Suppose that $G$ is a finite group with no non-trivial normal subgroups of odd order, and $\varphi$ is its $2$-automorphism centralizing a Sylow $2$-subgroup of $G$. Is it true that $\varphi^2$ is an inner automorphism of $G$?

(S. D. Berman). Prove that every automorphism of the centre of the integral group ring of a finite group induces a monomial permutation on the set of the class sums.

Is it true that every central unit of the integral group ring of a finite group is a product of a central element of the group and a symmetric central unit? (A unit is symmetric if it is fixed by the canonical antiinvolution that transposes the coefficients at the mutually inverse elements.)

a) Is it true that there exists a nilpotent group $G$ for which the lattice $\mathscr{L}(G)$ of all group topologies is not modular? (It is known that for abelian groups the lattice $\mathscr{L}(G)$ is modular and that there are groups for which this lattice is not modular: V. I. Arnautov, A. G. Topale, Izv. Akad. Nauk Moldova Mat., 1997, no. 1, 84–92 (Russian).)

b) Is it true that for every countable nilpotent non-abelian group $G$ the lattice $\mathscr{L}(G)$ of all group topologies is not modular?

Let $G$ be an infinite group admitting non-discrete Hausdorff group topologies, and $\mathscr{L}(G)$ the lattice of all group topologies on $G$.

a) Is it true that for any natural number $k$ there exists a non-refinable chain $\tau_0 < \tau_1 < \dots < \tau_k$ of length $k$ of Hausdorff topologies in $\mathscr{L}(G)$? (For countable nilpotent groups this is true (A. G. Topale, Deposited in VINITI, 25.12.98, no. 3849–V 98 (Russian)).)

b) Let $k, m, n$ be natural numbers and let $G$ be a nilpotent group of class $k$. Suppose that $\tau_0 < \tau_1 < \dots < \tau_m$ and $\tau'_0 < \tau'_1 < \dots < \tau'_n$ are non-refinable chains of Hausdorff topologies in $\mathscr{L}(G)$ such that $\tau_0 = \tau'_0$ and $\tau_m = \tau'_n$. Is it true that $m \leqslant n \cdot k$ and this inequality is best-possible? (This is true if $k = 1$, since for $G$ abelian the lattice $\mathscr{L}(G)$ is modular.)

c) Is it true that there exists a countable $G$ such that in the lattice $\mathscr{L}(G)$ there are a finite non-refinable chain $\tau_0 < \tau_1 < \dots < \tau_k$ of Hausdorff topologies and an infinite chain $\{\tau'_\gamma \mid \gamma \in \Gamma\}$ of topologies such that $\tau_0 < \tau'_\gamma < \tau_k$ for any $\gamma \in \Gamma$?

d) Let $G$ be an abelian group, $k$ a natural number. Let $\mathcal{A}_k$ be the set of all those Hausdorff group topologies on $G$ that, for every topology $\tau \in \mathcal{A}_k$, any non-refinable chain of topologies starting from $\tau$ and terminating at the discrete topology has length $k$. Is it true that $\mathcal{A}_k \cap \{\tau'_\gamma \mid \gamma \in \Gamma\} \neq \varnothing$ for any infinite non-refinable chain $\{\tau'_\gamma \mid \gamma \in \Gamma\}$ of Hausdorff topologies containing the discrete topology? (This is true for $k = 1$.)

A group $\Gamma$ is said to have Property $P_{\text{nai}}$ if, for any finite subset $F$ of $\Gamma \setminus \{1\}$, there exists an element $y_0 \in \Gamma$ of infinite order such that, for each $x \in F$, the canonical epimorphism from the free product $\langle x \rangle * \langle y_0 \rangle$ onto the subgroup $\langle x, y_0 \rangle$ of $\Gamma$ generated by $x$ and $y_0$ is an isomorphism. For $n \in \{2, 3, \dots\}$, does $PSL_n(\mathbb{Z})$ have Property $P_{\text{nai}}$? More generally, if $\Gamma$ is a lattice in a connected real Lie group $G$ which is simple and with centre reduced to $\{1\}$, does $\Gamma$ have Property $P_{\text{nai}}$?

Let $\Gamma_g$ be the fundamental group of a closed surface of genus $g \geqslant 2$. For each finite system $S$ of generators of $\Gamma_g$, let $\beta_S(n)$ denote the number of elements of $\Gamma_g$ which can be written as products of at most $n$ elements of $S \cup S^{-1}$, and let $\omega(\Gamma_g, S) = \limsup_{n \to \infty} \sqrt[n]{\beta_S(n)}$ be the growth rate of the sequence $(\beta_S(n))_{n\geqslant 0}$. Compute the infimum $\omega(\Gamma_g)$ of the $\omega(\Gamma_g, S)$ over all finite sets of generators of $\Gamma_g$.

It is easy to see that, for a free group $F_k$ of rank $k \geqslant 2$, the corresponding infimum is $\omega(F_k) = 2k - 1$ (M. Gromov, Structures métriques pour les variétés riemanniennes, Cedic/F. Nathan, Paris, 1981, Ex. 5.13). As any generating set of $\Gamma_g$ contains a subset of $2g - 1$ elements generating a subgroup of infinite index in $\Gamma_g$ with abelianization $\mathbb{Z}^{2g-1}$, hence a subgroup which is free of rank $2g - 1$, it follows that $\omega(\Gamma_g) \geqslant 4g - 3$.

Let $G$ denote the group of germs at $+\infty$ of orientation-preserving homeomorphisms of the real line $\mathbb{R}$. Let $\alpha \in G$ be the germ of $x \mapsto x + 1$. What are the germs $\beta \in G$ for which the subgroup $\langle \alpha, \beta \rangle$ of $G$ generated by $\alpha$ and $\beta$ is free of rank 2?

If $\beta$ is the germ of $x \mapsto x^k$ for an odd integer $k \geqslant 3$, it is known that $\langle \alpha, \beta \rangle$ is free of rank 2. The proofs of this rely on Galois theory (for $k$ an odd prime: S. White, J. Algebra, 118 (1988), 408–422; for any odd $k \geqslant 3$: S. A. Adeleke, A. M. W. Glass, L. Morley, J. London Math. Soc., 43 (1991), 255–268, and for any odd $k \neq \pm 1$ and any even $k > 0$ in several papers by S. D. Cohen and A. M. W. Glass).

(Well-known problem). Let $W^*(F_k)$ denote the von Neumann algebra of the free group of rank $k \in \{2, 3, \dots, \boldsymbol{\aleph}_0\}$. Is $W^*(F_k)$ isomorphic to $W^*(F_l)$ for $k \neq l$?

For a group $G$, recall that $W^*(G)$ is an appropriate completion of the group algebra $\mathbb{C}G$ (see e. g. S. Sakai, $C^*$-algebras and $W^*$-algebras, Springer, 1971, in particular Problem 4.4.44). It is known that either $W^*(F_k) \cong W^*(F_l)$ for all $k, l \in \{2, 3, \dots, \boldsymbol{\aleph}_0\}$, or that the $W^*(F_k)$ are pairwise non-isomorphic (F. Radulescu, Invent. Math., 115 (1994), 347–389, Corollary 4.7).

a) (Well-known problem). It is known that any recursively presented group embeds in a finitely presented group (G. Higman, Proc. Royal Soc. London Ser. A, 262 (1961), 455–475). Find an explicit and “natural” finitely presented group $\Gamma$ and an embedding of the additive group of the rationals $\mathbb{Q}$ in $\Gamma$.
b) Find an explicit embedding of $\mathbb{Q}$ in a finitely generated group; such a group exists by Theorem IV in (G. Higman, B. H. Neumann, H. Neumann, J. London Math. Soc., 24 (1949), 247–254).
c) Find an explicit and “natural” finitely presented group $\Gamma_n$ and an embedding of $GL_n(\mathbb{Q})$ in $\Gamma_n$.

Another phrasing of the same problems is: find a simplicial complex $X$ which covers a finite complex such that the fundamental group of $X$ is $\mathbb{Q}$ or, respectively, $GL_n(\mathbb{Q})$.

(Yu. I. Merzlyakov). It is a well-known fact that for the ring $R = \mathbb{Q}[x, y]$ the elementary group $E_2(R)$ is distinct from $SL_2(R)$. Find a minimal subset $A \subseteq SL_2(R)$ such that $\langle E_2(R), A \rangle = SL_2(R)$.

(Yu. I. Merzlyakov, J. S. Birman). Is it true that all braid groups $B_n$, $n \geqslant 3$, are conjugacy separable?

a) By definition the commutator length of an element $z$ of the derived subgroup of a group $G$ is the least possible number of commutators from $G$ whose product is equal to $z$. Does there exist a simple group on which the commutator length is not bounded?
b) Does there exist a finitely presented simple group on which the commutator length is not bounded?

(C. C. Edmunds, G. Rosenberger). We call a pair of natural numbers $(k, m)$ admissible if in the derived subgroup $F'_2$ of a free group $F_2$ there is an element $w$ such that the commutator length of the element $w^m$ is equal to $k$. Find all admissible pairs.

For the automorphism group $A_n = \operatorname{Aut} F_n$ of a free group $F_n$ of rank $n \geqslant 3$ find the supremum $k_n$ of the commutator lengths of the elements of $A'_n$.

It is easy to show that $k_2 = \infty$. On the other hand, the commutator length of any element of the derived subgroup of $\varinjlim A_n$ is at most 2 (R. K. Dennis, L. N. Vaserstein, K-Theory, 2, N 6 (1989), 761–767).

Following Yu. I. Merzlyakov we define the width of a verbal subgroup $V(G)$ of a group $G$ with respect to the set of words $V$ as the smallest $m \in \mathbb{N} \cup \{\infty\}$ such that every element of $V(G)$ can be written as a product of $\leqslant m$ values of words from $V \cup V^{-1}$. It is known that the width of any verbal subgroup of a finitely generated group of polynomial growth is finite. Is this statement true for finitely generated groups of intermediate growth?

Let $\text{IMA}(G)$ denote the subgroup of the automorphism group $\operatorname{Aut} G$ consisting of all automorphisms that act trivially on the second derived quotient $G/G''$. Find generators and defining relations for $\text{IMA}(F_n)$, where $F_n$ is a free group of rank $n \geqslant 3$.

We say that a family of groups $\mathscr{D}$ discriminates a group $G$ if for any finite subset $\{a_1, \dots, a_n\} \subseteq G \setminus \{1\}$ there exists a group $D \in \mathscr{D}$ and a homomorphism $\varphi : G \to D$ such that $a_j\varphi \neq 1$ for all $j = 1, \dots, n$. Is every finitely generated group acting freely on some $\Lambda$-tree discriminated by torsion-free hyperbolic groups?

We say that a group $G$ has the Noetherian Equation Property if every system of equations over $G$ in finitely many variables is equivalent to some finite part of it. Does an arbitrary hyperbolic group have the Noetherian Equation Property?

Does a free product of two groups have the Noetherian Equation Property (see 14.19) if this property is enjoyed by the factors?

Does a free pro-$p$-group have the Noetherian Equation Property (see 14.19)?

Prove that any irreducible system of equations $S(x_1, \dots, x_n) = 1$ with coefficients in a torsion-free linear group $G$ is equivalent over $G$ to a finite system $T(x_1, \dots, x_n) = 1$ satisfying an analogue of Hilbert’s Nullstellensatz, i. e. $\operatorname{Rad}_G(T) = \sqrt{T}$. This is true if $G$ is a free group (O. Kharlampovich, A. Myasnikov, J. Algebra, 200 (1998), 472–570).

Here both $S$ and $T$ are regarded as subsets of $G[X] = G * F(X)$, a free product of $G$ and a free group on $X = \{x_1, \dots, x_n\}$. By definition, $\operatorname{Rad}_G(T) = \{w(x_1, \dots, x_n) \in G[X] \mid w(g_1, \dots, g_n) = 1$ for any solution $g_1, \dots, g_n \in G$ of the system $T(X) = 1\}$, and $\sqrt{T}$ is the minimal normal isolated subgroup of $G[X]$ containing $T$.

Let $F_n$ be a free group with basis $\{x_1, \dots, x_n\}$, and let $\lvert \cdot \rvert$ be the length function with respect to this basis. For $\alpha \in \operatorname{Aut} F_n$ we put $\lVert \alpha \rVert = \max\{\lvert \alpha(x_1) \rvert, \dots, \lvert \alpha(x_n) \rvert\}$. Is it true that there is a recursive function $f : \mathbb{N} \to \mathbb{N}$ with the following property: for any $\alpha \in \operatorname{Aut} F_n$ there is a basis $\{y_1, \dots, y_k\}$ of $\operatorname{Fix}(\alpha) = \{x \mid \alpha(x) = x\}$ such that $\lvert y_i \rvert \leqslant f(\lVert \alpha \rVert)$ for all $i = 1, \dots, k$?

Let $\operatorname{Aut} F_n$ be the automorphism group of a free group of rank $n$ with norm $\lVert \cdot \rVert$ as in 14.23. Does there exist a recursive function $f : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ with the following property: for any two conjugate elements $\alpha, \beta \in \operatorname{Aut} F_n$ there is an element $\gamma \in \operatorname{Aut} F_n$ such that $\gamma^{-1}\alpha\gamma = \beta$ and $\lVert \gamma \rVert \leqslant f(\lVert \alpha \rVert, \lVert \beta \rVert)$?

Let $qG$ denote the quasivariety generated by a group $G$. Is it true that there exists a finitely generated group $G$ such that the set of proper maximal subquasivarieties in $qG$ is infinite?

A quasivariety $\mathfrak{M}$ is closed under direct $\mathbb{Z}$-wreath products if the direct wreath product $G \wr \mathbb{Z}$ belongs to $\mathfrak{M}$ for every $G \in \mathfrak{M}$ (here $\mathbb{Z}$ is an infinite cyclic group). Is the quasivariety generated by the class of all nilpotent torsion-free groups closed under direct $\mathbb{Z}$-wreath products?

Let $\Gamma$ be a group generated by a finite set $S$. Assume that there exists a nested sequence $F_1 \subset F_2 \subset \cdots$ of finite subsets of $\Gamma$ such that
$\qquad$ (i) $F_k \neq F_{k+1}$ for all $k \geqslant 1$,
$\qquad$ (ii) $\Gamma = \bigcup_{k \geqslant 1} F_k$,
$\qquad$ (iii) $\lim_{k \to \infty} |\partial F_k|/ |F_k| = 0$, where, by definition, $\partial F_k = \{ \gamma \in \Gamma \setminus F_k \mid$ there exists $s \in S$ such that $\gamma s \in F_k \}$, and
$\qquad$ (iv) there exist constants $c \geqslant 0$, $d \geqslant 1$ such that $|F_k| \leqslant ck^d$ for all $k \geqslant 1$.
Does it follow that $\Gamma$ has polynomial growth?

Let $\mathfrak{F}$ be a soluble Fitting formation of finite groups with Kegel’s property, that is, $\mathfrak{F}$ contains every finite group of the form $G = AB = BC = CA$ if $A, B, C$ are in $\mathfrak{F}$. Is $\mathfrak{F}$ a saturated formation?

Is there a soluble Fitting class of finite groups $\mathfrak{F}$ such that $\mathfrak{F}$ is not a formation and $A_{\mathfrak{F}} \cap B_{\mathfrak{F}} \subseteq G_{\mathfrak{F}}$ for every finite soluble group of the form $G = AB$?

Let $\operatorname{lFit} \mathfrak{X}$ be the local Fitting class generated by a set of groups $\mathfrak{X}$ and let $\Psi(G)$ be the smallest normal subgroup of a finite group $G$ such that $\operatorname{lFit}(\Psi(G) \cap M) = \operatorname{lFit} M$ for every $M \triangleleft\triangleleft\ G$ (K. Doerk, P. Hauck, Arch. Math., 35, no. 3 (1980), 218–227). We say that a Fitting class $\mathfrak{F}$ is saturated if $\Psi(G) \in \mathfrak{F}$ implies that $G \in \mathfrak{F}$. Is it true that every non-empty soluble saturated Fitting class is local?

Is the lattice of Fitting subclasses of the Fitting class generated by a finite soluble group finite?

Extending the classical definition of formations, let us define a formation of (not necessarily finite) groups as a nonempty class of groups closed under taking homomorphic images and subdirect products with finitely many factors. Must every first-order axiomatizable formation of groups be a variety?

Does there exist a finitely presented pro-$p$-group ($p$ being a prime) which contains an isomorphic copy of every countably based pro-$p$-group?

By definition, a locally compact group has the Kazhdan T-property if the trivial representation is an isolated point in the natural topological space of unitary representations of the group. Does there exist a profinite group with two dense discrete subgroups one of which is amenable, and the other has the Kazhdan T-property?

See (A. Lubotzky, Discrete groups, expanding graphs and invariant measures (Progress in Mathematics, Boston, Mass., 125), Birkhäuser, Basel, 1994) for further motivation.

Is every finitely presented group of prime exponent finite?

A group $G$ is called a $T$-group if every subnormal subgroup of $G$ is normal, while $G$ is said to be a $\overline{T}$-group if all of its subgroups are $T$-groups. Is it true that every non-periodic locally graded $\overline{T}$-group must be abelian?

Let $G(n)$ be one of the classical groups (special, orthogonal, or symplectic) of $(n \times n)$-matrices over an infinite field $K$ of non-zero characteristic, and $M(n)$ the space of all $(n \times n)$-matrices over $K$. The group $G(n)$ acts diagonally by conjugation on the space $M(n)^m = M(n) \oplus \dots \oplus M(n)$ ($m$ copies). Find generators of the algebra of invariants $K[M(n)^m]^{G(n)}$.

In characteristic 0 they were found in (C. Procesi, Adv. Math., 19 (1976), 306–381).

For every pro-$p$-group $G$ of $(2 \times 2)$-matrices for $p \neq 2$ an analogue of the Tits Alternative holds: either $G$ is soluble, or the variety of pro-$p$-groups generated by $G$ contains the group $\overline{\left\langle \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} \right\rangle} \leqslant SL_2(\mathbb{F}_p[[t]])$ (A. N. Zubkov, Algebra and Logic, 29, no. 4 (1990), 287–301). Is the same result true for matrices of size $\geqslant 3$ for $p \neq 2$?

Let $F(\mathfrak{V})$ be a free group of some variety of pro-$p$-groups $\mathfrak{V}$. Is there a uniform bound for the exponents of periodic elements in $F(\mathfrak{V})$?

Is a free pro-$p$-group representable as an abstract group by matrices over a commutative-associative ring with 1?

A group $G$ is said to be para-free if all factors $\gamma_i(G)/\gamma_{i+1}(G)$ of its lower central series are isomorphic to the corresponding lower central factors of some free group, and $\bigcap_{i=1}^\infty \gamma_i(G) = 1$. Is an arbitrary para-free group representable by matrices over a commutative-associative ring with 1?

Is a free pro-$p$-group representable by matrices over an associative-commutative profinite ring with 1?

A negative answer is equivalent to the fact that every linear pro-$p$-group satisfies a non-trivial pro-$p$-identity. This is known to be true in dimension 2 for $p \neq 2$ (A. N. Zubkov, Siberian Math. J., 28, no. 5 (1987), 742–747). It is also proved that a 2-dimensional linear pro-2-group in characteristic 2 satisfies a non-trivial pro-2-identity (D. E.-C. Ben-Ezra, E. Zelmanov, Trans. Amer. Math. Soc., 374, no. 6 (2021), 4093–4128).

Suppose that a finite group $G$ has the form $G = AB$, where $A$ and $B$ are nilpotent subgroups of classes $\alpha$ and $\beta$ respectively; then $G$ is soluble (O. H. Kegel, H. Wielandt, 1961). Although the derived length $\operatorname{dl}(G)$ of $G$ need not be bounded by $\alpha + \beta$ (see 5.17), can one bound $\operatorname{dl}(G)$ by a (linear) function of $\alpha$ and $\beta$?

Let $k(X)$ denote the number of conjugacy classes of a finite group $X$. Suppose that a finite group $G = AB$ is a product of two subgroups $A, B$ of coprime orders. Is it true that $k(AB) \leqslant k(A)k(B)$?

Note that one cannot drop the coprimeness condition and the answer is positive if one of the subgroups is normal, see 11.43.

Does there exist a (non-abelian simple) linearly right-orderable group all of whose proper subgroups are cyclic?

A finite group $G$ is said to be almost simple if $T \leqslant G \leqslant \operatorname{Aut}(T)$ for some nonabelian simple group $T$. By definition, a finite linear space consists of a set $V$ of points, together with a collection of $k$-element subsets of $V$, called lines ($k \geqslant 3$), such that every pair of points is contained in exactly one line. Classify the finite linear spaces which admit a line-transitive almost simple subgroup $G$ of automorphisms which acts transitively on points.

A. Camina, P. Neumann and C. E. Praeger have solved this problem in the case where $T$ is an alternating group. In (A. Camina, C. E. Praeger, Aequat. Math., 61 (2001), 221–232) it is shown that a line-transitive group of automorphisms of a finite linear space which is point-quasiprimitive (i. e. all of whose non-trivial normal subgroups are point-transitive) is almost simple or affine.

Is the lattice of all soluble Fitting classes of finite groups modular?

If an equation over a free group $F$ has no solution in $F$, is there a finite quotient of $F$ in which the equation has no solution?

Is $SL_3(\mathbb{Z})$ a factor group of the modular group $PSL_2(\mathbb{Z})$? Since the latter is isomorphic to the free product of two cyclic groups of orders 2 and 3, the question asks if $SL_3(\mathbb{Z})$ can be generated by two elements of orders 2 and 3.

(Z. I. Borevich). A subgroup $A$ of a group $G$ is said to be paranormal (respectively, polynormal) if $A^x \leqslant \langle A^u \mid u \in \langle A, A^x \rangle \rangle$ (respectively, $A^x \leqslant \bigcup_{u \in A^{\langle x \rangle}} A^u$) for any $x \in G$. Is every polynormal subgroup of a finite group paranormal?

(Well-known problems). Is there a finite basis for the identities of any
$\qquad$ a) abelian-by-nilpotent group?
$\qquad$ b) abelian-by-finite group?
$\qquad$ c) abelian-by-(finite nilpotent) group?

It is known that if a finitely generated group is residually torsion-free nilpotent, then the group is residually finite $p$-group, for every prime $p$. Is the converse true?

Conjecture: Let $G$ be a profinite group such that the set of solutions of the equation $x^n = 1$ has positive Haar measure. Then $G$ has an open subgroup $H$ and an element $t$ such that all elements of the coset $tH$ have order dividing $n$.

This is true in the case $n = 2$. It would be interesting to see whether similar results hold for profinite groups in which the set of solutions of some equation has positive measure.

Let $k(G)$ denote the number of conjugacy classes of a finite group $G$. Is it true that $k(G) \leqslant \lvert N \rvert$ for some nilpotent subgroup $N$ of $G$? This is true if $G$ is simple; besides, always $k(G) \leqslant \lvert S \rvert$ for some soluble subgroup $S \leqslant G$.

Let $J = N(\mathbb{Z}/p\mathbb{Z})$ be the Nottingham group.
$\qquad$ a) Prove that $J$ is finitely presented for $p > 2$.
$\qquad$ b) Prove that $J$ is finitely presented for $p = 2$.

Prove that if $G$ is an infinite pro-$p$-group with $G/\gamma_{2p+1}(G)$ isomorphic to $J/\gamma_{2p+1}(J)$ then $G$ is isomorphic to $J$, where $J$ is the Nottingham group.

Describe the hereditarily just infinite pro-$p$-groups of finite width.

Definitions: A pro-$p$-group $G$ has finite width $p^d$ if $\lvert \gamma_i(G)/\gamma_{i+1}(G) \rvert \leqslant p^d$ for all $i$, and $G$ is hereditarily just infinite if $G$ is infinite and each of its open subgroups has no closed normal subgroups of infinite index.

Suppose that $A$ is a periodic group of regular automorphisms of an abelian group.
$\qquad$ a) Is $A$ cyclic if $A$ has prime exponent?
$\qquad$ b) Is $A$ finite if $A$ is generated by elements of order 3?

Suppose that $G$ is a triply transitive group in which a stabilizer of two points contains no involutions, and a stabilizer of three points is trivial. Is it true that $G$ is similar to $PGL_2(P)$ in its natural action on the projective line $P \cup \{\infty\}$, for some field $P$ of characteristic 2? This is true under the condition that the stabilizer of two points is periodic.

Suppose that $H$ is a non-trivial normal subgroup of a finite group $G$ such that the factor-group $G/H$ is isomorphic to one of the simple groups $L_n(q)$, $n \geqslant 3$. Is it true that $G$ has an element whose order is distinct from the order of any element in $G/H$?

Determine all pairs $(\mathscr{S}, G)$, where $\mathscr{S}$ is a semipartial geometry and $G$ is an almost simple flag-transitive group of automorphisms of $\mathscr{S}$. A system of points and lines $(P, B)$ is a semipartial geometry with parameters $(\alpha, s, t, \mu)$ if every point belongs to exactly $t + 1$ lines (two different points belong to at most one line); every line contains exactly $s + 1$ points; for any anti-flag $(a, l) \in (P, B)$ the number of lines containing $a$ and intersecting $l$ is either $0$ or $\alpha$; and for any non-collinear points $a, b$ there are exactly $\mu$ points collinear with $a$ and with $b$.

Suppose that $H$ is a non-soluble normal subgroup of a finite group $G$. Does there always exist a maximal soluble subgroup $S$ of $H$ such that $G = H \cdot N_G(S)$?

What are the composition factors of non-soluble finite groups all of whose normalizers of Sylow subgroups are 2-nilpotent, in particular, supersoluble?

(M. F. Newman). Classify the finite 5-groups of maximal class; computer calculations suggest some conjectures (M. F. Newman, in: Groups–Canberra, 1989 (Lecture Notes Notes Math., 1456), Springer, Berlin, 1990, 49–62).

(Well-known problem). For a finite group $G$ let $\rho(G)$ denote the set of prime numbers dividing the order of some conjugacy class, and $\sigma(G)$ the maximum number of primes dividing the order of some conjugacy class. Is it true that $\lvert\rho(G)\rvert \leqslant 3\sigma(G)$?

A possible linear bound for $\lvert\rho(G)\rvert$ in terms of $\sigma(G)$ cannot be better than $3\sigma(G)$, since there is a family of groups $\{G_n\}$ such that $\lim_{n \to \infty} \lvert\rho(G_n)\rvert/\sigma(G_n) = 3$ (C. Casolo, S. Dolfi, Rend. Sem. Mat. Univ. Padova, 96 (1996), 121–130).

(Well-known problem). Let $G$ be a finite soluble group, $\pi(G)$ the set of primes dividing the order of $G$, and $\nu(G)$ the maximum number of primes dividing the order of some element. Does there exist a linear bound for $|\pi(G)|$ in terms of $\nu(G)$?

Suppose that $a$ is a non-trivial element of a finite group $G$ such that $\lvert C_G(a) \rvert \geqslant \lvert C_G(x) \rvert$ for every non-trivial element $x \in G$, and let $H$ be a nilpotent subgroup of $G$ which is normalized by $C_G(a)$. Is it true that $H \leqslant C_G(a)$? This is true if $H$ is abelian, or if $H$ is a $p'$-group for some prime number $p \in \pi(Z(C_G(a)))$.

(Well-known problem). Suppose that $F$ is an automorphism of order 2 of the polynomial ring $R_n = \mathbb{C}[x_1, \dots, x_n]$, $n > 2$. Does there exist an automorphism $G$ of $R_n$ such that $G^{-1}FG$ is a linear automorphism?

This is true for $n = 2$.

For every finite simple group find the minimum of the number of generating involutions satisfying an additional condition, in each of the following cases.
$\qquad$ a) The product of the generating involutions equals 1.
$\qquad$ b) (Malle–Saxl–Weigel). All generating involutions are conjugate.
$\qquad$ c) (Malle–Saxl–Weigel). The conditions a) and b) are simultaneously satisfied.
$\qquad$ d) All generating involutions are conjugate and two of them commute.

A group $G$ is called $n$-Engel if it satisfies the identity $[x, y, \dots, y] = 1$, where $y$ is taken $n$ times. Are there non-nilpotent finitely generated $n$-Engel groups?

Consider a free group $F$ of finite rank and an arbitrary group $G$. Define the $G$-closure $\text{cl}_G(T)$ of any subset $T \subseteq F$ as the intersection of the kernels of all those homomorphisms $\mu : F \to G$ of $F$ into $G$ that vanish on $T$: $$\text{cl}_G(T) = \bigcap \{ \text{Ker } \mu \mid \mu : F \to G; T \subseteq \text{Ker } \mu \}.$$ Groups $G$ and $H$ are called geometrically equivalent if for every free group $F$ and every subset $T \subset F$ the $G$- and $H$-closures of $T$ coincide: $\text{cl}_G(T) = \text{cl}_H(T)$. It is easy to see that if $G$ and $H$ are geometrically equivalent then they have the same quasiidentities. Is it true that if two groups have the same quasiidentities then they are geometrically equivalent? This is true for nilpotent groups.

Let $X$ be a regular algebraic variety over a field of arbitrary characteristic, and $G$ a finite cyclic group of automorphisms of $X$. Suppose that the fixed point variety $X^G$ of $G$ is a regular hypersurface of $X$ (of codimension 1). Is the quotient variety $X/G$ regular?

Conjecture: There is a function $f$ on the natural numbers such that, if $\Gamma$ is a finite, vertex-transitive, locally-quasiprimitive graph of valency $v$, then the number of automorphisms fixing a given vertex is at most $f(v)$. (By definition, a vertex-transitive graph $\Gamma$ is locally-quasiprimitive if the stabilizer in $\operatorname{Aut}(\Gamma)$ of a vertex $\alpha$ is quasiprimitive (see 14.46) in its action on the set of vertices adjacent to $\alpha$.)

To prove the conjecture above one need only consider the case where $\operatorname{Aut}(\Gamma)$ has the property that every non-trivial normal subgroup has at most two orbits on vertices (C. E. Praeger, Ars Combin., 19 A (1985), 149–163). The analogous conjecture for finite, vertex-transitive, locally-primitive graphs was made by R. Weiss in 1978 and is still open. For non-bipartite graphs, there is a “reduction” of Weiss’ conjecture to the case where the automorphism group is almost simple (see 14.46 for definition) (M. Conder, C. H. Li, C. E. Praeger, Proc. Edinburgh Math. Soc. (2), 43, no. 1 (2000), 129–138).

Let $k(G)$ denote the number of conjugacy classes of a finite group $G$. Is it true that $k(G) \leqslant k(P_1) \cdots k(P_s)$, where $P_1, \dots, P_s$ are Sylow subgroups of $G$ such that $\lvert G \rvert = \lvert P_1 \rvert \cdots \lvert P_s \rvert$?

Suppose that $\mathfrak{G} = \{G_1, G_2, \dots\}$ is a family of finite 2-generated groups which generates the variety of all groups. Is it true that a free group of rank 2 is residually in $\mathfrak{G}$?

Does there exist an absolute constant $c$ such that any finite $p$-group $P$ has an abelian section $A$ satisfying $\lvert A \rvert^c > \lvert P \rvert$?

By a result of A. Yu. Olshanskii (Math. Notes, 23 (1978), 183–185) we cannot require $A$ to be a subgroup. By a result of J. G. Thompson (J. Algebra, 13 (1969), 149–151) the existence of such a section $A$ would imply the existence of a class 2 subgroup $H$ of $P$ with $\lvert H \rvert^c > \lvert P \rvert$.

Let $p$ be a prime number and $X$ a finite set of powers of $p$ containing 1. Is it true that $X$ is the set of all lengths of the conjugacy classes of some finite $p$-group?

Suppose that $\mathfrak{H} \subseteq \mathfrak{F}_1 \subseteq \mathfrak{M}$ and $\mathfrak{H} \subseteq \mathfrak{F}_2 \subseteq \mathfrak{M}$ where $\mathfrak{H}$ and $\mathfrak{M}$ are local formations of finite groups and $\mathfrak{F}_1$ is a complement for $\mathfrak{F}_2$ in the lattice of all formations between $\mathfrak{H}$ and $\mathfrak{M}$. Is it true that $\mathfrak{F}_1$ and $\mathfrak{F}_2$ are local formations? This is true in the case $\mathfrak{H} = (1)$.

Suppose that $\mathfrak{F} = \mathfrak{MH} = \operatorname{lFit}(G)$ is a soluble one-generated local Fitting class of finite groups where $\mathfrak{H}$ and $\mathfrak{M}$ are Fitting classes and $\mathfrak{M} \neq \mathfrak{F}$. Is $\mathfrak{H}$ a local Fitting class?

Is the lattice of all totally local formations of finite groups modular? The definition of a totally local formation see in (L. A. Shemetkov, A. N. Skiba, Formatsii algebraicheskikh system, Moscow, Nauka, 1989 (Russian)).

Prove that the formation generated by a finite group has only finitely many $S_n$-closed subformations.

(Well-known problem). Describe the finite simple groups in which every element is a product of two involutions.

We say that an infinite simple group $G$ is a monster of the third kind if for every non-trivial elements $a, b$, of which at least one is not an involution, there are infinitely many elements $g \in G$ such that $\langle a, b^g \rangle = G$. (Compare with V. P. Shunkov’s definitions in Archive, 6.63, 6.64.) Is it true that every simple quasi-Chernikov group is a monster of the third kind? This is true for quasi-finite groups (A. I. Sozutov, Algebra and Logic, 36, no. 5 (1997), 336–348). (We say that a non-$\sigma$ group is quasi-$\sigma$ if all of its proper subgroups have the property $\sigma$.)

An element $g$ of a (relatively) free group $F_r(\mathfrak{M})$ of rank $r$ of a variety $\mathfrak{M}$ is said to be primitive if it can be included in a basis of $F_r(\mathfrak{M})$.
$\qquad$ a) Do there exist a group $F_r(\mathfrak{M})$ and a non-primitive element $h \in F_r(\mathfrak{M})$ such that for some monomorphism $\alpha$ of $F_r(\mathfrak{M})$ the element $\alpha(h)$ is primitive?
$\qquad$ b) Do there exist a group $F_r(\mathfrak{M})$ and a non-primitive element $h \in F_r(\mathfrak{M})$ such that for some $n > r$ the element $h$ is primitive in $F_n(\mathfrak{M})$?

Suppose that an endomorphism $\varphi$ of a free metabelian group of rank $r$ takes every primitive element to a primitive one. Is $\varphi$ necessarily an automorphism? This is true for $r \leqslant 2$.

Does there exist an infinite locally nilpotent $p$-group that is equal to its commutator subgroup and in which every proper subgroup is nilpotent?

We say that an element $u$ of a group $G$ is a test element if for any endomorphism $\varphi$ of $G$ the equality $\varphi(u) = u$ implies that $\varphi$ is an automorphism of $G$. Does a free soluble group of rank 2 and derived length $d > 2$ have any test elements?

(E. A. O’Brien, A. Shalev). Let $P$ be a finite $p$-group of order $p^m$ and let $m = 2n + e$ with $e = 0$ or 1. By a theorem of P. Hall the number of conjugacy classes of $P$ has the form $n(p^2 - 1) + p^e + a(p^2 - 1)(p - 1)$ for some integer $a \geqslant 0$, which is called the abundance of $P$.
$\qquad$ a) Is there a bound for the coclass of $P$ which depends only on $a$? Note that $a = 0$ implies coclass 1 and that all known examples with $a = 1$ have coclass $\leqslant 3$. (The group $P$ has coclass $r$ if $\lvert P \rvert = p^{c+r}$ where $c$ is the nilpotency class of $P$.)
$\qquad$ b) Is there an element $s \in P$ such that $\lvert C_P(s) \rvert \leqslant p^{f(a)}$ for some $f(a)$ depending only on $a$? We already know that we can take $f(0) = 2$ and it seems that $f(1) = 3$.

Note that A. Jaikin-Zapirain (J. Group Theory, 3, no. 3 (2000), 225–231) has proved that $\lvert P \rvert \leqslant p^{f(p,a)}$ for some function $f$ of $p$ and $a$ only.

Let $P$ be a finite $p$-group of abundance $a$ and nilpotency class $c$. Does there exist an integer $t = t(a)$ such that $\gamma_i(G) = \zeta_{c-i+1}(G)$ for $i \geqslant t$? This holds for $a = 0$ with $t = 1$, since $P$ has maximal class; it can be proved that for $a = 1$ one can take $t = 3$.

Let $p$ be a fixed prime. Do there exist finite $p$-groups of abundance $a$ for any $a \geqslant 0$?

(I. D. Macdonald). Every finite $p$-group has at least $p - 1$ conjugacy classes of maximum size. I. D. Macdonald (Proc. Edinburgh Math. Soc., 26 (1983), 233–239) constructed groups of order $2^n$ for any $n \geqslant 7$ with just one conjugacy class of maximum size. Are there any examples with exactly $p - 1$ conjugacy classes of maximum size for odd $p$?

Let $N(\mathbb{Z}/p\mathbb{Z})$ be the group defined in 12.24 (the so-called “Nottingham group”, or the “Wild group”). Find relations of $N(\mathbb{Z}/p\mathbb{Z})$ as a pro-$p$-group (it has two generators, e. g., $x + x^2$ and $x/(1 - x)$).

For each positive integer $r$ find the $p$-cohomological dimension $\operatorname{cd}_p(H_r)$ where $H_r$ is the closed subgroup of $N(\mathbb{Z}/p\mathbb{Z})$ consisting of the series $\displaystyle x \left( 1 + \sum_{i=1}^\infty a_i x^{p^{r_i}} \right)$, $a_i \in \mathbb{Z}/p\mathbb{Z}$.

(C. R. Leedham-Green, P. M. Neumann, J. Wiegold). For a finite $p$-group $P$, denote by $c = c(P)$ its nilpotency class and by $b = b(P)$ its breadth, that is, $p^b$ is the maximum size of a conjugacy class in $P$. Class-Breadth Problem: Is it true that $c \leqslant b + 1$ if $p \neq 2$?

Suppose that a finite $p$-group $P$ admits an automorphism of order $p^n$ having exactly $p^m$ fixed points. By (E. I. Khukhro, Russ. Acad. Sci. Sbornik Math., 80 (1995), 435–444) then $P$ has a subgroup of index bounded in terms of $p$, $n$ and $m$ which is soluble of derived length bounded in terms of $p^n$. Is it true that $P$ has also a subgroup of index bounded in terms of $p$, $n$ and $m$ which is soluble of derived length bounded in terms of $m$? There are positive answers in the cases of $m = 1$ (S. McKay, Quart. J. Math. Oxford, Ser. (2), 38 (1987), 489–502; I. Kiming, Math. Scand., 62 (1988), 153–172) and $n = 1$ (Yu. A. Medvedev, see 10.68).

Is it true that for any two different prime numbers $p$ and $q$ there exists a non-primary torsion locally soluble $\{p, q\}$-group that can be represented as the product of two of its $p$-subgroups?

We say that a metric space is a 2-end one (a narrow one), if it is quasiisometric to the real line $\mathbb{R}$ (respectively, to a subset of $\mathbb{R}$). All other spaces are said to be wide. Suppose that the Caley graph $\Gamma = \Gamma(G, A)$ of a group $G$ with a finite set of generators $A$ in the natural metric contains a 2-end subset, and suppose that there is $\varepsilon > 0$ such that the complement in $\Gamma$ to the $\varepsilon$-neighbourhood of any connected 2-end subset contains exactly two wide connected components. Is it true that the group $G$ in the word metric is quasiisometric to the Euclidean or hyperbolic plane?

A formation $\mathfrak{F}$ of finite groups is called superradical if it is $S_n$-closed and contains every finite group of the form $G = AB$ where $A$ and $B$ are $\mathfrak{F}$-subnormal $F$-subgroups.
$\qquad$ a) Find all superradical local formations.
$\qquad$ b) Prove that every $S$-closed superradical formation is a solubly saturated formation.

Is it true that in a Shunkov group (i. e. conjugately biprimitively finite group, see 6.57) having infinitely many elements of finite order every element of prime order is contained in some infinite locally finite subgroup? This is true under the additional condition that any two conjugates of this element generate a soluble subgroup (V. P. Shunkov, $M_p$-groups, Moscow, Nauka, 1990 (Russian)).

A group $G$ is saturated with groups from a class $\mathfrak{X}$ if every finite subgroup $K \leqslant G$ is contained in a subgroup $L \leqslant G$ isomorphic to some group from $\mathfrak{X}$. Is it true that a periodic group saturated with finite simple groups of Lie type of uniformly bounded ranks is itself a simple group of Lie type of finite rank?

(V. Lin). Let $B_n$ be the braid group on $n$ strings, and let $n > 4$.
$\qquad$ a) Does $B_n$ have any non-trivial non-injective endomorphisms with non-cyclic images?
$\qquad$ b) Is it true that every non-trivial endomorphism of the derived subgroup $[B_n, B_n]$ is an automorphism?
$\qquad$ c) Does $B_n$ have proper non-abelian torsion-free factor-groups?

Let $H$ be a proper subgroup of a group $G$ and let elements $a, b \in H$ have distinct prime orders $p, q$. Suppose that, for every $g \in G \setminus H$, the subgroup $\langle a, b^g \rangle$ is a finite Frobenius group with complement of order $pq$. Does the subgroup generated by the union of the kernels of all Frobenius subgroups of $G$ with complement $\langle a \rangle$ intersect $\langle a \rangle$ trivially? The case where all groups $\langle a, b^g \rangle$, $g \in G$, are finite is of special interest.

An infinite group $G$ is called a monster of the first kind if it has elements of order $> 2$ and for any such an element $a$ and for any proper subgroup $H$ of $G$, there is an element $g$ in $G \setminus H$, such that $\langle a, a^g \rangle = G$. A. Yu. Olshanskii showed that there are continuously many monsters of the first kind (see 6.63). Does there exist, for any such a monster, a torsion-free group which is a central extension of a cyclic group by the given monster?