A nonsoluble residually finite, locally finite group with a full prime-support Sylow basis

Problem

Must a locally finite group with a full Sylow basis be locally soluble?

G\text{ locally finite},\quad G=\langle P_p:p\in\pi(G)\rangle \quad\stackrel{?}{\Longrightarrow}\quad G\text{ locally soluble}.

Here the basis has the prime-support meaning: each $P_p$ is a maximal $p$ -subgroup and every finite join $\langle P_p:p\in J\rangle$ is a $J$ -group. Pairwise permutability is a stronger condition and is not assumed.

The prime-support formulation

For a periodic group $H$ , write

\pi(H)=\{p:p\text{ is prime and }p\mid\operatorname{ord}(h) \text{ for some }h\in H\}.

A $\Pi$ -group is a periodic group $H$ with $\pi(H)\subseteq\Pi$ . A Sylow $p$ -subgroup of an arbitrary group means a subgroup maximal among its $p$ -subgroups.

We use the definition of a Sylow $\Pi$ -basis given by Golberg (2, Definition 1) and reproduced by Starostin and Eidinov (3, p. 277): a family $(P_p)_{p\in\Pi}$ of Sylow $p$ -subgroups such that

\tag{2} \langle P_p:p\in J\rangle\text{ is a }J\text{-group} \qquad(J\subseteq\Pi\text{ finite}).

Fullness means

G=\langle P_p:p\in\Pi\rangle.

A stronger convention also requires

P_pP_q=P_qP_p,\qquad \langle P_p:p\in J\rangle\in\operatorname{Syl}_J(G) \quad(J\subseteq\Pi),

where $\operatorname{Syl}_J(G)$ denotes the maximal $J$ -subgroups; see Amberg and Sysak (1, p. 22). We use only (2) and maximality of the individual $P_p$ .

Theorem 1. There is a countable residually finite, locally finite group $G$ with a full Sylow basis $(P_p)_{p\in\{2,3,5\}}$ in the sense of (2), and an embedding

\operatorname{Sym}(5)\hookrightarrow G.

In particular, $G$ is not locally soluble.

Problem 3.12 of the Kourovka Notebook asks whether every locally finite group with a full Sylow basis is locally soluble. Theorem 1 gives a negative answer with the prime-support definition above. No assertion concerning the stronger pairwise-permutable formulation is needed in the proof.

We write $A\vee B=\langle A,B\rangle$ for subgroup joins. For finite groups $R,F$ , use the regular wreath product

R\wr F=R^F\rtimes F, \qquad {}^xd(z)=d(x^{-1}z).

We identify $F$ with its pure top subgroup and $R^F$ with the base subgroup. For $r\in R$ and $z\in F$ , let $\delta_z(r)\in R^F$ have value $r$ at $z$ and value $1$ elsewhere. Then

\tag{7} x\delta_z(r)x^{-1}=\delta_{xz}(r).

A finite nonsoluble seed

Put

S=\operatorname{Sym}(5),\qquad T=\langle t\rangle\cong C_3, \qquad F_0=S\wr T.

Let $a=(0\,1)$ and $b=(0\,1\,2\,3\,4)$ in $S$ . Define

\tag{9} A_2=\delta_1(\langle a\rangle),\qquad A_3=T, \qquad A_5=\delta_t(\langle b\rangle).

The notation $\delta_z(U)$ denotes the image of a subgroup $U\le S$ under $\delta_z$ .

Lemma 2. The subgroups in (9) satisfy

\begin{align*} \pi(F_0)&\subseteq\{2,3,5\},\tag{10a}\\ \pi(A_p)&\subseteq\{p\},\tag{10b}\\ \pi(A_p\vee A_q)&\subseteq\{p,q\}\quad(p\ne q),\tag{10c}\\ (A_p\vee A_q)\cap(A_p\vee A_r)&=A_p \quad(\{p,q,r\}=\{2,3,5\}).\tag{10d} \end{align*}

Moreover, $A_2\vee A_3\vee A_5$ contains a copy of $S$ .

Proof. The ambient group has order $120^3\cdot3$ , and the three selected subgroups have orders $2,3,5$ . Set $U=\langle a\rangle$ and $V=\langle b\rangle$ . Since $U\cap V=1$ and $t\ne1$ , the pair joins satisfy

\begin{align*} A_2\vee A_3&\le U^T\rtimes T,\\ A_3\vee A_5&\le V^T\rtimes T,\\ A_2\vee A_5&\le\delta_1(U)\delta_t(V). \end{align*}

These bounds prove (10c).

Write $w=(d,s)$ , with $d\in S^T$ and $s\in T$ . The same bounds give

\begin{align*} w\in(A_2\vee A_3)\cap(A_3\vee A_5) &\Rightarrow d(T)\subseteq U\cap V=1 \Rightarrow w\in A_3,\\ w\in(A_2\vee A_3)\cap(A_2\vee A_5) &\Rightarrow s=1,\quad d\in\delta_1(U) \Rightarrow w\in A_2,\\ w\in(A_2\vee A_5)\cap(A_3\vee A_5) &\Rightarrow s=1,\quad d\in\delta_t(V) \Rightarrow w\in A_5. \end{align*}

In the last two lines the other supported coordinate lies in $U\cap V$ . The reverse inclusions give (10d).

Finally,

t^{-1}\delta_t(b)t=\delta_1(b),

so the selected join contains both $\delta_1(a)$ and $\delta_1(b)$ . The adjacent transposition $a$ and the cycle $b$ generate $S$ : conjugating $a$ by powers of $b$ gives the adjacent transpositions along the cycle. Hence the injective coordinate map $\delta_1:S\to F_0$ has image in the selected join. ◻

Call a finite group $F$ together with subgroups $A_2,A_3,A_5$ a configuration if it satisfies (10a)–(10d). We do not require these three subgroups to generate $F$ . An embedding of configurations is an injective homomorphism $\iota:F\to F'$ satisfying

\tag{14} \iota^{-1}(A'_p)=A_p\qquad(p\in\{2,3,5\}).

A coordinate witness of another prime order

Lemma 3. For distinct primes $p,q$ , there is a finite $\{p,q\}$ -group $R$ , two cyclic $p$ -subgroups $C,C'\le R$ , and elements $c\in C$ , $c'\in C'$ such that

\operatorname{ord}(c'c^{-1})=q.

Proof. Take $R=C_q\wr C_p$ . Let $a$ and $b$ generate the top $C_p$ and the coordinate group $C_q$ , respectively. Set

C=\langle a\rangle, \qquad v=\delta_1(b), \qquad C'=vCv^{-1}, \qquad c=a, \qquad c'=vav^{-1}.

By (7),

c'c^{-1}=\delta_1(b)\delta_a(b^{-1}).

The two coordinates are distinct. The displayed element is nonidentity and has $q$ th power $1$ , so its order is $q$ . ◻

Let $L\le F$ and let $C,C'$ be as in Lemma 3. Define the coordinate subgroup

\tag{18} D=D(L;C,C')=\prod_{z\in F}D_z\le R^F, \qquad D_z=\begin{cases}C,&z\in L,\\ C',&z\notin L.\end{cases}

It is a $p$ -group. Left multiplication by $L$ preserves both $L$ and its complement, so $L$ normalizes $D$ .

Lemma 4. For $x\in F\setminus L$ , the subgroup $\langle D,x\rangle\le R\wr F$ contains an element of order $q$ .

Proof. Since $1\in L$ and $x\notin L$ , we have

\delta_1(c),\delta_x(c')\in D, \qquad \delta_x(c)=x\delta_1(c)x^{-1}\in\langle D,x\rangle.

Consequently,

\delta_x(c'c^{-1})\in\langle D,x\rangle.

The coordinate map is injective, so this element has order $q$ . ◻

Preserving the finite configuration

Lemma 5. Let $(F;A_2,A_3,A_5)$ be a configuration, let $p\in\{2,3,5\}$ , and let $x\in F\setminus A_p$ . There is a finite configuration $(F';A'_2,A'_3,A'_5)$ and an embedding $\iota:F\to F'$ satisfying (14) such that

\tag{21} \langle A'_p,\iota(x)\rangle\text{ contains an element of order }q \quad\text{for some }q\ne p.

The embedding has a homomorphic retraction $F'\to F$ .

Proof. Write $\{p,q,r\}=\{2,3,5\}$ , choosing $q,r$ so that

x\notin A_p\vee A_r.

Such a choice exists because

(A_p\vee A_q)\cap(A_p\vee A_r)=A_p.

For this ordering put

A=A_p,\qquad B=A_q,\qquad C_0=A_r, \qquad L=A\vee C_0.

Use Lemma 3 for $p,q$ , and form $D=D(L;C,C')$ from (18). In $F'=R\wr F$ , define

\tag{25} A'=D\rtimes A,\qquad B'=B,\qquad C'_0=C_0,

where the latter two subgroups lie in the pure top copy of $F$ . The group $A'$ is a $p$ -group, and $B',C'_0$ retain their original prime supports. The pair joins satisfy

\begin{align*} A'\vee C'_0&=D\rtimes(A\vee C_0),\tag{26a}\\ A'\vee B'&\le R^F\rtimes(A\vee B),\tag{26b}\\ B'\vee C'_0&=B\vee C_0.\tag{26c} \end{align*}

Thus their prime supports are contained in $\{p,r\}$ , $\{p,q\}$ , and $\{q,r\}$ , respectively. The ambient group $F'$ is a $\{2,3,5\}$ -group, because it is an extension of a direct power of the $\{p,q\}$ -group $R$ by $F$ .

Let $\rho:F'\to F$ be the top projection. The image of each pair join is the corresponding old pair join. If

w\in(A'\vee B')\cap(A'\vee C'_0),

then (26a) puts its base coordinate in $D$ , while

\rho(w)\in(A\vee B)\cap(A\vee C_0)=A.

Hence $w\in A'$ . Next, (26c) shows that an element of either

(A'\vee B')\cap(B'\vee C'_0),\qquad (A'\vee C'_0)\cap(B'\vee C'_0)

lies in the pure top group. Its top coordinate belongs to $B$ or $C_0$ , respectively, by the old intersection identities. This proves all three new intersection identities.

The pure top embedding $\iota$ satisfies (14), since an element of $F$ has trivial base coordinate, and $\rho\iota=1_F$ . Finally, Lemma 4 supplies an element of order $q$ in $\langle D,\iota(x)\rangle\le\langle A',\iota(x)\rangle$ . ◻

The prime-support arguments used here are closed under the required extensions. Explicitly, if $f:H\to K$ is a homomorphism, $K$ and $\ker f$ are $\Pi$ -groups, and $h\in H$ , choose positive $\Pi$ -numbers $m,n$ with

f(h)^m=1,\qquad (h^m)^n=1.

Then $h^{mn}=1$ , and $mn$ is again a $\Pi$ -number. Finite direct products are handled by multiplying the finitely many annihilating exponents.

The ascending sequence and its Sylow subgroups

Lemma 6. Every finite configuration embeds in a finite configuration such that (21) holds simultaneously for every $p\in\{2,3,5\}$ and every $x\in F\setminus A_p$ . The embedding satisfies (14) and has a homomorphic retraction.

Proof. List the finitely many pairs $(p,x)$ . Apply Lemma 5 successively to their images. Condition (14) preserves exclusion from the selected subgroup. An existing witness remains in the required generated subgroup under every later embedding, and its order is unchanged because the embeddings are injective. Compose the embeddings and their retractions. ◻

Starting from Lemma 2, apply Lemma 6 recursively to obtain

F_0\hookrightarrow F_1 \hookrightarrow F_2 \hookrightarrow\cdots.

Identify the stages with their images in the direct limit $H$ . For $p\in\{2,3,5\}$ , put

\tag{32} P_p=\bigcup_{n\ge0}A_{p,n},\qquad G=\langle P_2,P_3,P_5\rangle\le H.

Lemma 7. Each $P_p$ is a Sylow $p$ -subgroup of $H$ , and hence also of $G$ .

Proof. The chain $(A_{p,n})_n$ is increasing, so $P_p$ is a $p$ -group. Let $x\in H\setminus P_p$ . Choose $n$ with $x\in F_n$ . Then $x\notin A_{p,n}$ , and the next batch supplies

w\in\langle A_{p,n+1},x\rangle,\qquad \operatorname{ord}(w)=q\ne p.

Thus $\langle P_p,x\rangle$ is not a $p$ -group. No $p$ -subgroup of $H$ properly contains $P_p$ . ◻

Lemma 8. The groups $H$ and $G$ are residually finite.

Proof. Let $\iota_n:F_n\to F_{n+1}$ and $r_n:F_{n+1}\to F_n$ be the embedding and retraction retained by the $n$ th batch, so

r_n\iota_n=1_{F_n}.

Fix $n$ . For every $m$ , define $\theta_m:F_m\to F_n$ by

\theta_m=\begin{cases} \iota_{n-1}\cdots\iota_m,&m<n,\\ 1_{F_n},&m=n,\\ r_n\cdots r_{m-1},&m>n. \end{cases}

The retraction identities give

\theta_{m+1}\iota_m=\theta_m \qquad(m\ge0).

Thus $\theta_m$ induces a homomorphism $\rho_n:H\to F_n$ whose restriction to $F_n$ is the identity. If $1\ne x\in H$ , choose $n$ with $x\in F_n$ . Then $\rho_n(x)=x\ne1$ , and $F_n$ is finite. Restricting the same homomorphisms to $G$ proves the assertion for $G$ . ◻

Proof of Theorem 1. Residual finiteness follows from Lemma 8. The group $H$ is a countable union of finite groups. Every finitely generated subgroup of $H$ lies in one stage, so $H$ and $G$ are locally finite and countable. For distinct $p,q$ ,

P_p\vee P_q=\bigcup_{n\ge0}(A_{p,n}\vee A_{q,n}),

which is a $\{p,q\}$ -group by (10c). For the other subfamilies, writing $P_J=\langle P_p:p\in J\rangle$ ,

P_\varnothing=1,\qquad P_{\{p\}}=P_p,\qquad P_{\{2,3,5\}}=G,\qquad \pi(G)\subseteq\{2,3,5\}.

These exhaust the subsets of $\{2,3,5\}$ . Lemma 7 now gives a full basis satisfying (2).

The selected join at stage zero embeds in $G$ , because each $A_{p,0}$ lies in $P_p$ . Lemma 2 therefore embeds $\operatorname{Sym}(5)$ in $G$ . This finite subgroup is not soluble, so $G$ is not locally soluble. ◻

References

Preprint · Lean (GitHub)

B. Amberg and Y. Sysak, Products of locally cyclic groups, Arch. Math. 117 (2021), 19–28. doi:10.1007/s00013-021-01593-1.

P. A. Golberg, On a criterion for conjugacy of Sylow $\Pi$ -bases of an arbitrary group, Mat. Sb. (N.S.) 36(78), no. 2 (1955), 335–340 (Russian). Math-Net.Ru: sm5196.

A. I. Starostin and M. I. Eidinov, On Sylow bases of infinite groups, Sibirsk. Mat. Zh. 3, no. 2 (1962), 273–279 (Russian). Math-Net.Ru: smj4814.