Let $n$ be a positive integer. For a finite group $K$ and an automorphism $\phi$ of $K$ of order dividing $n$, let $X_{n,\phi}(K) := \{x \in K \mid x x^\phi \dots x^{\phi^{n-1}} = 1\}$. Let $c_n$ be the supremum of the ratios $|X_{n,\phi}(H)|/|H|$ over all finite groups $H$ and their automorphisms $\phi \in \text{Aut}(H)$ such that $\phi^n = \text{id}$ and $X_{n,\phi}(H) \neq H$.
$\qquad$ a) Let $n > 1$ be a positive integer such that $c_d < 1$ for all prime power divisors $d$ of $n$. Is it true that $c_n < 1$?
$\qquad$ b) For a finite group $G$ and a positive integer $n$, the generalized Hughes–Thompson subgroup is defined as $H_n(G) = \langle x \in G \mid x^n \neq 1 \rangle$. Suppose that $n$ is a positive integer for which there is a positive integer $k_n$ depending only on $n$ such that $|G : H_n(G)| \leqslant k_n$ for all finite groups $G$ with $H_n(G) \neq 1$. Is it true that then $c_n < 1$? This question is open even when $n \geqslant 5$ is prime.

Let $S$ be a finite simple group, and let $G$ be a finite group for which there exists a bijection $f : G \to S$ such that $|x|$ divides $|f(x)|$ for all $x \in G$. Must $G$ necessarily be simple?

Let $G = A_n$ or $S_n$ and let $H, K$ be soluble subgroups of $G$. For all sufficiently large $n$, can we always find an element $x \in G$ such that $H \cap K^x = 1$? Does this hold for all $n \geqslant 21$?

Note that the conclusion is false when $G = S_{20}$ and $H = K = (S_4 \wr S_4) \times S_4$.

Let $G$ be a finite group with trivial solvable radical and let $H_1, \dots, H_5$ be solvable subgroups of $G$. Then do there always exist elements $x_i \in G$ such that $\bigcap_i H_i^{x_i} = 1$? Cf. 17.41(b).

Let $p$ be a prime. Let $G$ be a transitive subgroup of the group of finitary permutations $\text{FSym}(\Omega)$ of a set $\Omega$, let $N$ be a normal subgroup of $G$, and let $S$ be a transitive Sylow $p$-subgroup of $G$.
$\qquad$ (a) Is it true that $S \cap N$ is a Sylow $p$-subgroup of $N$?
$\qquad$ (b) Is it true that $SN/N$ is a Sylow $p$-subgroup of $G/N$?
$\qquad$ (c) Are any two transitive Sylow $p$-subgroups of $G$ locally conjugate in $G$?

Two subgroups $X, Y$ of a group $G$ are said to be locally conjugate if there is a locally inner automorphism $\varphi$ of $G$ such that $X^\varphi = Y$. An automorphism $\varphi$ of $G$ is said to be locally inner if for every finite subset $A \subseteq G$ there is an element $g = g(A) \in G$ such that $a^\varphi = g^{-1} a g$ for all $a \in A$.

Let $p$ be a prime. A totally imprimitive $p$-group $H$ of finitary permutations is said to have the cyclic-block property if in the cycle decomposition of every element the support of every cycle is a block for $H$. Let $G$ be a transitive subgroup of the group of finitary permutations $\text{FSym}(\Omega)$ of a set $\Omega$. Does every transitive Sylow $p$-subgroup of $G$ contain a transitive subgroup which has the cyclic-block property?

(Well-known problem). A finite group $G$ is called an IYB-group if it is isomorphic to the permutation group of a finite involutive non-degenerate set-theoretic solution of the Yang–Baxter equation, or equivalently, $G$ is isomorphic to the multiplicative group of a finite left brace. Assume that the Sylow subgroups of a finite soluble group $G$ are IYB-groups. Is $G$ an IYB-group?

As in 17.57, let $r(m) = \{r + km \mid k \in \mathbb{Z}\}$ for integers $0 \leqslant r < m$; for $r_1(m_1) \cap r_2(m_2) = \varnothing$ let the class transposition $\tau_{r_1(m_1), r_2(m_2)}$ be the involution which interchanges $r_1 + tm_1$ and $r_2 + tm_2$ for each integer $t$ and fixes everything else, and let $\text{CT}(\mathbb{Z})$ be the group generated by all class transpositions.

Let $\text{CT}_k$ be the subgroup of $\text{CT}(\mathbb{Z})$ generated by the class transpositions $\tau_{r_1(k), r_2(k)}$ for $0 \leqslant r_1 \neq r_2 < k$. Since $\tau_{r_1(k), r_2(k)}$ permutes the residue classes modulo $k$, the group $\text{CT}_k$ is isomorphic to the symmetric group $S_k$. Let $\text{CT}_{(k)} = \langle \text{CT}_2, \text{CT}_3, \dots, \text{CT}_k \rangle$. Is it true that for $k > 3$ the group $\text{CT}_{(k)}$ is isomorphic to the symmetric group $S_N$, where $N$ is the least common multiple of the numbers $2, 3, \dots, k$?

Let $F$ be a non-abelian free pro-$p$ group of finite rank. Can one find a finite collection $U_1, \dots, U_n$ of open subgroups of $F$ including $F$ itself such that the only subgroup of $F$ which is contained in $U_i$ and is characteristic in $U_i$ for every $i$ is the trivial subgroup?

We call a group presentation finite if it represents a finite group. We say that a presentation is just finite if it is finite and is no longer finite on removal of any relation from it. Is it true that every finite group has a just finite presentation?

Note that if a group has a balanced presentation, then it is just finite. A similar argument can be applied for some $p$-groups using the Golod–Shafarevich inequality.

Can some or all groups of the following sorts be written as homomorphic images of nonprincipal ultraproducts of countable families of groups? This is Question 18 in (G. M. Bergman, Pacific J. Math., 274 (2015) 451–495).
$\qquad$ (a) Infinite finitely generated groups of finite exponent.
$\qquad$ (b) For an infinite set $X$, the group of those permutations of $X$ that move only finitely many elements.
It is known that no group of permutations containing an element with exactly one infinite orbit can be written as an image of such an ultraproduct (ibid.).

Suppose that $\mathscr{U}$ is a nonprincipal ultrafilter on $\omega$, and $B$ is a group such that every element $b \in B$ belongs to a subgroup of $B$ that is a homomorphic image of $\mathbb{Z}^\omega / \mathscr{U}$. Must $B$ then be a homomorphic image of an ultraproduct group $\prod_{i \in \omega} G_i / \mathscr{U}$ for some groups $G_i$?

This is Question 19 in (G. M. Bergman, Pacific J. Math., 274 (2015) 451–495). An affirmative answer would imply that every torsion group was such a homomorphic image for every $\mathscr{U}$, and so would give positive answers to both parts of 21.11.

It is known that the group $\mathbb{Z}^\omega$ has a subgroup whose dual is free abelian of rank $2^{\aleph_0}$ (see 17.24 in Archive). Does $\mathbb{Z}^\omega$ have a subgroup whose dual is free abelian of still larger rank (the largest possible being $2^{2^{\aleph_0}}$)? This is Question 11 in (G. M. Bergman, Portugaliae Math., 69 (2012) 69–84).

Suppose $\alpha$ is an endomorphism of a group $G$ such that for every group $H$ and every homomorphism $f : G \to H$, there exists an endomorphism $\beta_f$ of $H$ such that $\beta_f f = f\alpha$. Must $\alpha$ then be either an inner automorphism of $G$ or the trivial endomorphism? This is Question 5 in (G. M. Bergman, Publ. Matem., 56 (2012), 91–126).

Suppose $B$ is a subgroup of the symmetric group $S_\Omega$ on an infinite set $\Omega$. Will the amalgamated free product $S_\Omega \ast_B S_\Omega$ of two copies of $S_\Omega$ with amalgamation of $B$ be embeddable in $S_\Omega$? This is a weakened form of the group case of Question 4.4 in (G. M. Bergman, Indag. Math., 18 (2007), 349–403).

It is known that $S_\Omega \ast_B S_\Omega$ need not be so embeddable by a map respecting $B$ (Algebra Number Theory, 3 (2009), 847–879, 10).

Let the width of a group (respectively, a monoid) $H$ with respect to a generating set $X$ mean the supremum over $h \in H$ of the least length of a group word (respectively, a monoid word) in elements of $X$ expressing $h$. A group (or monoid) is said to have finite width if its width with respect to every generating set is finite. (A common finite bound for these widths is not required.) Do there exist groups $G$ having finite width as groups, but not as monoids? This is Question 9 in (G. M. Bergman, Bull. London Math. Soc., 38 (2006), 429–440).

If $\mathfrak{X}$ is a class of groups, let $\mathbf{H}(\mathfrak{X})$ denote the class of homomorphic images of groups in $\mathfrak{X}$, let $\mathbf{S}(\mathfrak{X})$ denote the class of groups isomorphic to subgroups of groups in $\mathfrak{X}$, let $\mathbf{P}(\mathfrak{X})$ denote the class of groups isomorphic to (unrestricted) direct products of families of groups in $\mathfrak{X}$, and let $\mathbf{P}_f(\mathfrak{X})$ denote the class of groups isomorphic to direct products of finite families of groups in $\mathfrak{X}$. By Birkhoff’s theorem, $\mathbf{H}(\mathbf{S}(\mathbf{P}(\mathfrak{X})))$ is the variety of groups generated by $\mathfrak{X}$.

If $\mathfrak{M}$ is a class of metabelian groups, must $\mathbf{H}(\mathbf{S}(\mathbf{P}_f(\mathfrak{M}))) \subseteq \mathbf{S}(\mathbf{H}(\mathbf{P}(\mathbf{S}(\mathfrak{M}))))$? This is Question 27 in (G. M. Bergman, Algebra Universalis, 26 (1989), 267–283).

Suppose that $G$ is a finite group, and $A_1, A_2, A_3$ are subsets of $G$ such that the multiplication map $A_1 \times A_2 \times A_3 \to G$ is bijective. Must the subgroup $\langle A_2 \rangle$ generated by $A_2$ have order divisible by the cardinality $|A_2|$? This is Question 8 in (G. M. Bergman, J. Iranian Math. Soc., 1 (2020), 157–161).

It is known (ibid.) that the corresponding statement is true for the subgroups $\langle A_1 \rangle$ and $\langle A_3 \rangle$. Moreover, $|A_2|$ will at least divide the order of the least subgroup containing $A_2$ and closed under conjugation by members of $A_1$, and similarly of the least subgroup containing $A_2$ and closed under conjugation by members of $A_3$.

Suppose that $S$ and $M$ are groups of finite Morley rank, $S$ is an infinite group, and $M$ is a non-trivial connected group definably and faithfully acting on $S$. This action is said to be irreducible if $M$ does not leave invariant any definable non-trivial proper subgroup of $S$. Prove that if $S$ is a simple group such that every proper definable subgroup of $S$ is nilpotent, and the action of $M$ on $S$ is irreducible, then this action is equivalent to the action of $S$ on itself by conjugation.

Prove that a simple group of finite Morley rank without involutions cannot act definably, faithfully, and irreducibly on a connected group other than on itself acting by conjugation.

Prove that a simple (that is, without proper non-trivial connected normal subgroups) algebraic group $M$ over an algebraically closed field cannot act definably, faithfully, and irreducibly on a simple group of finite Morley rank other than $M/Z(M)$.

Is the (standard, restricted) wreath product $G \wr H$ of two finitely generated Hopfian groups Hopfian?

The same question where $G$ is assumed to be abelian or nilpotent is equivalent to Kaplansky’s direct finiteness conjecture; see (H. Bradford, F. Fournier-Facio, Math. Z., 308, no. 4 (2024), Paper no. 58).

A graph is called a cograph if it has no induced subgraph isomorphic to a path with 4 vertices. A graph is said to be chordal if it has no induced cycles with $n$ vertices for every $n \geqslant 4$. For a finite group $G$, the enhanced power graph $\mathcal{E}(G)$ is the graph with vertex set $G$ and edges $\{x, y\}$ for all $x \neq y \in G$ such that $\langle x, y \rangle$ is cyclic.
$\qquad$ (a) For a given integer $n \geqslant 4$, determine the set of all finite nonabelian simple groups $G$ such that $\mathcal{E}(G)$ has no induced cycles with $n$ vertices.
$\qquad$ (b) Determine the set of all finite nonabelian simple groups $G$ such that $\mathcal{E}(G)$ is chordal.

In (Preprint, 2025, https://arxiv.org/abs/2510.18073) we proved that if the enhanced power graph of a given finite group is a cograph, then it is also chordal. Also the finite nonabelian simple groups whose enhanced power graph is a cograph are described, and additional information is obtained on finite nonabelian simple groups whose enhanced power graph has no induced cycles with 4 vertices.

For a finite group $G$, the power graph $\mathcal{P}(G)$ is the graph with vertex set $G$ and edges $\{x, y\}$ for all $x \neq y \in G$ such that either $x \in \langle y \rangle$ or $y \in \langle x \rangle$. Is it true that, for every finite group $G$, if $\mathcal{P}(G)$ is a cograph, then $\mathcal{P}(G)$ is chordal? Cf. 21.23.

This holds if every element of $G$ has prime power order (D. Bubboloni, F. Fumagalli, C. E. Praeger, Preprint, 2025, https://arxiv.org/abs/2510.18073) and if $G$ is a nonabelian simple group (J. Cameron, P. Manna, R. Mehatari, J. Algebra, 591 (2022), 59–74; J. Brachter, E. Kaja, J. Algebr. Comb., 58 (2023), 1095–1124).

(T. Breuer, R. M. Guralnick). Let $G$ be a finite simple group and let $p_1, p_2$ be any (not necessarily distinct) prime divisors of $|G|$. Then can we always find Sylow $p_i$-subgroups $H_i$ such that $G = \langle H_1, H_2 \rangle$?

(F. Lisi, L. Sabatini). Let $G$ be a non-trivial finite group and let $p_1, \dots, p_k$ be the distinct prime divisors of $|G|$. For each $i$, let $H_i$ be a Sylow $p_i$-subgroup of $G$. Is it true that there exists an element $x \in G$ such that for all $i$ the subgroup $H_i \cap H_i^x$ is inclusion-minimal in $\{H_i \cap H_i^g \mid g \in G\}$?

(M. Larsen, A. Shalev, P. H. Tiep). A permutation on a set $\Omega$ is called a derangement if it has no fixed points in $\Omega$. Let $G$ be a finite simple transitive permutation group. Is it true that every element in $G$ is the product of two derangements?

Let $G$ be a finite simple transitive permutation group, and let $\delta(G)$ be the proportion of derangements in $G$. Is it true that $\delta(G) \geqslant 89/325$?

Note that $\delta(G) = 89/325$ for the action of the Tits group $G = {}^2F_4(2)'$ on the cosets of a maximal parabolic subgroup of the form $2^2.[2^8].S_3$ (Forum Math. Sigma, 13 (2025), paper no. e98, 62 pp.).

Let $G \leqslant \text{Sym}(\Omega)$ be a finite primitive permutation group with a regular suborbit (that is, $G$ has a trivial 2-point stabiliser). Then is it true that for all $\alpha, \beta \in \Omega$, there exists $\gamma \in \Omega$ such that the 2-point stabilisers $G_{\alpha, \gamma}$ and $G_{\beta, \gamma}$ are both trivial?

(Well-known question). A discrete group $G$ is said to have the Haagerup property (also known as Gromov’s a-T-menability property) if there exists a metrically proper isometric action of $G$ on a (possibly infinite-dimensional) Hilbert space. Are all 1-relator groups Haagerup groups?

Conjecture: If $N$ is a finite soluble group, then any regular subgroup in the holomorph $\text{Hol}(N)$ of $N$ is also soluble.

Is the following problem decidable, and if so, what is its complexity? Given a finite group $G$, is there a finite group $H$ such that the derived subgroup of $H$ is isomorphic to $G$?

Does an analogue of Dunwoody’s theorem hold for totally disconnected locally compact groups, that is, must a tdlc group of rational discrete cohomological dimension at most 1 be topologically isomorphic to the fundamental group of a graph of profinite groups?

(Well-known problem). A group $G$ is a unique product group if, for any nonempty finite subsets $A, B$ of $G$, there exists an element of $G$ which can be written uniquely as $ab$ with $a \in A$ and $b \in B$. A group $G$ is locally invariant orderable if $G$ admits a partial order $<$ such that for all $g, h \in G$ with $h \neq 1$, we have either $gh > g$ or $gh^{-1} > g$. Does there exist a unique product group which is not locally invariant orderable?

Let $G$ be a finite group, $w$ a multilinear commutator group-word, and $p$ a prime. Suppose that $p$ divides the order $|xy|$ whenever $x$ is a $w$-value of $p'$-order in $G$ and $y$ is a $w$-value in $G$ of order divisible by $p$. Is it true that then the verbal subgroup $w(G)$ must be $p$-nilpotent?

Without the assumption that $w$ be multilinear, the answer is negative. An affirmative answer has been obtained in several special cases (J. Algebra, 609 (2022), 926–936).

Kropholler’s hierarchy (see 15.45) is closed under finite extensions, that is, $(\mathbf{H}_\alpha \mathfrak{F})\mathfrak{F} \subseteq \mathbf{H}_\alpha \mathfrak{F}$ for every $\alpha$ (P. Kropholler, J. Pure Appl. Algebra, 90 (1993), 55–67). Let a hierarchy of tdlc groups $\mathbf{H} \mathfrak{K}$ be defined analogously to Kropholler’s hierarchy in 15.45, with $\mathfrak{K}$ being the class of profinite groups and with the cell stabilisers of the admissible action required to be open. Is it true that $\mathbf{H} \mathfrak{K}$ is closed under profinite extensions, that is, $(\mathbf{H}_\alpha \mathfrak{K})\mathfrak{K} \subseteq \mathbf{H}_\alpha \mathfrak{K}$ for every $\alpha$?

By definition, a constructible totally disconnected, locally compact (tdlc) group is the result of a sequence of profinite extensions and ascending HNN-extensions starting from the trivial group. As in the discrete case, soluble constructible tdlc groups have type $FP_\infty$ (G. C. Cook, I. Castellano, J. Algebra, 543 (2020), 54–97). Are soluble tdlc groups of type $FP_\infty$ constructible?

(S. Harper, C. Donoven). The spread of a group $G$ is the greatest nonnegative integer $k$ such that for all nontrivial elements $x_1, \dots, x_k \in G$ there exists $y \in G$ such that $\langle x_1, y \rangle = \dots = \langle x_k, y \rangle = G$, or is $\infty$ in case there is no such maximum. Does there exist a group with spread equal to 1?

It is known that there exist residually finite, locally finite, characteristically simple groups with finitely many orbits under automorphisms (A. B. Apps, J. Algebra, 81 (1983), 320–339). Are there any locally finite, characteristically simple groups with finitely many orbits under automorphisms that are not residually finite?

Let $G$ be a subgroup of $\text{GL}(n, \mathbb{Q})$ with finitely many orbits under automorphisms. Is $G$ a virtually soluble group?

A group is said to be self-similar if it admits a faithful state-closed representation by automorphisms of a regular one-rooted $m$-tree for some $m$. Can a torsion-free finitely presented metabelian group which is self-similar contain a subgroup isomorphic to the restricted wreath product $H = \mathbb{Z} \wr \mathbb{Z}$?

It is known that $\mathbb{Z}\wr\mathbb{Z}$ itself is self-similar (A. C. Dantas, T. M. G. Santos, S. N. Sidki, J. Algebra, 567 (2021), 564–581).

Let $T_{d,c}$ denote the class of $d$-generated, torsion-free nilpotent groups having class $c$. It is known that $T_{d,2}$-groups are self-similar for all $d$, that $T_{2,3}$-groups are self-similar, and that there are $T_{4,3}$-groups that are not self-similar (A. Berlatto, T. Santos, Preprint, 2025, https://arxiv.org/abs/2509.16947). Are there $T_{3,3}$-groups that are not self-similar? (See 21.41 for the definition of self-similarity.)

Conjecture: Suppose that for a fixed positive integer $k$ at least half of the elements of a finite group $G$ have order $k$. Then $G$ is solvable.

Let $W_n = A_5 \wr \dots \wr A_5$ be the $n$-times iterated permutational wreath product of $A_5$ in its natural action (so $W_n$ acts on $5^n$ points), and let $W = \varprojlim W_n$ be the inverse limit (infinite iterated wreath product of $A_5$). Does $W$ contain a finitely generated dense subgroup of subexponential growth?

(Well-known problem). Does there exist a finitely presented (infinite) simple group requiring more than two generators? Cf. 6.44.

(Well-known problem). Does there exist a finitely presented (infinite) simple group of finite cohomological dimension greater than 2?

(Well-known problem). Does there exist a finitely presented group $G$ such that $G \cong G \times H$ for some non-trivial group $H$?

The first finitely generated example was constructed in (J. M. Tyrer Jones, J. Austral. Math. Soc., 17 (1974), 174–196). A finitely presented group that surjects onto its own direct square was constructed in (G. Baumslag, C. F. Miller, III, Bull. London Math. Soc., 20, no. 3 (1988), 239–244).

A quasimorphism on a group $G$ is a function $f : G \to \mathbb{R}$ such that the quantity $\sup_{g,h} |f(g) + f(h) - f(gh)|$ is finite. A quasimorphism is homogeneous if it restricts to a homework on every cyclic subgroup of $G$.

Let $G$ be a group admitting an unbounded homogeneous quasimorphism $G \to \mathbb{R}$ that is not a homomorphism. Must $G$ contain a non-abelian free subgroup?

An isometric action of a group $G$ on a metric space $S$ is called acylindrical if for every $\varepsilon > 0$ there exist $R, N > 0$ such that for every two points $x, y$ with $d(x, y) \geqslant R$, there are at most $N$ elements $g \in G$ satisfying $d(x, gx) \leqslant \varepsilon$ and $d(y, gy) \leqslant \varepsilon$. A group is said to be acylindrically hyperbolic if it is not virtually cyclic and admits an acylindrical action on a hyperbolic space with unbounded orbits. Is the automorphism group of a finitely generated acylindrically hyperbolic group also acylindrically hyperbolic?

Does every finite 3-group $T$ have a nontrivial characteristic subgroup $C$ such that if $T$ is a Sylow 3-subgroup of a finite group $G$, then $T \cap G' = T \cap H'$, where $H = N_G(C)$?

Such a characteristic subgroup is known to exist in $p$-groups for $p \geqslant 5$ (G. Glauberman, Math. Z., 117 (1970), 46–56), and for $p = 3$ there are two characteristic subgroups $K_1, K_2$ such that $T \cap G' = (T \cap H_1')(S \cap H_2')$, where $H_i = N_G(K_i)$ (G. Glauberman, J. Algebra, 648 (2024), 62–86). The group $S_4$ shows that no such characteristic subgroups can be found in some Sylow 2-subgroups.

Let $p$ be a prime, and $P$ a finite $p$-group.
$\qquad$ (a) Suppose that $P$ has an abelian subgroup of order $p^n$. For which $n$ does $P$ necessarily have a normal abelian subgroup of order $p^n$?
$\qquad$ (b) Suppose that $P$ has an elementary abelian subgroup of order $p^n$. For which $n$ does $P$ necessarily have a normal elementary abelian subgroup of order $p^n$?

Let $L$ be a finite non-abelian simple group, and let $D$ be a conjugacy class of involutions in $L$. Consider the complete graph $\Gamma$ with vertex set $D$. Define an equivalence relation $\sim$ (graph coloring) on the set of edges as follows: $(a, b) \sim (c, d)$ if and only if $|ab| = |cd|$. An automorphism of the coloured graph $\Gamma$ is a permutation $\tau \in S_D$ such that $(a, b) \sim (a^\tau, b^\tau)$ for every edge $(a, b)$. Is it true that the automorphism group of $\Gamma$ is a subgroup of $\text{Aut}(L)$?

In the notation of 21.52, let $\text{Aut}_t(\Gamma)$ be the set of permutations $\tau \in S_D$ such that $(a, b) \sim (a^\tau, b^\tau)$ whenever $|ab| = t$ for $a, b \in D$. Clearly, $\text{Aut}(\Gamma) = \bigcap_t \text{Aut}_t(\Gamma)$. Is it true that for every finite simple group $G$ we have $\text{Aut}(\Gamma) = \text{Aut}_2(\Gamma) \cap \text{Aut}_p(\Gamma)$, where $\{2, p\}$ are the two minimal prime divisors of $|G|$?

Let $G$ be a finite soluble group with triality, which means that $G$ admits a group of automorphisms $S$ isomorphic to the symmetric group of degree 3 given by the presentation $S = \langle \sigma, \rho \mid \sigma^2 = \rho^3 = 1; \sigma\rho\sigma = \rho^2 \rangle$ such that $m \cdot m^\rho \cdot m^{\rho^2} = 1$ for all $m$ in the set of commutators $M(G) := \{[g, \sigma] \mid g \in G\}$.

Suppose in addition that $G = [G, S]$, the group $G$ is generated by $d$ elements of $M(G)$ and their images under $S$, and $x^n = 1$ for all $x \in M(G)$. Is it true that the Fitting height of $G$ is bounded in terms of $d$ and $n$?

An affirmative answer would provide a reduction of the analogue of the Restricted Burnside Problem for Moufang loops to the nilpotent case.

Let $q$ be a power of a prime $p$, and let $m_n(q)$ be the maximum $p$-length of $p$-solvable subgroups of $\text{GL}(n, q)$. Is it true that
$$\lim_{n \to \infty} m_n(q)/\log_2 n = 1?$$

Let $\ell(X)$ denote the composition length of a finite group $X$. Let $A$ be a finite nilpotent group acting by automorphisms on a finite soluble group $G$. Let $c(G, A)$ be the number of trivial $A$-modules in a given $A$-composition series of $G$. (Note that $c(G, A) = \ell(C_G(A))$ if $(|A|, |G|) = 1$.)

Conjecture: there are absolute constants $C_1$ and $C_2$ such that the Fitting height of $G$ is at most $C_1\ell(A) + C_2c(G, A)$.

Let $\mathfrak{X}$ be a non-empty class of finite groups of odd order closed under taking subgroups, homomorphic images, and extensions. Let $H$ be an $\mathfrak{X}$-maximal subgroup of a finite group $G$, and $N$ a normal subgroup of $G$. Must $H \cap N$ be an $\mathfrak{X}$-maximal subgroup of $N$?

We say that a product $XY = \{xy \mid x \in X, y \in Y\}$ of two subsets $X, Y$ of a group $G$ is direct if for every $z \in XY$ there are unique $x \in X, y \in Y$ such that $z = xy$. Is there an infinite group $G$ such that every subset $A \subseteq G$ satisfies the following property: all the maximal subsets $B$ for which the product $AB$ is direct have the same cardinality?

Note that for checking the property for a given infinite group $G$, it suffices to consider only those subsets $A \subseteq G$ for which $|A| = |G \setminus A|$. Indeed, the property is equivalent to $A^{-1}A \cap B B^{-1} = \{1\}$ and $A^{-1}AB = G$, and these imply $|G| = |A||B|$, since $G$ is infinite. Now, if $|A| < |G \setminus A|$, then $|A| < |G|$, and so $|B| = |G|$; and if $|A| > |G \setminus A|$, then $A^{-1}A = G$, and so $|B| = 1$, for all $B$ satisfying the property.

For a finite group $G$, let $\chi_1(G)$ denote the totality of the degrees of all irreducible complex characters of $G$ with allowance for their multiplicities. Suppose that $H$ is a finite group with $\chi_1(H) = \chi_1(G)$.
$\qquad$ a) If $G$ is an almost simple group, must $H$ be isomorphic to $G$?
$\qquad$ b) If $G$ is a quasisimple group, must $H$ be isomorphic to $G$?

Let $G$ be a finite group, $\mathbb{Z}_{(p)}$ the localization at $p$, and $\mathbb{F}_p$ the field of $p$ elements. Let $\mathcal{X}$ be the class of $\mathbb{F}_p G$-modules obtained by reduction of simple $\mathbb{Q}G$-modules. Is it true that $\mathbb{Z}_{(p)} G$ is semiperfect if and only if each projective indecomposable $\mathbb{F}_p G$-module can be written as an $\mathbb{N}$-linear combination of modules in $\mathcal{X}$ inside the Grothendieck group of $\mathbb{F}_p G$?

For a fixed (finitely generated free)-by-cyclic group $G = F_n \rtimes \mathbb{Z}$, is there an algorithm that, given a finite subset $S$ of $G$, finds a finite presentation for the subgroup $H = \langle S \rangle$? Cf. 4.8.

Is the uniform subgroup membership problem decidable for (finitely generated free)-by-cyclic groups? That is, for a fixed group $G = F_n \rtimes \mathbb{Z}$, is there an algorithm that, given elements $w, h_1, \dots, h_k \in G$, decides whether or not $w$ belongs to the subgroup $H = \langle h_1, \dots, h_k \rangle$?

(E. Zelmanov). Let $F$ be a field of characteristic $p > 0$, and let $\Gamma$ be the principal congruence subgroup of $\text{Aut}(F[x_1, \dots, x_n])$ consisting of all automorphisms that send each variable $x_i$ to $x_i$ modulo terms of higher degree. Then $\Gamma$ is a residually $p$ group. Does $\Gamma$ satisfy a pro-$p$ identity?

Is it true that if a normal subgroup $A$ of a Sylow $p$-subgroup of a $p$-soluble finite group $G$ has exponent $p^e$, then the normal closure of $A$ in $G$ has $(p, e)$-bounded (or even $e$-bounded) $p$-length?

Suppose that $\phi$ is an automorphism of a finite soluble group $G$. Must $G$ contain a subgroup of index bounded in terms of $|\phi|$ and $|C_G(\phi)|$ whose Fitting height is bounded
$\qquad$ a) in terms of $|\phi|$?
$\qquad$ b) or even in terms of the composition length of $\langle \phi \rangle$?

Suppose that $A$ is a nilpotent group of automorphisms of a finite soluble group $G$. Is the Fitting height of $G$ bounded in terms of $|A|$ and $|C_G(A)|$?

Suppose that $\phi$ is an automorphism of a finite soluble group $G$, and let $r$ be the (Prüfer) rank of the fixed-point subgroup $C_G(\phi)$. Is the Fitting height of $G$ bounded terms of $|\phi|$ and $r$?

An affirmative answer to this question would imply an affirmative answer to 13.8(b).

A finite group $G$ is said to be semi-abelian if it has a sequence of subgroups $1 = G_0 \leqslant G_1 \leqslant \dots \leqslant G_n = G$ such that for every $i$ the subgroup $G_{i+1}$ is isomorphic to a quotient of a semidirect product $A_i \rtimes G_i$ for some abelian group $A_i$.

Conjecture: Semi-abelian finite groups are monomial.

Is there an algorithm deciding if a given one-relator group is hyperbolic?

A group $G$ is called an orientable Poincaré duality group of dimension $n$ over a ring $R$ if it is of type FP over $R$ and $H^i(G; RG) = 0$ for $i \neq n$, while $H^n(G; RG) = R$ as an $RG$-module, where the action on $R$ is trivial. (Note that $G$ is not required to be finitely presented.)

If $G$ is an orientable Poincaré duality group of dimension $n$ over all fields, is it an orientable Poincaré duality group over the integers?

For a ring $R$, we say that a group $G$ is of type $\mathbf{FL}(R)$ if the trivial $RG$-module $R$ admits a finite resolution by finitely generated free modules. If $R$ is a field, we define the Euler characteristic of $G$ over $R$ to be the alternating sum of $R$-ranks of homology groups $H_i(G; R)$.

Does there exist a group of type $\mathbf{FL}(\mathbb{F}_i)$ for two fields $\mathbb{F}_1$ and $\mathbb{F}_2$ such that the Euler characteristics of the group over the fields $\mathbb{F}_1$ and $\mathbb{F}_2$ differ?

We say that a group is a Tarski monster if it is finitely generated, not cyclic, and all of its proper non-trivial subgroups are isomorphic to each other.
$\qquad$ a) Do there exist amenable torsion-free Tarski monsters?
$\qquad$ b) Do there exist amenable Tarski monsters of prime exponent?

Is the conjugacy problem in $\text{CT}(\mathbb{Z})$ algorithmically decidable? See the definition of $\text{CT}(\mathbb{Z})$ in 17.57.

Is it algorithmically decidable whether a given element $g \in \text{CT}(\mathbb{Z})$ (see the definition of $\text{CT}(\mathbb{Z})$ in 17.57)
$\qquad$ a) permutes a nontrivial partition of $\mathbb{Z}$ into residue classes?
$\qquad$ b) has only finite cycles?
$\qquad$ c) has no finite cycles?

Given two distinct sets $\mathcal{P}_1$ and $\mathcal{P}_2$ of odd primes none of which is a subset of the other, is it true that $$\langle \text{CT}_{\mathcal{P}_1}(\mathbb{Z}), \text{CT}_{\mathcal{P}_2}(\mathbb{Z}) \rangle \subsetneq \text{CT}_{\mathcal{P}_1 \cup \mathcal{P}_2}(\mathbb{Z})?$$ See the definition of $\text{CT}_{\mathcal{P}}(\mathbb{Z})$ in 17.60.

Let $\sigma = (\sigma_{ij})$, $1 \leqslant i \neq j \leqslant n$, be an irreducible elementary net (carpet) of order $n \geqslant 3$ over a field $K$ (see 19.48). The net $\sigma$ is said to be closed if the elementary net subgroup $E(\sigma)$ does not contain new elementary transvections. The net $\sigma$ is said to be completable if its diagonal can be supplemented with subgroups to a complete net. Completable elementary nets are closed. It is known that over fields of characteristic 0 and 2 there exist irreducible closed elementary nets that are not completable (V. A. Koibaev, Trudy Inst. Mat. Mekh. Ural Div. Ross. Akad. Nauk, 17, no. 4 (2011), 134–141 (Russian); V. A. Koibaev, Siberian Math. J., 62, no. 2 (2021), 262–266).

Do there exist irreducible closed elementary nets of order $n \geqslant 3$ over a field of odd characteristic that are not completable?

Let $d$ be an integer that is not divisible by $n$-th powers of primes, let $x^n - d$ be an irreducible polynomial over $\mathbb{Q}$, let $\theta = \sqrt[n]{d}$, and let $K = \mathbb{Q}(\theta)$ be the radical extension of degree $n$ of the field $\mathbb{Q}$. The multiplicative group $K^*$ of the field $K$ is canonically embedded into the group $\text{Aut}_{\mathbb{Q}}(K)$ of all invertible $\mathbb{Q}$-linear mappings of the $\mathbb{Q}-space$ $K$; let $T$ be the image of $K^*$ under this embedding. In the natural basis $1, \theta, \theta^2, \dots, \theta^{n-1}$ of the $\mathbb{Q}-space$ $K$ the group $\text{Aut}_{\mathbb{Q}}(K)$ corresponds to $G = \text{GL}(n, \mathbb{Q})$, and the subgroup $T$ to a subgroup $T(d)$ (unsplit maximal torus). Every subgroup $H$ of $G$ containing $T(d)$ and some one-dimensional transformation is rich in elementary transvections (V. A. Koibaev, St. Petersbg. Math. J., 21, no. 5 (2010), 731–742) and thus defines a net $\sigma = \sigma(H)$ (V. A. Koibaev, A. V. Shilov, J. Math. Sci. New York, 171, no. 3 (2010), 380–385). Let $E(s)$ denote the subgroup generated by all transvections in the net group $G(\sigma)$. Is it true that $H \leqslant N_G(E(\sigma))$?

Let $p$ be a prime and let $G$ be a pro-$p$ group. Suppose that all of the (continuous Galois) cohomology groups $H^n(G, \mathbb{F}_p)$ of $G$ with coefficients in the field of $p$ elements are finite. Does it necessarily follow that the cohomology ring $H^*(G, \mathbb{F}_p)$ is finitely generated?

Let $G$ be a finitely generated group with a fixed finite generating set $S$ and the corresponding word metric $L_S(*)$. An element $g$ is said to be distorted in $G$ if $L_S(g^n)/n \to 0$ as $n \to \infty$; this notion is independent of the choice of the generating set $S$. For any, not necessarily finitely generated, group $H$, an element $g \in H$ is said to be distorted if there is a finitely generated subgroup $G$ of $H$ containing $g$ in which $g$ is distorted. Do there exist finitely generated left-orderable groups in which every nontrivial element is distorted?

Note that it is straightforward to construct countable (not finitely generated) left orderable groups with this property using HNN-extensions and applying results of V. V. Bludov and A. M.W. Glass.

Do there exist finitely generated left-orderable groups with only one nontrivial conjugacy class?

A positive answer to this question implies a positive answer to 21.78. Note that D. Osin constructed torsion-free finitely generated groups with only one nontrivial conjugacy class; see 9.10.

(J. Wiegold). Let $\Gamma$ be a finite simple group and let $\mathcal{N}_n(\Gamma)$ denote the set of normal subgroups of the free group $F_n$ of rank $n$ whose quotient is isomorphic to $\Gamma$.

Conjecture: $\text{Aut}(F_n)$ acts transitively on $\mathcal{N}_n(\Gamma)$ for $n \geqslant 3$.

This is not true for $n = 2$ (B. H. Neumann, H. Neumann, Math. Nachr., 4 (1951), 106–125).

Conjecture: For $n \geqslant 3$, there are no finite simple characteristic quotients of the free group $F_n$.

This is not true for $n = 2$ (W. Y. Chen, A. Lubotzky, P. H. Tiep, to appear in Comment. Math. Helvetici, 2025).

The function $d_n(\sigma, \tau) = (1/n) \cdot |\{x \in \{1, \dots, n\} \mid \sigma(x) \neq \tau(x)\}|$ is a distance on the symmetric group $S_n$. For a finitely generated group $G$, an almost-homomorphism is a sequence of set-theoretic maps $f_n : G \to S_n$ satisfying $d_n(f_n(g)f_n(h), f_n(gh)) \to 0$ as $n \to \infty$ for all $g, h \in G$. An almost-homomorphism $\{f_n\}$ is said to be close to a homomorphism if there is a sequence of group homomorphisms $\rho_n : G \to S_n$ such that $d_n(\rho_n(g), f_n(g)) \to 0$ as $n \to \infty$ for all $g \in G$. The group $G$ is said to be permutation stable if every almost-homomorphism of $G$ is close to a homework.

Conjecture: Metabelian groups are permutation-stable.

For $\sigma \in S_n$ and $\tau \in S_m$, where $n \leqslant m$, let
$$d_n^{\text{flex}}(\sigma, \tau) = (1/n) \cdot (|\{x \in \{1, \dots, n\} \mid \sigma(x) \neq \tau(x)\}| + (m - n)).$$ An almost-homomorphism $\{f_n\}$ is said to be flexibly close to a homomorphism if there is a sequence of group homomorphisms $\rho_n : G \to S_{m_n}$ with $n \leqslant m_n$ such that $d_n^{\text{flex}}(\rho_n(g), f_n(g)) \to 0$ as $n \to \infty$ for all $g \in G$. The group $G$ is said to be flexibly permutation-stable if every almost-homomorphism of $G$ is flexibly close to a homomorphism.

Is $\text{SL}_n(\mathbb{Z})$ flexibly permutation-stable?

Is a flexibly permutation-stable group always permutation-stable? See the definitions in 21.83, 21.84.

(M. Gromov, B.Weiss). A group $G$ is said to be sofic if for every finite set $F \subseteq G$ containing 1 and every $\varepsilon > 0$ there exist $n \in \mathbb{N}$ and a map $\phi : F \to S_n$ such that $\phi(1) = 1$, $d(\phi(gh), \phi(g)\phi(h)) < \varepsilon$ for all $g, h$ such that $gh \in F$, $\phi(g)$ does not have fixed points for every $g \in F \setminus \{1\}$.

Is every group sofic?

Assume that a finite group $G$ has a family of $d$-generator subgroups whose indices have no common divisor. Is it true that $G$ can be generated by $d+1$ elements?

Is there a finite non-abelian group $G$ of odd order, with $k(G)$ conjugacy classes, such that $k(G)/|G| = 1/17$?

For $n > 39$, is it true that the number of conjugacy classes in the symmetric group $S_n$ of degree $n$ is never a divisor of the order of $S_n$? In other words, is it true that, for $n > 39$, the number $p(n)$ of integer partitions of $n$ is never a divisor of $n!$?

Let $\Gamma$ be a graph of diameter $d$. For $i \in \{1, 2, \dots, d\}$, let $\Gamma_i$ be the graph on the same vertex set as $\Gamma$ with vertices $u, w$ adjacent in $\Gamma_i$ if and only if $d_\Gamma(u, w) = i$. Does there exist a $Q$-polynomial distance-regular graph $\Gamma$ of diameter 3 such that $\Gamma_2$ and $\Gamma_3$ are strongly regular?

(W. Willems). Conjecture: The sum of squares of the degrees of the irreducible $p$-Brauer characters of a finite group $G$ is at least the $p'$-part of $|G|$.

Conjecture: The number of irreducible $p$-Brauer characters of a finite group $G$ is bounded above by the maximum of the number of conjugacy classes $k(H)$ in $p'$-subgroups $H$ of $G$.

(M. Herzog, J. Schönheim). Let $G$ be a group and let $k \geqslant 2$. Let $H_1, \dots, H_k$ be subgroups of $G$, and $g_1, \dots, g_k$ elements of $G$ such that the cosets $g_1 H_1, \dots, g_k H_k$ form a partition of $G$. Is it true that $|G : H_i| = |G : H_j|$ for some $i \neq j$?

The Gruenberg–Kegel graph (or the prime graph) $GK(G)$ of a finite group $G$ is a labelled graph with vertex set consisting of all prime divisors of the order of $G$ in which different vertices $p$ and $q$ are adjacent if and only if $G$ contains an element of order $pq$. Let $\overline{GK}(G)$ denote the abstract graph obtained from $GK(G)$ by removing all labels. A finite group $G$ is said to be recognizable by the isomorphism type of its Gruenberg–Kegel graph if there are no finite groups $H \not\cong G$ with $\overline{GK}(H)$ isomorphic to $\overline{GK}(G)$.

Are there infinitely many (pairwise non-isomorphic) finite groups which are recognizable by the isomorphism type of the Gruenberg–Kegel graph?

Is there an almost simple but not simple group which is recognizable by the isomorphism type of its Gruenberg–Kegel graph (see 21.94)?

Is it true that a periodic group containing an involution is locally finite if the centralizer of every element of even order is locally finite?

(M. Tărnăuceanu). Is it true that for every positive rational number $r$ there exists a finite group $G$ such that $|\text{Aut}(G)|/|G| = r$?

A similar question is answered in the positive for graphs, monoids, partial groups, and posets (R. Molinier, Preprint, 2025, https://arxiv.org/abs/2504.21059).

It is also known that the set $\{|\text{Aut}(G)|/|G|: G$ is a finite abelian group$\}$ is dense in $[0, +\infty)$ (M. Tărnăuceanu, Elemente Math. (2025), https://ems.press/journals/em/articles/14298544).

Let $w$ be a multilinear commutator word, and assume that $G$ is a group where the set of $w$-values is covered by finitely many cyclic subgroups. Is it true that the verbal subgroup $w(G)$ is finite-by-cyclic?

This is true for lower central words (G. Cutolo, C. Nicotera, J. Algebra, 324, no. 7 (2010), 1616–1624).

Conjecture: If $G$ is a transitive permutation group on a finite set $\Omega$, then for any distinct $\alpha, \beta$ in $\Omega$ there is an element $g \in G$ with $\alpha^g = \beta$ whose number of fixed points is different from 1.

Suppose that $A$ and $G$ are finite groups such that $A$ acts coprimely on $G$ by automorphisms. Let $C = C_G(A)$ be the fixed-point subgroup, and let $C'$ denote its derived subgroup. Is it true that the number of $A$-invariant irreducible characters $\chi$ of $G$ whose restriction $\chi_C$ is never zero is exactly $|C/C'|$?

This would follow if one could show that $\chi_C$ is never zero if and only if the Glauberman–Isaacs correspondent $\chi^*$ of $\chi$ is linear.

Which finite almost simple groups are the automorphism groups of regular polytopes of rank 3? In other words, which finite almost simple groups are generated by three involutions two of which commute?

This question has been answered for finite simple groups; see 7.30.

Let $\mathfrak{V}$ be a variety generated by a finite group, and let $f(n)$ be the order of the free group in $\mathfrak{V}$ on $n$ generators. Is it true that the sequence $\sqrt[n]{\log f(n)}$ has a limit as $n \to \infty$, and this limit is an integer?

(V. V. Uspenskii). A Hausdorff topological group $G$ is called minimal if it does not admit a strictly coarser Hausdorff group topology. A topological group is called Raikov complete if its two-sided uniform structure is complete. It is known that a finite direct product of Raikov complete minimal topological groups is again minimal. Is it true that an arbitrary Cartesian product of Raikov complete minimal topological groups remains minimal?

It is known that an arbitrary Cartesian product of centre-free minimal topological groups is minimal (M. Megrelishvili, Topology Appl., 62, no. 1 (1995), 1–19).

For a group word $w(x_1, \dots, x_n)$ on $n$ letters, define $e_0(x_1, \dots, x_n) = x_1$ and $e_{k+1}(x_1, \dots, x_n) = w(e_k(x_1, \dots, x_n), \dots, x_n)$ for all $k \in \mathbb{N}$. A group $G$ is said to satisfy the Engel type iterated identity $w$ if for all $x_1, \dots, x_n \in G$ there exists $m \in \mathbb{N}$ such that $e_m(x_1, \dots, x_n) = 1$.

Conjecture: For every non-trivial word $w$, if a finitely generated branch group $G$ (see 15.12) satisfies the iterated identity $w$, then $G$ is a torsion group.

(D. Segal). A group word $w$ is said to be concise in a class $\mathcal{C}$ of groups if for every group $G$ in $\mathcal{C}$ such that the set $G_w$ of word values of $w$ in $G$ is finite, the verbal subgroup $w(G) = \langle G_w \rangle$ is also finite (see also 2.45). Is every word concise in the class of residually finite groups?

A first order formula $\phi(x)$ in the group language with one free variable is said to be concise in a class $\mathcal{C}$ of groups if for every group $G$ in $\mathcal{C}$ such that the set $G_\phi$ of elements in $G$ satisfying $\phi$ is finite, the subgroup $\phi(G)$ generated by $G_\phi$ is also finite (cf. 21.105 and 2.45). Is every formula with one free variable concise in the class of residually finite groups?

A sequence $\{F_n\}$ of pairwise disjoint finite subsets of a topological group is called expansive if for every open subset $U$ there is a number $m$ such that $F_n \cap U \neq \varnothing$ for all $n > m$. Suppose that a countable group $G$ can be partitioned into countably many dense subsets. Is it true that in $G$ there exists an expansive sequence? Cf. 15.80.

For a finite group $G$ let $\text{Cod}(G)$ denote the set of irreducible character codegrees of $G$ (see 20.78). Define $\sigma(G) = \max\{|\pi(m)|: m \in \text{Cod}(G)\}$, where $\pi(m)$ denotes the set of prime divisors of an integer $m$. It is proved that there exists a constant $k$ such that $|\pi(G)| \leqslant k \cdot \sigma(G)$ for every finite group $G$ (Y. Yang, G. Qian, J. Algebra, 478 (2017), 215–219), but the estimate provided for $k$ is very crude. Can the constant $k$ be taken as 4?

Conjecture: The derived length of a finite solvable group $G$ does not exceed $|\text{Cod}(G)| - 1$. (See 20.78 for the notation.)

The Fitting height of $G$ is known to be at most $\min\{|\text{Cod}(G)| - 1, |\text{Cod}(G)|/2 + 1\}$ (G. Qian, Y. Zeng, J. Group Theory, to appear).

Let $S$ be a nonabelian finite simple group, and $x$ a nonidentity automorphism of $S$. Let $\alpha(x)$ be the smallest number of conjugates of $x$ in $G = \langle x, \text{Inn}\,S \rangle$ that generate $G$. The values of $\alpha(x)$ had been studied in (R. Guralnick, J. Saxl, J. Algebra, 268, no. 2 (2003), 519–571).
$\qquad$ a) (R. Guralnick, J. Saxl). Conjecture: If $S$ is an exceptional group of Lie type, then $\alpha(x) \leqslant 5$ for every nonidentity automorphism $x$ of $S$.
$\qquad$ b) For each exceptional group $S$ of Lie type, find the largest value of $\alpha(x)$.

Let $S$ be a finite simple nonabelian group that is not isomorphic to any group ${}^2B_2(q)$. A nonidentity automorphism $x$ of $S$ is called a $\tau$-automorphism if every two conjugates of $x$ in $\langle x, \text{Inn}(S) \rangle$ generate a subgroup of order not divisible by 3. If $S$ admits a $\tau$-automorphism, we call $S$ a $\tau$-group.
$\qquad$ a) List all $\tau$-groups up to isomorphism.
$\qquad$ b) Do $\tau$-automorphisms of odd order exist?

A nonempty class $\mathfrak{X}$ of finite groups is said to be complete if $\mathfrak{X}$ is closed under taking subgroups, homomorphic images, and extensions. The symmetric boundary of a complete class $\mathfrak{X}$ other than the class of all finite groups is defined as the largest integer $n$ such that $S_n \in \mathfrak{X}$. Every positive integer $n \neq 3$ coincides with the symmetric boundary of some complete class. It is proved (mod CFSG, D. O. Revin, Algebra i Analiz, 37, no. 1 (2025), 141–176 (Russian)) that, for every complete class $\mathfrak{X}$, there exists a nonnegative integer $m$ with the following property: for every finite group $G$ and each conjugacy class $D$ of $G$, if every $m$ elements of $D$ generate a subgroup belonging to $\mathfrak{X}$, then $\langle D \rangle \in \mathfrak{X}$. The smallest such $m$ is called the Baer–Suzuki width of $\mathfrak{X}$ denoted by $\text{BS}(\mathfrak{X})$. It is also proved (mod CFSG, ibid.) that, for a complete class $\mathfrak{X}$ of symmetric boundary $n$, the value of $\text{BS}(\mathfrak{X})$ is at least $n$ and is bounded above in terms of $n$. For every positive integer $n \neq 3$, let $f_+(n)$ and $f_-(n)$ be respectively the maximum and the minimum of $\text{BS}(\mathfrak{X})$, where $\mathfrak{X}$ runs over all complete classes of symmetric boundary $n$.
$\qquad$ a) Find $f_+(n)$ for $n = 4, 5, 6$. It is known that $f_+(1) = 2, f_+(2) = 3$, and $f_+(n) = 2(n - 1)$ for $n \geqslant 7$.
$\qquad$ b) Is it true that $f_-(n) = n$ for all $n \neq 3$? This is known to be true for $n = 1, 2, 4$.

Let $G$ be a finite group and $p$ be a prime. Let $\Psi_{p,G}$ be the class function of $G$ which vanishes on all $p$-singular elements of $G$ and whose value at each $p$-regular element $x$ of $G$ is the number of $p$-elements of $C_G(x)$.
$\qquad$ a) Is it true that $\Psi_{p,G}$ is a character of $G$?
$\qquad$ b) If yes, can $\Psi_{p,G}$ be afforded by a projective $RG$-module, where $R$ is a complete discrete valuation ring of characteristic zero such that the field of fractions of $R$ is a splitting field for $G$ and its subgroups, and the residue field $R/J(R)$ is a splitting field of characteristic $p$ for $G$ and its subgroups?

It is known that $\Psi_{p,G}$ is a character when $G \cong S_n$ for any positive integer $n$ and any prime $p$ (T. Scharf, J. Algebra, 139, no. 2 (1991), 446–457).

A finite group $G$ is called weakly ab-maximal if $|H : [H, H]| \leqslant |G : [G, G]|$ for all $H \leqslant G$. Do weakly ab-maximal groups have bounded derived length?

It is known that weakly ab-maximal groups are direct products of weakly ab-maximal $p$-groups (F. Lisi, L. Sabatini, J. Group Theory, 27 (2024), 1203–1217).

Let $C_1, \dots, C_n$ be (left or right) cosets of a finite group $G$ such that $U := C_1 \cup \dots \cup C_n$ is not $G$. Is it always true that $|G \setminus U| \geqslant |G|/2^n$?

A group is boundedly acyclic if its bounded cohomology with trivial real coefficients vanishes in all positive degrees. Is every branch group boundedly acyclic?

(a) Does there exist a finitely generated simple group that is of exponential growth but not of uniformly exponential growth?
(b) Does there exist a finitely generated hereditarily just-infinite group that is of exponential growth but not of uniformly exponential growth?

(A. Thom). Is there any group which is not isomorphic to the quotient of a residually finite group by an amenable normal subgroup?

Does there exist a group $G$ that contains a family $(G_n)_{n \in \mathbb{N}}$ of finite-index subgroups such that for every $n$ there is a homomorphism $f_n : G_n \to \mathbb{Z}$ whose kernel is of type $F_n$, but not of type $F_{n+1}$?

A pro-$p$ group is (relatively) strictly finitely presented if it is the pro-$p$ completion of a group that is finitely presented (respectively, finitely presented in some finitely-based variety of groups). A pro-$p$ group is finitely axiomatizable if it is determined up to isomorphism by a single sentence in the first-order language of group theory.
$\qquad$ a) Does there exist a (relatively) strictly finitely presented pro-$p$ group that is not finitely axiomatizable in the class of all pro-$p$ groups?
$\qquad$ b) In particular, is every finitely generated free pro-$p$ group finitely axiomatizable?

See (A. Nies, K. Tent, D. Segal, Proc. London Math. Soc. (3), 123 (2021), 597–635; D. Segal, Preprint, 2025, https://arxiv.org/abs/2505.04816).

Let $p$ be a prime number. A group $\Gamma$ is called $p$-Jordan if there exist constants $J$ and $e$ such that any finite subgroup $G \subset \Gamma$ contains a normal abelian subgroup of order coprime to $p$ and of index at most $J \cdot |G_{(p)}|^e$. (For example by the results of Brauer–Feit and Larsen–Pink, for any field $K$ of characteristic $p$ the group $\text{GL}_n(K)$ is $p$-Jordan with $e = 3$.) Let the $p$-Jordan exponent $e(\Gamma)$ of the group $\Gamma$ be the infimum of all constants $e$ for which the above bound holds for some constant $J = J(e)$.
$\qquad$ a) Is it true that this infimum is always attained?
$\qquad$ b) Is it true that $e(\Gamma) \leqslant 3$ for any $p$-Jordan group $\Gamma$?

Let $w$ be a group word, and $G$ a profinite group. Is it true that the cardinality of the set of $w$-values in $G$ is either finite or at least continuum?

Is it true that the extension of the A. Agrachev–R. Gamkrelidze construction of groups from pre-Lie rings in (J. Soviet Math., 17 (1981), 1650–1675) suggested in Definition 66 of (A. Smoktunowicz, J. Pure Appl. Algebra, 229, no. 12 (2025), 108128) produces groups from pre-Lie rings?

If this extended construction does give a group, then it also gives a brace, and so an affirmative answer to this question would have consequences for the theory of set-theoretic solutions of the Yang–Baxter equation and for the theory of braces.

A group $G$ is said to be virtually special if $G$ has a finite-index subgroup isomorphic to the fundamental group of a special complex (in the sense of F. Haglund, D. T. Wise, Geom. Funct. Anal., 17, no. 5 (2008), 1551–1620; cf 20.60.) A group $G$ is called a $\text{CAT}(0)$ group if it acts properly discontinuously and cocompactly by isometries on a $\text{CAT}(0)$ metric space.
$\qquad$ a) Is every $\text{CAT}(0)$ free-by-cyclic group virtually special?
$\qquad$ b) A weaker question: does every $\text{CAT}(0)$ free-by-cyclic group virtually embed into a right-angled Artin group?

(M. Bridson). Let $F_m$ be a free group of rank $m$ and let $\phi \in \text{Aut}(F_m)$ be a polynomially growing automorphism of maximal degree $m - 1$, which means that for some (equivalently, any) free basis $\{x_1, \dots, x_m\}$ of $F_m$, the sequence $\max_i |\phi^n(x_i)|$ grows at the rate of $n^{m-1}$, where $|g|$ denotes the minimal length of $g$ in the $x_i$ and their inverses.
$\qquad$ a) Is the free-by-cyclic group $F_m \rtimes_\phi \mathbb{Z}$ virtually special?
$\qquad$ b) In particular, are the Hydra groups
$$G_m = F_m \rtimes \mathbb{Z} = \langle a_1, \dots, a_m, t \mid t^{-1}a_1t = a_1, t^{-1}a_it = a_i a_{i-1} \text{ for all } i > 1 \rangle$$ virtually special?

(N. Brady). Do there exist finitely presented subgroups of right-angled Artin groups whose Dehn functions are super-exponential, or sub-exponential but not polynomial?

Such subgroups are known to exist in general $\text{CAT}(0)$ groups, whereas the only Dehn functions currently realized for subgroups of right-angled Artin groups are exponential and polynomial of arbitrary degree.

Let $G$ be a right-angled Artin group. Is the stable commutator length $\text{scl}(g)$ a rational number for every $g \in [G, G]$? For free groups this is true by Calegari’s Rationality Theorem. (See 18.40 for the definition of $\text{scl}(g)$.)

Two groups $G_1$ and $G_2$ are said to be commensurable if there exist finite index subgroups $H_1 \leqslant G_1$ and $H_2 \leqslant G_2$ (not necessarily of the same index) such that $H_1 \cong H_2$. Let $A[F_4]$ and $A[H_4]$ denote the Artin groups of spherical types $F_4$ and $H_4$, respectively. Are these two groups commensurable? This is the most difficult case in the classification of Artin groups of spherical type up to commensurability.

If two Artin groups of spherical type are quasi-isometric, must they be commensurable (see 21.128)? (This is not true for right-angled Artin groups.)

Conjecture: Let $G$ be a finite additive abelian group with $|G|$ odd. Then any subset $A$ of $G$ with $|A| = n > 2$ can be written as $\{a_1, \dots, a_n\}$ in such a way that all the sums $a_1 + a_2, a_2 + a_3, \dots, a_{n-1} + a_n, a_n + a_1$ are distinct.

Construct a homework of a subgroup of a Golod group (see 9.76) onto an infinite AT-group as defined in (A. V. Rozhkov, Math. Notes, 40, no. 5 (1986), 827–836). Cf. 13.55.

Based on the development of E. S. Golod’s construction (see, for example, Discrete Math. Appl., 23, no. 5–6 (2013), 491–501), for each prime number $p$, construct a finitely generated residually finite $p$-group with a non-trivial finite centre.

Such groups with infinite and trivial centres are known (see 9.76 and 11.101).

Does a group need to have a subnormal abelian series if every countable subgroup of it has such a series?

For a finite group $G$, let the type of $G$ be the function on positive integers whose value at $n$ is the number of solutions of the equation $x^n = 1$ in $G$.
$\qquad$ a) Is it true that a group having the same type as a group with trivial solvable radical must also have trivial solvable radical? Note that there are solvable and nonsolvable groups with the same type (see 12.37).
$\qquad$ b) Is it true that a group having the same type as an almost simple group must be isomorphic to it? This is true for a group having the same type as a simple group, as follows from the affirmative answer to 12.39.

For a finite group $G$, let $\chi_1(G)$ denote the totality of the degrees of all irreducible complex characters of $G$ with allowance for their multiplicities. Suppose that $H$ is a finite group with $\chi_1(H) = \chi_1(G)$. If $G$ has trivial solvable radical, must $H$ also have trivial solvable radical?

Let $G$ be a profinite group with fewer than $2^{\aleph_0}$ conjugacy classes of elements of infinite order. Must $G$ be a torsion group?
This holds in the case when $G$ is finitely generated (J. S. Wilson, Arch. Math., 120 (2023), 557–563).

If the $p$-th powers in a finite $p$-group form a subgroup, must that subgroup be powerful? That is, for $p \neq 2$, if the $p$-th powers in a $p$-group of exponent $p^2$ form a subgroup, must that subgroup be abelian? For a 2-group of exponent 8, if the squares form a subgroup, must that subgroup be abelian?

Let $G$ be an infinite finitely presented group such that every subgroup of infinite index is free. Must $G$ be isomorphic to either a free group or a surface group?

Let $G$ be a hyperbolic group which is virtually compact special in the sense of Haglund–Wise. Suppose that the set of second Betti numbers of the finite-index subgroups of $G$ is bounded. Must $G$ be virtually either a free group or a surface group?

Let $G$ be a torsion-free group of type $F_\infty$ of infinite cohomological dimension. Must $G$ contain a copy of Thompson’s group $F$?

Let $G = G_1 \amalg_H G_2$ be a free pro-$p$ product of coherent pro-$p$ groups with polycyclic amalgamation. Is $G$ coherent?

For abstract groups this is known to be true. A group is said to be coherent if each of its finitely generated subgroups is finitely presented, and in the question the coherency is used in the pro-$p$ sense.

A group $G$ is said to be invariably generated by $a$ and $b$ if $G$ is generated by the conjugates $a^g$, $b^h$ for every $g, h$. Let $p \neq q$ be fixed primes. Does every finite group embed into a finite group invariably generated by an element of order $p$ and an element of order $q$?

(Well-known problem). Is Thompson’s group $F$ automatic?

(M. Brin, M. Sapir). Conjecture: Every subgroup of Thompson’s group $F$ is either elementary amenable or else contains a subgroup isomorphic to $F$.

(M. Bridson). Is Thompson’s group $F$ quasi-isometric
$\qquad$ a) to $F \times \mathbb{Z}$?
$\qquad$ b) to $F \times F$?

(Well-known problem). A classifying space for a group $G$ is a connected CW-complex with fundamental group $G$ and all higher homotopy groups trivial. A group is of type $F_n$ if it has a classifying space with finite $n$-skeleton. For example, type $F_1$ is equivalent to finite generation, and type $F_2$ is equivalent to finite presentability. Type $F_\infty$ means type $F_n$ for all $n$.

For $n \geqslant 3$, does every group of type $F_{n-1}$ embed as a subgroup of a group of type $F_n$? Or even in a group of type $F_\infty$?

(V. M. Kopytov, N. Ya. Medvedev). A subgroup $H$ of a right-orderable group $G$ is said to be right-relatively convex if it is convex under some right ordering on $G$. Is the lattice of right-relatively convex subgroups of a right-orderable group always a sublattice of the lattice of its subgroups?

(V. M. Kopytov, N. Ya. Medvedev). Is it true that the lattice of right-relatively convex subgroups (see 21.147) of a right-orderable group is distributive if and only if it is a chain?

(V. M. Kopytov, N. Ya. Medvedev). Are there order automorphisms of Dlab groups that are not induced by conjugation by elements of a (possibly bigger) Dlab group?

Let $G$ be an extension of a normal elementary abelian subgroup $A$ by an elementary abelian group $B \cong G/A$ such that $A$ contains an element $a$ with $C_B(a) = 1$. Is it true that the rank of the subgroup $Z(\langle a, B \rangle) \cap (\langle a, B \rangle)'$ is at most the rank of $B$?