An element $g$ of a group $G$ is a non-near generator of $G$ if for every subset $S \subseteq G$ such that $|G : \langle g, S \rangle| < \infty$ it follows that $|G : \langle S \rangle| < \infty$. The set of all non-near generators of $G$ forms a characteristic subgroup of $G$ called the lower near Frattini subgroup of $G$, denoted by $\lambda(G)$. A subgroup $M \leqslant G$ is nearly maximal in $G$ if it is maximal with respect to being of infinite index in $G$. The intersection of all nearly maximal subgroups forms a characteristic subgroup called the upper near Frattini subgroup of $G$, denoted by $\mu(G)$. In general, $\lambda(G) \leqslant \mu(G)$. If $\lambda(G) = \mu(G)$, then this subgroup is called the near Frattini subgroup of $G$, denoted by $\psi(G)$.
$\qquad$ a) Is it true that $\psi(G) = 1$ if $G$ is the knot group of any product of knots?
$\qquad$ b) Is it true that $\psi(G) = 1$ if $G$ is a cable knot group?

For a subgroup $L$ of a group $G$, let $L_G$ denote the largest normal subgroup of $G$ contained in $L$. If $M \triangleleft G$, then we say that $G$ nearly splits over $M$ if there is a subgroup $N \leqslant G$ such that $|G : N| = \infty$, $|G : MN| < \infty$, and $(M \cap N)_G = 1$.

Let $G$ be any group, and $H$ a normal subgroup of prime order. Is it true that $\psi(G) \cap H = 1$ if and only if $G$ nearly splits over $H$?

Let $G = A \ast_H B$ be the generalized free product of groups $A$ and $B$ with amalgamated subgroup $H$. Is $\psi(G) = 1$ in the following cases?
$\qquad$ a) $H$ is finite cyclic and $H_G = 1$.
$\qquad$ b) $H$ is finite cyclic and either $\lambda(A) \cap H_G = 1$ or $\lambda(B) \cap H_G = 1$.
$\qquad$ c) $H_G = 1$ and $H$ satisfies the minimum condition on subgroups.
$\qquad$ d) $\lambda(G) \cap H = 1$.
$\qquad$ e) $A$ and $B$ are free groups, $H$ is finitely generated and at least one of $|A : H|$ or $|B : H|$ is infinite.
$\qquad$ f) $H$ is infinite cyclic and is a retract of $A$ and $B$.
$\qquad$ g) $G$ nearly splits over $H$, and $H$ is a normal subgroup of $G$ of prime order.

a) If $G$ is the free product of infinitely many finitely generated free groups with cyclic amalgamation, then is $\psi(G) = 1$?
b) If $G$ is the free product of infinitely many finitely generated free abelian groups with cyclic amalgamation, then is $\psi(G) = 1$?

If $G$ is the free product of infinitely many finitely generated abelian groups with amalgamated subgroup $H$, then is $\psi(G)$ equal to the torsion subgroup of $H$?

Let $G = A \ast_H B$ be the generalized free product of groups $A$ and $B$ with amalgamated subgroup $H$.
$\qquad$ a) Conjecture: If both $A$ and $B$ are nilpotent, then $\mu(G) \leqslant H$.
$\qquad$ b) Conjecture: If $G$ is residually finite, and $H$ satisfies a nontrivial identical relation, then $\lambda(G) \leqslant H$.

The group of virtual pure braids $VP_n$, $n \geqslant 2$, is generated by elements $\lambda_{ij}$, $1 \leqslant i \neq j \leqslant n$, and is defined by the relations $\lambda_{ij}\lambda_{kl} = \lambda_{kl}\lambda_{ij}$, $\lambda_{ki}\lambda_{kj}\lambda_{ij} = \lambda_{ij}\lambda_{kj}\lambda_{ki}$, where different letters denote different indices.
$\qquad$ a) Construct a normal form of words in the group $VP_n$ for $n \geqslant 4$.
$\qquad$ b) Is the group $VP_n$ linear for $n \geqslant 4$? It is known that the group $VP_3$ is linear.

A word in an alphabet $A = \{a_1, a_2, \dots, a_n\}$ is called a palindrome if it reads the same from left to right and from right to left. Let $k$ be a non-negative integer. A word in the alphabet $A$ is called an almost $k$-palindrome if it can be transformed into a palindrome by changing $\leqslant k$ letters in it. (So an almost 0-palindrome is a palindrome.) Let elements of a free group $F_2 = \langle x, y \rangle$ be represented as words in the alphabet $\{x^{\pm 1}, y^{\pm 1}\}$. Do there exist positive integers $m$ and $c$ such that every element in $F_2$ is a product of $\leqslant c$ almost $m$-palindromes?

It is known that for $m = 0$ there is no such a number $c$.

Let $G$ be a finitely generated linear group.
$\qquad$ a) Is it true that the non-abelian tensor square $G \otimes G$ is a linear group?
$\qquad$ b) In particular, is this true for the braid group $B_n$ for $n > 3$?

It is known that $B_3 \otimes B_3$ is linear, and there is a countable group $G$ such that $G \otimes G$ is not linear. See the definition of the non-abelian tensor square in (R. Brown, J.-L. Loday, Topology, 26, no. 3 (1987), 311–335).

a) Is it possible to embed a finitely generated non-abelian free pro-$p$ group as an open subgroup of a simple totally disconnected locally compact group?
b) If the answer is yes, can we require that the simple envelope is also compactly generated?

Does there exist a constant $c$ such that the number of conjugacy classes in a finite group $G$ is always at least $c \log_2 |G|$?

A finite group $G$ is called conjugacy-expansive if for every normal subset $N$ and conjugacy class $C$ of $G$ the normal set $NC$ contains at least as many conjugacy classes of $G$ as $N$ does. Is it true that every finite simple group is conjugacy-expansive?

The well-known Baer–Suzuki theorem states that if every two conjugates of an element $a$ of a finite group $G$ generate a finite $p$-subgroup, then $a$ is contained in a normal $p$-subgroup. Does such a theorem hold in the class of periodic groups for $p > 2$? A counterexample is known for $p = 2$; see 11.11 a).

Let $G$ be a finite group such that the quasivariety generated by all its Sylow subgroups contains only finitely many subquasivarieties. Is it true that the quasivariety generated by $G$ also contains only finitely many subquasivarieties?

For a class of groups $\mathcal{M}$, let $L(\mathcal{M})$ denote the class of all groups $G$ in which the normal closure $\langle a \rangle^G $ of any element $a \in G$ is contained in $\mathcal{M}$. Let $qA$ be the quasivariety generated by a finite nilpotent group $A$. Can the quasivariety $L(qA)$ contain a non-nilpotent group?

(C. Drutu and M. Sapir). Is every free-by-cyclic group of the form $F_n \rtimes \mathbb{Z}$ linear?

(Well-known problem). Suppose that $G$ is a finitely presented group such that the set of first Betti numbers over all the finite index subgroups of $G$ is unbounded above. (Here, the first Betti number is the maximum $n$ for which there is a surjective homomorphism onto $\mathbb{Z}^n$.)
$\qquad$ a) Must $G$ have a finite index subgroup with a surjective homomorphism to a non-abelian free group?
$\qquad$ b) Can $G$ even be soluble?

Given an infinite set $\Omega$, define an algebra $A$ (the reduced incidence algebra of finite subsets) as follows. Let $V_n$ be the set of functions from the set of $n$-element subsets of $\Omega$ to the rationals $\mathbb{Q}$. Now let $A = \bigoplus V_n$, with multiplication as follows: for $f \in V_n$, $g \in V_m$, and $|X| = m + n$, let $(fg)(X) = \sum f(Y)g(X \setminus Y)$, where the sum is over the $n$-element subsets $Y$ of $X$. If $G$ is a permutation group on $\Omega$, let $A^G$ be the algebra of $G$-invariants in $A$.

Suppose that $G$ has no finite orbits on $\Omega$. It is known that then $A^G$ is an integral domain (M. Pouzet, Theor. Inf. App., 42, no. 1 (2008), 83–103). Is it true that the quotient of $A^G$ by the ideal generated by constant functions is also an integral domain?

A finite transitive permutation group $G$ is said to have the road closure property if, given any orbit $O$ of $G$ on 2-sets, and any proper block of imprimitivity for $G$ acting on $O$, the graph with edge set $O \setminus B$ is connected. Such a group must be primitive, and basic (not contained in a wreath product with the product action); it cannot have an imprimitive subgroup of index 2. In addition, such a group cannot be one of the permutation groups arising from triality (whose socle is $D_4(q)$ and intersects the point stabiliser in the parabolic subgroup corresponding to the three leaves in the Coxeter–Dynkin diagram for $D_4$).

Classify the basic primitive groups $G$ which do not have the road closure property. In particular, is it true that such a group either has a subgroup of index at most 3 or is almost simple?

For a finite group $G$, let $\text{End}(G)$ denote the semigroup of endomorphisms of $G$, and $\text{PIso}(G)$ the semigroup of partial isomorphisms of $G$ (isomorphisms between subgroups of $G$). If $G$ is abelian, then $|\text{End}(G)| = |\text{PIso}(G)|$. Is the converse true?

Can a non-discrete, compactly generated, topologically simple, locally compact group be amenable and non-compact? (A positive answer implies a negative answer to 19.107.)

Let $\mathcal{S}$ denote the class of nondiscrete compactly generated, topologically simple totally disconnected locally compact groups. Two topological groups are called locally isomorphic if they contain isomorphic open subgroups. Is the number of local isomorphism classes of groups in $\mathcal{S}$ uncountable?

For a group $G$, let $\text{Tor}_1(G)$ be the normal closure of all torsion elements of $G$, and then by induction let $\text{Tor}_{i+1}(G)$ be the inverse image of $\text{Tor}_1(G/\text{Tor}_i(G))$. The torsion length of $G$ is defined to be either the least positive integer $l$ such that $G/\text{Tor}_l(G)$ is torsion-free, or $\omega$ if no such integer exists (since $G/\bigcup \text{Tor}_i(G)$ is always torsion-free).

Does there exists a finitely generated, or even finitely presented, soluble group with torsion length greater than 2?

For a group $G$, let $\text{Tor}(G)$ be the normal closure of all torsion elements of $G$. Does there exist a finitely presented group $G$ such that $G/\text{Tor}(G)$ is not finitely presented? Such a group must necessarily be non-hyperbolic.

Let $G$ and $H$ be finite groups of the same order with $\sum_{g \in G} \phi(|g|) = \sum_{h \in H} \phi(|h|)$, where $\phi$ is the Euler totient function. Suppose that $G$ is simple. Is $H$ necessarily simple?

(Y. O. Hamidoune). Suppose that $A$ and $B$ are finite subsets of a group $G$ such that $|A| \geqslant 2$ and $|B| \geqslant 2$, and let $A \cdot_2 B$ denote the set of elements of $G$ which can be expressed in the form $ab$ for at least two different $(a, b) \in A \times B$. Is it true that $|A| + |B| - \frac{1}{2}|AB| - \frac{1}{2}|A \cdot_2 B| \leqslant \max\{2, |gH| \mid H \leqslant G, g \in G, gH \subseteq A \cdot_2 B\}$?

(Well-known question). A finitely generated group $G$ that acts on a tree in such a way that all vertex and edge stabilizers are infinite cyclic groups is called a generalized Baumslag–Solitar group. The Bass–Serre theory gives finite presentations of such groups. Is the isomorphism problem soluble for generalized Baumslag–Solitar groups?

Let $G$ be a group, and $\phi$ an automorphism of $G$. Elements $x, y \in G$ are said to be $\phi$-conjugate if $x = z^{-1} y \phi(z)$ for some $z \in G$. The $\phi$-conjugacy is an equivalence relation; the number of $\phi$-conjugacy classes is denoted by $R(\phi)$.

Conjecture: If a finitely generated residually finite group $G$ has an automorphism $\phi$ such that $R(\phi)$ is finite, then $G$ has a soluble subgroup of finite index.

The conjecture is proved if $\phi$ has prime order (E. Jabara, J. Algebra, 320, no. 10 (2008), 3671–3679). If $G$ is infinitely generated, then $G$ does not have to be almost solvable (K. Dekimpe, D. Gonçalves, Bull. London Math. Soc., 46, no. 4 (2014), 737–746), but if it is of finite upper rank, then it has to be almost solvable (E. Troitsky, J. Group Theory, 28, no. 1 (2025), 151–164).

Let $\Phi \in \text{Out}\,G = \text{Aut}\,G/\text{Inn}\,G$. Two automorphisms $\varphi, \psi \in \Phi$ are said to be isogradient if $\varphi = \hat{g}^{-1} \psi \hat{g}$ for some inner automorphism $\hat{g}$. The number of isogradiency classes in $\Phi$ is denoted by $S(\Phi)$.

Conjecture: If a finitely generated residually finite group $G$ has an outer automorphism $\Phi$ such that $S(\Phi)$ is finite, then $G$ has a soluble subgroup of finite index.

An element $g$ of a finite group $G$ is said to be vanishing if $\chi(g) = 0$ for some ordinary irreducible complex character $\chi \in \text{Irr}(G)$. Must a finite group and a finite simple group be isomorphic if they have equal orders and the same set of orders of vanishing elements?

Let $\omega(G, S)$ be the exponential growth rate of a group $G$ with a finite generating set $S$ (see 14.7). Let $\text{MCG}(\Sigma_g)$ be the mapping class group of the orientable surface $\Sigma_g$ of genus $g$. Is there a constant $C > 1$ such that $\omega(\text{MCG}(\Sigma_g), S) \geqslant C$ for every $g > 0$ and every finite generating set $S$ of $\text{MCG}(\Sigma_g)$?

Let $p \geqslant 673$ be a prime and $r \geqslant 2$ be an integer. Let $B$ be the free group in the Burnside variety of exponent $p$ on $a_1, \dots, a_r$. Must every non-identical one-variable equation $w(a_1, \dots, a_r, x) = 1$ over $B$ have at most finitely many solutions in $B$?

Conjecture: Let $G$ be a finite group, $p$ a prime number, and $P$ a Sylow $p$-subgroup of $G$. Suppose that an irreducible ordinary character $\chi$ of $G$ has degree divisible by $p$. If the restriction $\chi_P$ of $\chi$ to $P$ has a linear constituent, then $\chi_P$ has at least $p$ different linear constituents.

Let $G$ be a finitely generated group such that for every element $g \in G$ the set of commutators $\{[x, g] \mid x \in G\}$ is a subgroup of $G$. Is it true that $G$ is residually nilpotent?

Let $G$ be a finite group of order $n$. Is it true that for every factorization $n = a_1 \cdots a_k$ there exist subsets $A_1, \dots, A_k$ such that $|A_1| = a_1, \dots, |A_k| = a_k$ and $G = A_1 \cdots A_k$?

Let $G$ be a periodic group and let $\mathscr{I} = \{x \in G \mid x^2 = 1 \neq x\}$ be the set of its involutions. Let $D$ be a non-empty set of odd integers greater than 1; then $G$ is called a group with $D$-involutions if $G = \langle \mathscr{I} \rangle$ and for $x, y \in \mathscr{I}$ the order of $xy$ is in the set $\{1, 2\} \cup D$ and all these values actually occur. It is clear that if $G$ is a group with $D$-involutions, then $\mathscr{I}$ is a single conjugacy class.

Conjecture: If $G$ is a group with $\{3, 5\}$-involutions, then $G \cong A_5$ or $G \cong PSU(3, 4)$.

a) Does there exist an absolute constant $k$ such that for any nilpotent injector $H$ of an arbitrary finite group $G$ there are $k$ conjugates of $H$ the intersection of which is equal to the Fitting subgroup $F(G)$ of $G$?
b) Can one choose $k = 3$ as such a constant? This is true for finite soluble groups (D. S. Passman, Trans. Amer. Math. Soc., 123, no. 1 (1966), 99–111; A. Mann, Proc. Amer. Math. Soc., 53, no. 1 (1975), 262–264).

Suppose that $H$ is a subgroup of a finite soluble group $G$ that covers all Frattini chief factors of $G$ and avoids all complemented chief factors of $G$. Is it true that there are elements $x, y \in G$ such that $H \cap H^x \cap H^y = H_G$, where $H_G$ is the largest normal subgroup of $G$ contained in $H$? This is true if $H$ is a prefrattini subgroup of $G$.

A group is said to be $\mathfrak{X}$-critical if it does not belong to $\mathfrak{X}$ but all its proper subgroups belong to $\mathfrak{X}$. Let $\mathfrak{F}$ be a soluble hereditary formation of finite groups for which all $\mathfrak{F}$-critical groups are either groups of prime order or $\mathfrak{U}$-critical groups, where $\mathfrak{U}$ is the formation of all supersoluble finite groups. Must $\mathfrak{F}$ be a saturated formation?

Does Thompson's group $F$ (see 12.20) have the Howson property, that is, is the intersection of any two finitely generated subgroups of $F$ finitely generated?

Let $\phi$ be an automorphism of a free group $F_n$ of finite rank, and let $F_n \rtimes_\phi \mathbb{Z}$ be the split extension of $F_n$ by $\mathbb{Z}$ with $\mathbb{Z}$ acting as $\langle \phi \rangle$. Is the group $F_n \rtimes_\phi \mathbb{Z}$ conjugacy separable?

Suppose that $H$ is a word-hyperbolic subgroup of a word-hyperbolic group $G$ such that the inclusion of $H$ to $G$ extends to a continuous $H$-equivariant map $j : \partial H \to \partial G$ between their hyperbolic boundaries. If such an extension exists, it is unique and $j$ is called the Cannon–Thurston map.
$\qquad$ a) Is it true that for every point $p \in \partial G$ its full preimage $j^{-1}(p)$ is finite?
$\qquad$ b) Moreover, is it true that there is a number $N = N(G, H) < \infty$ such that for every $p \in \partial G$ the full preimage $j^{-1}(p)$ consists of at most $N$ points?

Suppose that $\varphi$ is an automorphism of a finite soluble group $G$. Is the Fitting height of $G$ bounded in terms of $|\varphi|$ and $|C_G(\varphi)|$?

By definition a profinite group has finite rank at most $r$ if every subgroup of it can be (topologically) generated by $r$ elements. Suppose that for every element $g$ of a profinite group $G$ there is a closed subgroup $E_g$ of finite rank such that for every $x \in G$ all sufficiently long Engel commutators $[x, g, \dots, g]$ belong to $E_g$, that is, for every $x \in G$ there is a positive integer $n(x, g)$ such that $[x, {}_n g] \in E_g$ whenever $g$ is repeated $\geqslant n(x, g)$ times. Is it true that $G$ has a normal subgroup $N$ of finite rank with locally nilpotent quotient $G/N$?

Let $G$ be a group generated by 3 class transpositions (see the definition in 17.57), and let $m$ be the least common multiple of the moduli of the residue classes interchanged by the generators of $G$. Assume that $G$ does not setwisely stabilize any union of residue classes modulo $m$ except for $\varnothing$ and $\mathbb{Z}$, and assume that the integers $0, 1, \dots, 42$ all lie in the same orbit under the action of $G$ on $\mathbb{Z}$. Is the action of $G$ on $\mathbb{N} \cup \{0\}$ necessarily transitive?

The bound 42 cannot be replaced by a smaller number, since the finite group $\langle \tau_{0(2), 1(2)}, \tau_{0(3), 2(3)}, \tau_{0(7), 6(7)} \rangle$ acts transitively on the set $\{0, \dots, 41\}$, as well as on the set of residue classes modulo 42.

Does the group $\text{CT}(\mathbb{Z})$ have finitely generated infinite periodic subgroups? (See the definition of $\text{CT}(\mathbb{Z})$ in 17.57).

Let $K$ be a finite extension of degree $n$ of a field $k$ of odd characteristic. The multiplicative group $K^*$ embeds into the group of all $k$-linear automorphisms $\text{Aut}_k(K)$ by the rule $t(x) = tx$ for all $x \in K$. For a fixed basis of $K$ over $k$, the group $\text{Aut}_k(K)$ is isomorphic to $\text{GL}(n, k)$. The image of $K^*$ is called a nonsplit maximal torus corresponding to the extension $K/k$ and is denoted by $T(K/k)$. A subgroup of $\text{GL}(n, k)$ is said to be rich in transvections if it contains all elementary transvections.

Let $H$ be a subgroup of $\text{GL}(n, k)$ which contains $T(K/k)$ and a one-dimensional transformation. Is $H$ rich in transvections?

A system of additive subgroups $\sigma_{ij}$, $1 \leqslant i, j \leqslant n$, of a field $K$ is called a net (or a carpet) of order $n$ if $\sigma_{ir}\sigma_{rj} \subseteq \sigma_{ij}$ for all $i, r, j$. A net that does not contain the diagonal is called an elementary net. A net $\sigma = (\sigma_{ij})$ is said to be irreducible if all $\sigma_{ij}$ are nontrivial. A net $\sigma$ is said to be closed if the elementary net subgroup $E(\sigma)$ does not contain additional elementary transvections.
Let $R$ be a principal ideal domain with $1 \in R$, let $k$ be the field of fractions of $R$, and $K$ an algebraic extension of the field $k$.

Let $\sigma = (\sigma_{ij})$ be an irreducible elementary net over $K$ such that all $\sigma_{ij}$ are $R$-modules. Is the net $\sigma$ closed?

A skew brace is a set $B$ equipped with two operations $+$ and $\cdot$ such that $(B, +)$ is an additively written (but not necessarily abelian) group, $(B, \cdot)$ is a multiplicatively written group, and $a \cdot (b + c) = ab - a + ac$ for any $a, b, c \in B$.

Let $A$ be a skew brace with left-orderable multiplicative group. Is the additive group of $A$ left-orderable?

A finite graph is said to be integral if all eigenvalues of its adjacency matrix are integers.
$\qquad$ a) Let $G$ be a finite group generated by a normal subset $R$ consisting of involutions. Is it true that the Cayley graph $Cay(G, R)$ is integral?
$\qquad$ b) Let $A_n$ be the alternating group of degree $n$, let $S = \{(123), (124), \dots, (12n)\}$ and $R = S \cup S^{-1}$. Is it true that the Cayley graph $Cay(A_n, R)$ is integral?

A finite group is monomial if each irreducible character of it is induced from a linear character of some subgroup. Monomial groups are soluble (Taketa). The group is normally (subnormally) monomial if each irreducible character is induced from a linear character of some normal (respectively, subnormal) subgroup. Metabelian groups are normally monomial, and abelian-by-nilpotent groups are subnormally monomial. In the other direction, it is known that there exist normally monomial groups of arbitrarily large derived length.
$\qquad$ a) For a given prime $p$, do there exist normally monomial finite $p$-groups of arbitrarily large derived length?
$\qquad$ b) Do there exist subnormally monomial groups of arbitrarily large nilpotence length?

The Gruenberg–Kegel graph (or the prime graph) $GK(G)$ of a finite group $G$ has vertex set consisting of all prime divisors of the order of $G$, and different vertices $p$ and $q$ are adjacent in $GK(G)$ if and only if the number $pq$ is an element order in $G$. Is there a finite non-solvable group $G$ such that $GK(G)$ does not contain 3-cocliques and is not isomorphic to the Gruenberg–Kegel graph of any finite solvable group?

It is known that there are no examples of such groups $G$ among almost simple groups.

Let $G$ be a group generated by elements $x, y, z$ such that $x^3 = y^2 = z^2 = (xy)^3 = (yz)^3 = 1$ and $g^{12} = 1$ for all $g \in G$. Is it true that $|G| \leqslant 12$?

What are the chief factors of a finite group in which every 2-maximal subgroup is not $m$-maximal for any $m \geqslant 3$?

A subgroup $H$ of a group $G$ is said to be $m$-maximal if there is a chain of subgroups $H = H_0 < H_1 < \dots < H_{m-1} < H_m = G$ in which $H_i$ is maximal in $H_{i+1}$ for every $i$. Note that for every $m \geqslant 3$ there is a finite group in which some 2-maximal subgroup is $m$-maximal.

Suppose that in a finite group $G$ every maximal subgroup $M$ is supersoluble whenever $\pi(M) = \pi(G)$, where $\pi(G)$ is the set of all prime divisors of the order of $G$.
$\qquad$ a) What are the non-abelian composition factors of $G$?
$\qquad$ b) Determine the exact upper bounds for the nilpotency length, the rank, and the $p$-length of $G$ if $G$ is soluble.

A $\star$-commutator is a commutator $[x, y]$ of two elements $x, y$ of coprime prime-power orders. Is a finite group $G$ soluble if $|ab| \geqslant |a||b|$ for any $\star$-commutators $a$ and $b$ of coprime orders?

What are the non-abelian composition factors of a finite group in which every maximal subgroup is simple or $p$-nilpotent for some fixed odd prime $p \in \pi(G)$?

What are the non-abelian composition factors of a finite group in which every maximal subgroup is simple or $p$-decomposable for some fixed odd prime $p \in \pi(G)$?

Does the group of isometries of $\mathbb{Q}^3$ have a free non-abelian subgroup such that every non-trivial element acts without nontrivial fixed points in $\mathbb{Q}^3$?

The answer is yes if the field of rational numbers $\mathbb{Q}$ is replaced by the field $\mathbb{R}$ of real numbers (see G. Tomkowicz, S. Wagon, The Banach–Tarski Paradox, Cambridge Univ. Press, 2016.)

Let $R$ be a principal ideal domain, let $d$ be a positive integer, and let $X_1, \dots, X_m$ be members of $\text{SL}(d, R)$ (or of $\text{GL}(d, R)$). Is it decidable whether or not these $m$ matrices freely generate a free group? If it is, design an effective algorithm for computing the answer.

Let $\mathfrak{A} = \{\mathfrak{A}_r \mid r \in \Phi\}$ be an elementary carpet of type $\Phi$ over a commutative ring $K$ (see 7.28), and let $\Phi(\mathfrak{A}) = \langle xr(\mathfrak{A}_r) \mid r \in \Phi\rangle$ be its carpet subgroup. Define the closure of the carpet $\mathfrak{A}$ to be the set of additive subgroups $\overline{\mathfrak{A}} = \{\overline{\mathfrak{A}}_r \mid r \in \Phi\}$, where $\overline{\mathfrak{A}}_r = \{t \in K \mid xr(t) \in \Phi(\mathfrak{A})\}$. Is the closure $\overline{\mathfrak{A}}$ of a carpet $\mathfrak{A}$ always a carpet?

Let $\mathfrak{A} = \{\mathfrak{A}_r \mid r \in \Phi\}$ be an elementary carpet of type $\Phi$ of rank $l \geqslant 2$ (see 7.28). For $p \in \Phi$, define a set of additive subgroups $\mathfrak{B}_p = \sum c_{ij,rs}\mathfrak{A}_r^i\mathfrak{A}_s^j$, where the sum is taken over all natural numbers $i, j$ and roots $r, s \in \Phi$ such that $ir + js = p$. It is known that the set $\mathfrak{B} = \{\mathfrak{B}_p \mid p \in \Phi\}$ is a carpet called the derived carpet of $\mathfrak{A}$. It is also known that for $\Phi = A_l$ the set $\mathfrak{B}$ is a closed (admissible) carpet, which means that its carpet subgroup does not contain new root elements. Is every derived carpet of type $\Phi$ over a commutative ring closed (admissible)?

Let $\mathfrak{A} = \{\mathfrak{A}_r \mid r \in \Phi\}$ be an elementary carpet of type $\Phi$ over a commutative ring $K$ (see 7.28) and let $\mathfrak{A}_r^2 = \{t^2 \mid t \in \mathfrak{A}_r\}$. Are the inclusions $\mathfrak{A}_r^2\mathfrak{A}_{-r} \subseteq \mathfrak{A}_r$, $r \in \Phi$, sufficient for the carpet $\mathfrak{A}$ to be closed (admissible)?

Let $G$ be a group, and $(g_1, \dots, g_n)$ a tuple of its elements. The type of this tuple in $G$, denoted $Tp^G(g_1, \dots, g_n)$, is the set of all first order formulas in free variables $x_1, \dots, x_n$ in the standard group theory language which are true on $(g_1, \dots, g_n)$ in $G$. Two groups $G$ and $H$ are called isotypic if for every tuple of elements $\bar{h} = (h_1, \dots, h_n)$ in $H$ there is a tuple $\bar{g} = (g_1, \dots, g_n)$ in $G$, such that $Tp(\bar{h}) = Tp(\bar{g})$ and vice versa, for every tuple $\bar{g}$ in $G$ there is a tuple $\bar{h}$ in $H$ such that $Tp(\bar{h}) = Tp(\bar{g})$. Is it true that every two isotypic finitely generated groups are isomorphic?

It is well known that varieties of groups form a free semigroup $N$ (A. L. Shmel’kin, DAN SSSR, 149 (1963), 543–545 (Russian)); B. H. Neuman, H. Neumann, P. M. Neumann, Math. Z., 80 (1962), 44–62). Varieties of linear representations over an infinite field of zero characteristic also form a free semigroup $M$, and $N$ acts freely on $M$ (B. I. Plotkin, Siberian Math. J., 13, no. 5 (1972), 713–729)

A similar theorem holds for Lie algebras (V. A. Parfenov, Algebra Logika, 6, no. 4 (1967), 61–73 (Russian); L. A. Simonyan, Siberian Math. J., 29, no. 2 (1988), 276–283). Are there other varieties of algebras $\Theta$ where the same situation takes place?

We say that a variety $\Theta$ is of Tarski type if any two non-abelian $\Theta$-free groups of finite rank are elementarily equivalent.
$\qquad$ a) Find examples of Tarski type varieties distinct from the variety of all groups.
$\qquad$ b) Is it true that the Burnside variety $\mathfrak{B}_n$ of all groups of exponent $n$, where $n$ is big enough, is of Tarski type?
$\qquad$ c) Is it true that the $n$-Engel variety $\mathfrak{E}_n$ of all groups satisfying the identity $[[[x, y], y], \dots, y] \equiv 1$ ($n$ copies of $y$), where $n$ is big enough, is of Tarski type?

Let $G \leqslant \text{Sym}(\Omega)$, where $\Omega$ is finite. The 2-closure $G^{(2)}$ of the group $G$ is defined to be the largest subgroup of $\text{Sym}(\Omega)$ containing $G$ which has the same orbits as $G$ in the induced action on $\Omega \times \Omega$. Is it true that if $G$ is solvable, then every composition factor of $G^{(2)}$ is either a cyclic or an alternating group?

For a finite group $G$ and a permutation group $K$, let $b_G(K)$ denote the number of conjugacy classes of regular subgroups of $K$ isomorphic to $G$. Does there exist a function $f$ such that $b_G(K) \leqslant n^{f(r)}$ for every abelian group $G$ of order $n$ and rank $r$, and every group $K$ such that $K^{(2)} = K$? (See 19.67 for the definition of $K^{(2)}$.)

(P. Wesolek). The acronym tdlc stands for totally disconnected and locally compact, and tdlcsc for tdlc and second countable. Let $\text{Res}(G)$ denote the intersection of all open normal subgroups of a topological group $G$. The class of elementary tdlcsc groups is defined as the smallest class $\mathcal{E}$ of tdlcsc groups such that (1) $\mathcal{E}$ contains all second countable profinite groups and countable discrete groups; (2) $\mathcal{E}$ is closed under taking closed subgroups, Hausdorff quotients, directed unions of open subgroups, and group extensions.

This class admits a well-behaved, ordinal-valued decomposition rank $\xi$ defined recursively as follows: $\xi(\{1\}) = 1$, and if $G \in \mathcal{E}$ is a union of an increasing sequence $(O_i)$ of compactly generated open subgroups, then $\xi(G) = \sup_i \{\xi(\text{Res}(O_i))\} + 1$.

What is the supremum of the decomposition ranks of elementary tdlcsc groups? In particular, is it countable?

(P. Wesolek and P.-E. Caprace). Let $\mathcal{S}$ denote the class of nondiscrete compactly generated, topologically simple tdlc groups. Let $G$ be a non-elementary tdlcsc group. Is there a compactly generated closed subgroup $H \leqslant G$ such that $H$ has a continuous quotient in $\mathcal{S}$?

(G. Willis). Can a group $G$ in $\mathcal{S}$ as defined in 19.70 be such that every element of $G$ normalizes a compact open subgroup of $G$?

Let $G$ be a group in $\mathcal{S}$ as defined in 19.70. Can every element of $G$ have trivial contraction group?

(P.-E. Caprace and N. Monod). Is there a compactly generated, locally compact group that is topologically simple, but not abstractly simple?

A subgroup $H$ of a group $G$ is called pronormal if $H$ and $H^g$ are conjugate in $\langle H, H^g \rangle$ for every $g \in G$. A subgroup $H$ of a group $G$ is called abnormal if $g \in \langle H, H^g \rangle$ for every $g \in G$. Does there exist an infinite group that does not contain nontrivial proper pronormal subgroups?

The question is equivalent to the following: does there exist an infinite simple group that does not contain proper abnormal subgroups?

Let $P$ be a finite 2-group of exponent $2^e$ such that the rank of every abelian subgroup is at most $r$. Is it true that $|P| \leqslant 2^{r(e+1)}$? This bound would be sharp (for a direct product of quaternion groups).

A semigroup presentation is called tree-like if all relations have the form $a = bc$ where $a, b, c$ are letters and no two relations share the left-hand side or the right-hand side. Is it decidable whether the semigroup given by a finite tree-like presentation contains an idempotent?

This is equivalent to the question whether the closure of a finitely generated subgroup of R. Thompson’s group $F$ contains an isomorphic copy of $F$.

The Stable Small Cancellation Conjecture: If $F$ is a free group of rank $r \geqslant 2$, then, for any fixed $n > 0$, there exists a generic subset $S$ of $F^n$ such that, for any automorphism $\phi$ of $F$, the set $\phi(S)$ satisfies the small cancellation condition $C'(1/6)$.

Does there exist an infinite simple subgroup of $\text{SL}(2, \mathbb{Q})$?

This is a special case of a question which Serge Cantat asked me about “non algebraic” simple subgroups of $\text{GL}(n, \mathbb{C})$. It is also a special case of 15.57.

(T. Springer). Let $G$ be a group. Suppose that for every integer $n > 0$ the group $G$ has a unique (up to isomorphism) irreducible complex linear representation of dimension $n$. (Note that $G = \text{SL}(2, \mathbb{Q})$ has these properties, by a theorem of Borel–Tits (Ann. Math., 97 (1973), 499–571).) Is it true that $G$ has a normal subgroup $N$ such that $G/N$ is isomorphic to $\text{SL}(2, \mathbb{Q})$?

For a periodic group $G$, let $\pi_e(G)$ denote the set of orders of elements of $G$. A periodic group $G$ is said to be an $OC_n$-group if $\pi_e(G) = \{1, 2, \dots, n\}$. Is it true that every $OC_7$-group is isomorphic to the alternating group $A_7$? For finite groups, the answer is affirmative.

(Well-known problem). Is the conjugacy problem in the braid group $B_n$ in the class NP (that is, decidable in nondeterministic polynomial time with respect to the maximum of the lengths $|u|, |v|$, where $u, v$ are the input braid words)? A stronger question: given two conjugate elements of $B_n$ represented by braid words of lengths $\leqslant m$, is there a conjugator whose length is bounded by a polynomial function of $m$?

Let $h^*(G)$ denote the generalized Fitting height of a finite group $G$ defined as the minimum number $k$ such that $F^*_k(G) = G$, where $F^*_1(G) = F^*(G)$ is the generalized Fitting subgroup of $G$, and by induction $F^*_{i+1}(G)$ is the inverse image of $F^*(G/F^*_i(G))$. If $G$ is soluble, then $h^*(G) = h(G)$ is the Fitting height of $G$. Does every finite group $G$ contain a soluble subgroup $K$ such that $h^*(G) = h(K)$?

An element $g$ of a group $G$ is almost Engel if there is a finite set $\mathcal{E}(g)$ such that for every $x \in G$ all sufficiently long commutators $[x, {}_n g]$ belong to $\mathcal{E}(g)$, that is, for every $x \in G$ there is a positive integer $n(x, g)$ such that $[x, {}_n g] \in \mathcal{E}(g)$ whenever $n \geqslant n(x, g)$. By a linear group we understand a subgroup of $\text{GL}(m, F)$ for some field $F$ and a positive integer $m$. Is the set of almost Engel elements in a linear group always a subgroup?

Let $\mathbb{P}$ be the set of all primes, and let $\sigma = \{\sigma_i \mid i \in I\}$ be some partition of $\mathbb{P}$ into disjoint subsets. A finite group $G$ is said to be $\sigma$-primary if $G$ is a $\sigma_i$-group for some $i$; $\sigma$-nilpotent if $G$ is a direct product of $\sigma$-primary groups; $\sigma$-soluble if every chief factor of $G$ is $\sigma$-primary. A subgroup $A$ of a finite group $G$ is said to be $\sigma$-subnormal in $G$ if there is a chain $A = A_0 \leqslant A_1 \leqslant \dots \leqslant A_n = G$ such that for every $i$ either $A_{i-1} \trianglelefteq A_i$ or $A_i/(A_{i-1})_{A_i}$ is $\sigma$-primary, where $(A_{i-1})_{A_i}$ is the largest normal subgroup of $A_i$ contained in $A_{i-1}$. Suppose that a subgroup $A$ of a finite group $G$ is $\sigma$-subnormal in $\langle A, A^x \rangle$ for all $x \in G$. Is it true that then $A$ is $\sigma$-subnormal in $G$?

Suppose that every Schmidt subgroup of a finite group $G$ is $\sigma$-subnormal in $G$ (see 19.84). Is it true that then there is a normal $\sigma$-nilpotent subgroup $N \leqslant G$ such that $G/N$ is cyclic?

Suppose that a finite group $G$ has a $\sigma_i$-Hall subgroup for every $i \in I$. Suppose that a subgroup $A \leqslant G$ is such that $A \cap H$ is a $\sigma_i$-Hall subgroup of $A$ for every $i \in I$ and every $\sigma_i$-Hall subgroup $H$ of $G$ (see 19.84). Is it true that then $A$ is $\sigma$-subnormal in $G$?

Suppose that for every Sylow subgroup $P$ of a finite group $G$ and every maximal subgroup $V$ of $P$ there is a $\sigma$-soluble subgroup $T$ such that $VT = G$. Is it true that then $G$ is $\sigma$-soluble?

Suppose that for every Sylow subgroup $P$ of a finite group $G$ and every maximal subgroup $V$ of $P$ there is a $\sigma$-nilpotent subgroup $T$ such that $VT = G$. Is it true that then $G$ is $\sigma$-nilpotent?

A permutation group is subdegree-finite if every orbit of every point stabiliser is finite. Let $\Omega$ be countably infinite, and let $G$ be a closed and subdegree-finite subgroup of $\text{Sym}(\Omega)$ regarded as a topological group under the topology of pointwise convergence.

Conjecture: If the minimal degree of $G$ is infinite, then there is some subset of $\Omega$ whose setwise stabiliser in $G$ is trivial.

Note that this conjecture implies Tom Tucker’s well-known Infinite Motion Conjecture for graphs.

A skew brace is a set $B$ equipped with two operations $+$ and $\cdot$ such that $(B, +)$ is an additively written (but not necessarily abelian) group, $(B, \,\cdot\,)$ is a multiplicatively written group, and $a \cdot (b + c) = ab - a + ac$ for any $a, b, c \in B$.
$\qquad$ a) Is there a skew brace with soluble additive group but non-soluble multiplicative group?
$\qquad$ b) Is there a skew brace with non-soluble additive group but nilpotent multiplicative group?
$\qquad$ c) Is there a finite skew brace with soluble additive group but non-soluble multiplicative group?
$\qquad$ d) Is there a finite skew brace with non-soluble additive group but nilpotent multiplicative group?

Let $G$ be a finite group with an abelian Sylow $p$-subgroup $A$. Suppose that $B$ is a strongly closed elementary abelian subgroup of $A$. Without invoking the Classification Theorem for Finite Simple Groups (CFSG), prove that $G$ has a normal subgroup $N$ such that $B = \Omega_1(A \cap N)$.

For $p = 2$, this is a corollary of a theorem of Goldschmidt. For $p$ odd, this has been proved by Flores and Foote (Adv. Math., 222 (2009), 453–484), but their proof relies on CFSG. A CFSG-free proof in the special case when $p = 3$ and $A$ has 3-rank 3 would already be quite interesting. This case arises in Aschbacher’s treatment of the $e(G) = 3$ problem, and a proof would provide an alternative to part of his argument. (For this application, one could assume that all proper simple sections of $G$ are known.)

Let $A$ be the algebra of $3 \times 3$ skew-Hermitian matrices over the real octonions, where multiplication is given by bracket product. Determine $\text{Aut}(A)$.

The question might be of interest for octonion algebras over a different base field or ring. The question was inspired by an observation of John Faulkner concerning a possible connection between $A$ or a related algebra and the Dwyer–Wilkerson 2-compact group $\text{BDI}(4)$.

Conjecture: There exists a function $f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ such that, if $G$ is a finite $p$-group with $d$ generators and $G$ has no epimorphic images isomorphic to the wreath product $C_p \wr C_p$, then each factor of the $p$-lower central series of $G$ has order bounded above by $f(p, d)$.

It is interesting to compare this conjecture with the celebrated characterization of Aner Shalev of finitely generated $p$-adic analytic pro-$p$-groups.

By definition two elements $g_1$ and $g_2$ of a free metabelian group $F$ with basis $\{x_1, \dots, x_n\}$ have the same distribution if the equations $g_1(x_1, \dots, x_n) = g$ and $g_2(x_1, \dots, x_n) = g$ have the same number of solutions in any finite metabelian group $G$ for every $g \in G$. Is it true that elements $g_1$ and $g_2$ have the same distribution if and only if they are conjugate by some automorphism of $F$?

Let $G_\Gamma$ be a partially commutative soluble group of derived length $n \geqslant 3$ with defining graph $\Gamma$ (the definition is similar to the case of $n = 2$, see 17.104). Is it true that the centralizer of any vertex of the graph is generated by its adjacent vertices?

Let $M$ be a free metabelian group of finite rank $r \geqslant 2$, and let $P$ be the set of its primitive elements. (An element of a relatively free group is said to be primitive if it can be included in a basis of the group.)
$\qquad$ a) Is $P$ a first-order definable set?
$\qquad$ b) Is the set of bases of the group $M$ a first-order definable set?

Let $G$ be an $m$-generated group the elementary theory of which $\text{Th}(G)$ coincides with the elementary theory $\text{Th}(F)$ of a free soluble group $F$ of derived length $n \geqslant 3$ of finite rank $r \geqslant 2$. Must the groups $G$ and $F$ be isomorphic?

The answer is affirmative if $m \leqslant r$ or $n = 2$.

A connected graph $\Sigma$ is a symmetrical extension of a graph $\Gamma$ by a graph $\Delta$ if there exist a vertex-transitive group $G$ of automorphisms of $\Sigma$ and an imprimitivity system $\sigma$ of $G$ on the set of vertices of $\Sigma$ such that the quotient graph $\Sigma/\sigma$ is isomorphic to $\Gamma$ and blocks of $\sigma$ generate in $\Sigma$ subgraphs isomorphic to $\Delta$.
$\qquad$ (a) Let $\Gamma$ be a locally finite Cayley graph of a finitely presented group, and $\Delta$ a finite graph. Are there only finitely many symmetrical extensions of $\Gamma$ by $\Delta$?
$\qquad$ (b) Let $\Gamma$ be a locally finite graph which has the property of $k$-contractibility for some positive integer $k$ (see the definition in (V. I. Trofimov, Proc. Steklov Inst. Math., 279, suppl. 1 (2012), 107–112); note that any $\Gamma$ from (a) is such a graph) and let $\Delta$ be a finite graph. Are there only finitely many symmetrical extensions of $\Gamma$ by $\Delta$?

Is it true that for any positive integer $t$ there is a positive integer $n(t)$ such that any finite group with at least $n(t)$ conjugate classes of soluble maximal subgroups of Fitting height at most $t$ is itself a soluble group of Fitting height at most $t$?

Suppose that a finite group $G$ admits a factorization $G = AB = BC = CA$, where $A, B, C$ are abnormal supersoluble subgroups. Is $G$ supersoluble?

The maximum length of a chain of nested centralizers of a group is called its c-dimension. Let $G$ be a locally finite group of finite c-dimension $k$, and let $S$ be the preimage in $G$ of the socle of $G/R$, where $R$ is the locally solvable radical of $G$. Is it true that the factor group $G/S$ contains an abelian subgroup of index bounded by a function of $k$?

A subgroup $H$ of a free group $F$ is called inert if $r(H \cap K) \leqslant r(K)$ for every $K \leqslant F$; and compressed if $r(H) \leqslant r(K)$ for every $H \leqslant K \leqslant F$. Is it true that compressed subgroups are inert?

Is it true that an intersection of compressed subgroups is compressed? It is known that arbitrary intersections of inert subgroups are inert. (For the definitions, see 19.102)

Is there an algorithm which decides whether a given subgroup of $F$ is inert?

An algorithm to decide whether a given $H$ is compressed is known.

(For the definitions, see 19.102)

Is it true that the fixed subgroups of endomorphisms of $F$ are inert (see 19.102)?

Let $G$ be a uniformly locally finite group (which means that there is a function $f$ on the natural numbers such that any subgroup generated by $n$ elements has size at most $f(n)$). Suppose that for any two definable subgroups $H$ and $K$, the intersection $H \cap K$ has finite index either in $H$ or in $K$. Is $G$ necessarily nilpotent-by-finite? Or even finite-by-abelian-by-finite?

Let $\mathcal{AE}$ be the smallest class of locally compact groups such that (1) $\mathcal{AE}$ contains the compact groups and the discrete amenable groups; (2) $\mathcal{AE}$ is closed under taking closed subgroups, Hausdorff quotients, directed unions of open subgroups, and group extensions.
Is every amenable locally compact group an element of $\mathcal{AE}$?

(A negative answer would follow from a positive answer to 19.21.)

Let $P$ be a finite $p$-group, where $p$ is an odd prime. Let $\chi$ be a complex irreducible character of $P$. If $\chi(x) \neq 0$ for some $x \in P$, is it true that the order of $x$ must divide $|P|/\chi(1)^2$?

A subgroup $H$ of a finite group $G$ is called pronormal if for any $g \in G$ the subgroups $H$ and $H^g$ are conjugate by an element of $\langle H, H^g \rangle$. A maximal subgroup of a maximal subgroup is called second maximal. Is it true that in a non-abelian finite simple group $G$ all maximal subgroups are Hall subgroups if and only if every second maximal subgroup of $G$ is pronormal in $G$?

(G. M. Bergman and A. Magidin). Do there exist varieties of groups in which the relatively free group of rank 2 is finite, and the relatively free group of rank 3 is infinite?

Do there exist infinite groups all of whose proper subgroups are cyclic of order $p$ that do not satisfy any nontrivial group identity except $x^p = 1$ and its consequences?

Note that there exist infinite groups all of whose proper subgroups are infinite cyclic that do not satisfy any nontrivial group identity (due to A. Olshanskii; see P. Zusmanovich, J. Algebra, 388 (2013), 268–286, Remark after Theorem 6.1).