementary abelian. They are normal in $H^+$ : base conjugation preserves $N_i\trianglelefteq A(I)$ , and top conjugation only permutes the coordinates within $P_j$ . Moreover,

\tag{36} B=\bigvee_{(j,i)\in J\times I}E_{j,i}.

Indeed, the generators $x_i$ generate each coordinate copy of $A(I)$ ; each orbit is finite and every base element has finite orbit support.

The action of $H$ on $E_{j,i}$ factors through $P_j$ . Lemma 4 gives a finite ascending normal series from $H$ to $H E_{j,i}$ . Choose a well-order of $J\times I$ of type $\theta$ . Lemma 6 gives a series from $H$ to $HB=H^+$ of length $\omega\theta$ , and

\lvert \omega\theta\rvert\le\aleph_0\lvert J\times I\rvert\le\kappa.

This proves the first assertion.

For the second, the first part of Lemma 6 gives an elementary normal series from $1$ to $B$ of length $\theta$ , with every term normal in $H^+$ . Pull an elementary normal series of $H$ back along $H^+\to H$ and append it. If its old length was $\delta$ , the new length is $\theta+\delta$ and

\lvert \theta+\delta\rvert\le\kappa+\kappa=\kappa.

◻

The tower and the two radicals

Start with $H_0=C_2$ and its identity quotient. Apply (31b) recursively, retaining the quotient family (34), to obtain

\tag{39} H_0\hookrightarrow H_1 \hookrightarrow H_2\hookrightarrow\cdots.

Each arrow is the pure top inclusion. At every stage, the finite-quotient family has at most $\kappa$ members, because

\lvert \{S\subseteq I:S\text{ finite}\}\rvert=\kappa, \qquad \kappa\cdot\kappa=\kappa.

Let $G_\kappa$ be the direct limit, identifying each $H_n$ with its image, so

\tag{41} G_\kappa=\bigcup_{n<\omega}H_n.

All inclusions are injective. The group $G_\kappa$ is therefore nontrivial, and every element belongs to a $2$ -group $H_n$ , so $G_\kappa$ is a $2$ -group.

Lemma 13. Every cyclic subgroup of $G_\kappa$ has a continuous ascending normal series to $G_\kappa$ of cardinal length at most $\kappa$ .

Proof. The elementary group $H_0=C_2$ has an elementary normal series of length $1$ . Induction using Lemma 12 gives an elementary normal series in every $H_n$ with cardinal length at most $\kappa$ . Let $g\in H_n$ . Since $H_n$ is a $2$ -group, $\langle g\rangle$ is finite. Lemma 5 gives a series from $\langle g\rangle$ to $H_n$ with cardinal length at most $\kappa$ .

Each subsequent inclusion $H_m\le H_{m+1}$ has such a series by Lemma 12. Lemma 3, applied to the tail of (39), gives a series from $H_n$ to $G_\kappa$ with the same cardinal bound. Concatenating the two series preserves that bound, since $\kappa+\kappa=\kappa$ . ◻

Lemma 14. If $g\in G_\kappa$ and $\langle g\rangle$ is $\kappa$ -ascendant in $G_\kappa$ , then $g=1$ .

Proof. Suppose $g\ne1$ and choose $n$ with $g\in H_n$ . Intersect an alleged short series in $G_\kappa$ with $H_{n+1}$ . This gives a series of the same length from $\langle g\rangle$ to $H_{n+1}$ . By Lemma 11, some surjection

H_{n+1}\twoheadrightarrow A(I)\wr P_j

sends $g$ to a nonidentity pure top element. The image series contradicts Proposition 10. ◻

Proof of Theorem 1. The construction gives a nontrivial $2$ -group. Lemmas 13 and 14 prove its two ascent properties. For $\mu>\kappa$ , the whole group is $\mu$ -overnilpotent, so (2) gives $\mathrm R_\mu(G_\kappa)=G_\kappa$ .

Let $N\trianglelefteq G_\kappa$ be $\kappa$ -overnilpotent and $g\in N$ . Append one normal step to a short series inside $N$ :

\langle g\rangle\ \underset{|\gamma|<\kappa}{\nearrow}\ N \trianglelefteq G_\kappa, \qquad |\gamma+1|<\kappa.

Here the arrow denotes that series. Lemma 14 gives $g=1$ , so

N=1\quad\text{for every such }N, \qquad \mathrm R_\kappa(G_\kappa)=\bigvee\{1\}=1.

Taking $\kappa=\mu_1<\mu_2$ proves the asserted separation. ◻

References

Preprint · Lean (GitHub)

S. M. Vovsi, On ascendent elements in groups, Dokl. Akad. Nauk SSSR 203, no. 3 (1972), 517–519 (Russian). Math-Net.Ru: dan36783.

S. M. Vovsi, Radical classes of $\mathfrak m$ -hypernilpotent groups, Izv. Vyssh. Uchebn. Zaved. Mat., no. 1 (1975), 21–26 (Russian). Math-Net.Ru: ivm6285.