Given a group $G$ of finite order $n$, does there necessarily exist a bijection $f$ from $G$ onto a cyclic group of order $n$ such that for each element $x \in G$, the order of $x$ divides the order of $f(x)$?

Let the group $G = AB$ be the product of two Chernikov subgroups $A$ and $B$ each of which has an abelian subgroup of index at most 2. Is $G$ soluble?

Let the group $G = AB$ be the product of two central-by-finite subgroups $A$ and $B$. By a theorem of N. Chernikov, $G$ is soluble-by-finite. Is $G$ metabelian-by-finite?

(A. Rhemtulla, S. Sidki).
a) Is a group of the form $G = ABA$ with cyclic subgroups $A$ and $B$ always soluble? (This is known to be true if $G$ is finite.)
b) The same question when in addition $A$ and $B$ are conjugate in $G$.

Let the (soluble) group $G = AB$ with finite torsion-free rank $r_0(G)$ be the product of two subgroups $A$ and $B$. Is the equation $r_0(G) = r_0(A) + r_0(B) - r_0(A \cap B)$ valid in this case? It is known that the inequality $\leqslant$ always holds.

Is every finitely generated one-relator group residually amenable?

Let $n \geqslant 665$ be an odd integer. Is it true that the group of outer automorphisms $\text{Out}(B(m, n))$ of the free Burnside group $B(m, n)$ for $m > 2$ is complete, that is, has trivial center and all its automorphisms are inner?

Let $\mathcal{X}$ be a class of finite simple groups such that $\pi(\mathcal{X}) = \text{char}(\mathcal{X})$. A formation of finite groups $\mathfrak{F}$ is said to be $\mathcal{X}$-saturated if a finite group $G$ belongs to $\mathfrak{F}$ whenever the factor group $G/\Phi(O_{\mathcal{X}}(G))$ is in $\mathfrak{F}$, where $O_{\mathcal{X}}(G)$ is the largest normal subgroup of $G$ whose composition factors are in $\mathcal{X}$. Is every $\mathcal{X}$-saturated formation $\mathcal{X}$-local in the sense of Förster? (See the definition in (P. Förster, Publ. Sec. Mat. Univ. Autònoma Barcelona, 29, no. 2–3 (1985), 39–76).)

Does there exist a subgroup-closed saturated formation $\mathfrak{F}$ of finite groups properly contained in $\mathfrak{E}_\pi$, where $\pi = \text{char}(\mathfrak{F})$, satisfying the following property: if $G \in \mathfrak{F}$, then there exists a prime $p$ (depending on the group $G$) such that the wreath product $C_p \wr G$ belongs to $\mathfrak{F}$, where $C_p$ is the cyclic group of order $p$?

A formation $\mathfrak{F}$ of finite groups satisfies the Wielandt property for residuals if whenever $U$ and $V$ are subnormal subgroups of $\langle U, V \rangle$ in a finite group $G$, then the $\mathfrak{F}$-residual $\langle U, V \rangle_{\mathfrak{F}}$ of $\langle U, V \rangle$ coincides with $\langle U_{\mathfrak{F}}, V_{\mathfrak{F}} \rangle$. Does every Fitting formation $\mathfrak{F}$ satisfy the Wielandt property for residuals?

Let $G$ and $H$ be subgroups of the automorphism group $\text{Aut}(F_n)$ of a free group $F_n$ of rank $n \geqslant 2$. Is it true that the free product $G \ast H$ embeds in the automorphism group $\text{Aut}(F_m)$ for some $m$?

Let $G$ be a finitely generated group of intermediate growth. Is it true that there is a positive integer $m$ such that every element of the derived subgroup $G'$ is a product of at most $m$ commutators?

This assertion is valid for groups of polynomial growth, since they are almost nilpotent by Gromov’s theorem. On the other hand, there are groups of exponential growth (for example, free groups) for which this assertion is not true.

(D. B. A. Epstein). Is it true that the group
$$H = (\mathbb{Z}_3 \times \mathbb{Z}) \ast (\mathbb{Z}_2 \times \mathbb{Z}) = \langle x, y, z, t \mid x^3 = z^2 = [x, y] = [z, t] = 1 \rangle$$ cannot be defined by three relators in the generators $x, y, z, t$?

It is known that the relation module of the group $H$ has rank 3 (K. W. Gruenberg, P. A. Linnell, J. Group Theory, 11, no. 5 (2008), 587–608). An affirmative answer would give a solution of the relation gap problem.

For an automorphism $\varphi \in \text{Aut}(G)$ of a group $G$, let $[e]_\varphi = \{g^{-1} g^\varphi \mid g \in G\}$.
Conjecture: If $[e]_\varphi$ is a subgroup for every $\varphi \in \text{Aut}(G)$, then the group $G$ is nilpotent. If in addition $G$ is finitely generated, then $G$ is abelian.

For an automorphism $\varphi \in \text{Aut}(G)$ of a group $G$, let $[e]_\varphi = \{g^{-1} g^\varphi \mid g \in G\}$. Is it true that if a group $G$ has trivial center, then there is an inner automorphism $\varphi$ such that $[e]_\varphi$ is not a subgroup?

Is it true that any definable endomorphism of any ordered abelian group is of the form $x \to rx$, for some rational number $r$?

Is there a torsion-free group which is finitely presented in the quasi-variety of torsion-free groups but not finitely presentable in the variety of all groups?

Mal’cev proved that the set of sentences that hold in all finite groups is not computably enumerable, although its complement is. Is it true that both the set of sentences that hold in almost all finite groups and its complement are not computably enumerable?

Is any torsion-free, relatively free group of infinite rank not $\aleph_1$-homogeneous? This is true for group varieties in which all free groups are residually finite (O. Belegradek, Arch. Math. Logic, 51 (2012), 781–787).

Characters $\varphi$ and $\psi$ of a finite group $G$ are said to be semiproportional if they are not proportional and there is a normal subset $M$ of $G$ such that $\varphi|_M$ is proportional to $\psi|_M$ and $\varphi|_{G \setminus M}$ is proportional to $\psi|_{G \setminus M}$.
Conjecture: If $\varphi$ and $\psi$ are semiproportional irreducible characters of a finite group, then $\varphi(1) = \psi(1)$.

(B. H. Neumann and H. Neumann). Fix an integer $d \geqslant 2$. If $\mathfrak{V}$ is a variety of groups such that all $d$-generated groups in $\mathfrak{V}$ are finite, must $\mathfrak{V}$ be locally finite?

(H. Neumann). Is the Kostrikin variety of all locally finite groups of given prime exponent $p$ determined by finitely many identities?

The normal covering number of the symmetric group $S_n$ of degree $n$ is the minimum number $\gamma(S_n)$ of proper subgroups $H_1, \dots, H_{\gamma(S_n)}$ of $S_n$ such that every element of $S_n$ is conjugate to an element of $H_i$, for some $i = 1, \dots, \gamma(S_n)$. Write $n = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$ for primes $p_1 < \dots < p_r$ and positive integers $\alpha_1, \dots, \alpha_r$.

Conjecture:
$$\gamma(S_n) = \begin{cases} \frac{n}{2}(1 - \frac{1}{p_1}) & \text{if } r = 1 \text{ and } \alpha_1 = 1 \\ \frac{n}{2}(1 - \frac{1}{p_1}) + 1 & \text{if } r = 1 \text{ and } \alpha_1 \geqslant 2 \\ \frac{n}{2}(1 - \frac{1}{p_1})(1 - \frac{1}{p_2}) + 1 & \text{if } r = 2 \text{ and } \alpha_1 + \alpha_2 = 2 \\ \frac{n}{2}(1 - \frac{1}{p_1})(1 - \frac{1}{p_2}) + 2 & \text{if } r \geqslant 2 \text{ and } \alpha_1 + \dots + \alpha_r \geqslant 3 \end{cases}$$

This is the strongest form of the conjecture. We would be also interested in a proof that this holds for $n$ sufficiently large. The result for $r \leqslant 2$, which includes the first three cases above, is proved for $n$ odd (D. Bubboloni, C. E. Praeger, J. Combin. Theory (A), 118 (2011), 2000–2024). When $r \geqslant 3$ we know that $cn \leqslant \gamma(S_n) \leqslant \frac{2}{3}n$ for some positive constant $c$ (D. Bubboloni, C. E. Praeger, P. Spiga, J. Algebra, 390 (2013) 199–215). We showed that the conjectured value for $\gamma(S_n)$ is an upper bound, by constructing a normal covering for $S_n$ with this number of conjugacy classes of maximal subgroups, and gave further evidence for the truth of the conjecture in other cases (D. Bubboloni, C. E. Praeger, P. Spiga, Int. J. Group Theory, 3, no. 2 (2014), 57–75).

For a group $G$, a function $\phi: G \to \mathbb{R}$ is a quasimorphism if there is a least non-negative number $D(\phi)$ (called the defect) such that $|\phi(gh) - \phi(g) - \phi(h)| \leqslant D(\phi)$ for all $g, h \in G$. A quasimorphism is homogeneous if in addition $\phi(g^n) = n\phi(g)$ for all $g \in G$. Let $\phi$ be a homogeneous quasimorphism of a free group $F$. For any quasimorphism $\psi$ of $F$ (not required to be homogeneous) with $|\phi - \psi| < $ const, we must have $D(\psi) \geqslant D(\phi)/2$. Is it true that there is some $\psi$ with $D(\psi) = D(\phi)/2$?

Let $\text{form}(G)$ be the formation generated by a finite group $G$. Suppose that $G$ has a unique composition series $1 \triangleleft G_1 \triangleleft G_2 \triangleleft G$ and the consecutive factors of this series are $\mathbb{Z}_p, X, \mathbb{Z}_q$, where $p$ and $q$ are primes and $X$ is a non-abelian simple group. Is it true that $\text{form}(G)$ has infinitely many subformations if and only if $p = q$?

Suppose that a finite group $G$ has a normal series $1 \lhd G_1 \lhd G_2 \lhd G$ such that the groups $G_1$ and $G_2/G_1$ are elementary abelian $p$-groups, $G/G_2 \cong A_5 \times A_5$ (where $A_5$ is the alternating group of degree 5), $G_1$ and $G_2/G_1$ are minimal normal subgroups of $G$ and $G/G_1$, respectively. Is it true that $\text{form}(G)$ (see 18.25) has finitely many subformations?

Does there exist an algorithm that determines whether there are finitely many subformations in $\text{form}(G)$ (see 18.25) for a given finite group $G$?

Is it true that any subformation of every one-generator formation $\text{form}(G)$ (see 18.25) is also one-generator?

Let $\mathfrak{F}$ be a Fitting class of finite soluble groups which contains every soluble group $G = AB$, where $A$ and $B$ are abnormal $\mathfrak{F}$-subgroups of $G$. Is $\mathfrak{F}$ a formation?

A subgroup $H$ of a group $G$ is called $\mathbb{P}$-subnormal in $G$ if either $H = G$, or there is a chain of subgroups $H_0 \subset H_1 \subset \dots \subset H_n = G$ such that $|H_i : H_{i-1}|$ is a prime for all $i = 1, \dots, n$. Must a finite group be soluble if every Shmidt subgroup of it is $\mathbb{P}$-subnormal?

Let $\pi$ be a set of primes. Is it true that in any $D_\pi$-group $G$ (see 3.62) there are three Hall $\pi$-subgroups whose intersection coincides with $O_\pi(G)$?

Is every Hall subgroup of a finite group pronormal in its normal closure?

A group in which the derived subgroup of every 2-generated subgroup is cyclic is called an Alperin group. Is there a bound for the derived length of finite Alperin groups?

G. Higman proved that finite Alperin groups are soluble, and finite Alperin $p$-groups have bounded derived length, see 17.46.

Let $G$ be a topological group and $S = G_0 \leqslant \dots \leqslant G_n = G$ a subnormal series of closed subgroups. The infinite-length of $S$ is the cardinality of the set of infinite factors $G_i/G_{i-1}$. The virtual length of $G$ is the supremum of the infinite lengths taken over all such series of $G$. It is known that if $G$ is a pronilpotent group with finite virtual length then every closed subnormal subgroup is topologically finitely generated (N. Gavioli, V. Monti, C. M. Scoppola, J. Austral. Math. Soc., 95, no. 3 (2013), 343–355). Let $G$ be a pro-$p$ group (or more generally a pronilpotent group). If every closed subnormal subgroup of $G$ is topologically finitely generated, is it true that $G$ has finite virtual length?

A family $\mathcal{F}$ of group homomorphisms $A \to B$ is separating if for every nontrivial $a \in A$ there is $f \in \mathcal{F}$ such that $f(a) \neq 1$, and discriminating if for any finitely many nontrivial elements $a_1, \dots, a_n \in A$ there is $f \in \mathcal{F}$ such that $f(a_i) \neq 1$ for all $i = 1, \dots, n$. Let $G$ be a relatively free group of rank 2 in the variety of metabelian groups. Let $T$ be a metabelian group in which the centralizer of every nontrivial element is abelian. If $T$ admits a separating family of surjective homomorphisms $T \to G$ must it also admit a discriminating family of surjective homomorphisms $T \to G$?

A positive answer would give a metabelian analogue of a classical theorem of B. Baumslag.

There are groups of cardinality at most $2^{\aleph_0}$, even nilpotent of class 2, that cannot be embedded in $S := \text{Sym}(\mathbb{N})$ (V. A. Churkin, Algebra and model theory (Novosibirsk State Tech. Univ.), 5 (2005), 39–43 (Russian)). One condition which might characterize subgroups of $S$ is the following. Consider the metric $d$ on $S$ where for all $x \neq y$ we define $d(x, y) := 2^{-k}$ when $k$ is the least element of the set $\mathbb{N}$ such that $k^x \neq k^y$. Then $(S, d)$ is a separable topological group, and so each subgroup of $S$ (not necessarily a closed subgroup of $(S, d)$) is a separable topological group under the induced metric. Is it true that every group $G$ for which there is a metric $d'$ such that $(G, d')$ is a separable topological group is (abstractly) embeddable in $S$?

Note that any such group $G$ can be embedded as a section in $S$.

(Well-known problem). Is every locally graded group of finite rank almost locally soluble?

Let $E_\pi$ denote the class of finite groups that contain a Hall $\pi$-subgroup. Does the inclusion $E_{\pi_1} \cap E_{\pi_2} \subseteq E_{\pi_1 \cap \pi_2}$ hold for arbitrary sets of primes $\pi_1$ and $\pi_2$?

Conjecture: Let $G$ be a hyperbolic group. Then every 2-dimensional rational homology class is virtually represented by a sum of closed surface subgroups, that is, for any $\alpha \in H_2(G; \mathbb{Q})$ there are finitely many closed oriented surfaces $S_i$ and injective homomorphisms $\rho_i : \pi_1(S_i) \to G$ such that $\sum_i [S_i] = n\alpha$, where $[S_i]$ denotes the image of the fundamental class of $S_i$ in $H_2(G)$.

The commutator length $\text{cl}(g)$ of $g \in [G, G]$ is the least number of commutators in $G$ whose product is $g$, and the stable commutator length is $\text{scl}(g) := \lim_{n \to \infty} \text{cl}(g^n)/n$.

Conjecture: Let $G$ be a hyperbolic group. Then the stable commutator length takes on rational values on $[G, G]$.

Let $F$ be a free group of rank 2.
$\qquad$ a) It is known that $\text{scl}(g) = 0$ only for $g = 1$, and $\text{scl}(g) \geqslant 1/2$ for all $1 \neq g \in [F, F]$. Is 1/2 an isolated value?
$\qquad$ b) Are there any intervals $J$ in $\mathbb{R}$ such that the set of values of $\text{scl}$ on $[F, F]$ is dense in $J$?
$\qquad$ c) Is there some $T \in \mathbb{R}$ such that every rational number $\geqslant T$ is a value of $\text{scl}(g)$ for $g \in [F, F]$?

For an orientation-preserving homeomorphism $f : S^1 \to S^1$ of a unit circle $S^1$, let $\tilde{f}$ be its lifting to a homeomorphism of $\mathbb{R}$; then the rotation number of $f$ is defined to be the limit $\lim_{n \to \infty}(\tilde{f}^n(x) - x)/n$ (which is independent of the point $x \in S^1$). Let $F$ be a free group of rank 2 with generators $a, b$. For $w \in F$ and $r, s \in \mathbb{R}$, let $R(w, r, s)$ denote the maximum value of the (real-valued) rotation number of $w$, under all representations from $F$ to the universal central extension of $\text{Homeo}^+(S^1)$ for which the rotation number of $a$ is $r$, and the rotation number of $b$ is $s$.
$\qquad$ a) If $r, s$ are rational, must $R(w, r, s)$ be rational?
$\qquad$ b) Let $R(w, r^-, s^-)$ denote the supremum of the rotation number of $w$ (as above) under all representations for which $a$ and $b$ are conjugate to rotations through $r$ and $s$, respectively (in the universal central extension of $\text{Homeo}^+(S^1)$). Is $R(w, r^-, s^-)$ always rational if $r$ and $s$ are rational?
$\qquad$ c) Weak Slippery Conjecture: Let $w$ be a word containing only positive powers of $a$ and $b$. A pair of values $(r, s)$ is slippery if there is a strict inequality $R(w, r', s') < R(w, r^-, s^-)$ for all $r' < r$, $s' < s$. Is it true that then $R(w, r^-, s^-) = h_a(w)r + h_b(w)s$, where $h_a(w)$ counts the number of $a$’s in $w$, and $h_b(w)$ counts the number of $b$’s in $w$?
$\qquad$ d) Slippery Conjecture: More precisely, is it always (without assuming $(r, s)$ being slippery) true that if $w$ has the form $w = a^{\alpha_1} b^{\beta_1} \dots a^{\alpha_m} b^{\beta_m}$ (all positive) and $R(w, r, s) = p/q$ where $p/q$ is reduced, then $|p/q - h_a(w)r - h_b(w)s| \leqslant m/q$?

a) Do there exist elements $u, v$ of a 2-generator free group $F(a, b)$ such that $u, v$ are not conjugate in $F(a, b)$ but for any matrices $A, B \in \text{GL}(3, \mathbb{C})$ we have $\text{trace}(u(A, B)) = \text{trace}(v(A, B))$?

b) The same question if we replace $\text{GL}(3, \mathbb{C})$ by $\text{SL}(3, \mathbb{C})$.

Let $G$ be a finite non-abelian group and $V$ a finite faithful irreducible $G$-module. Suppose that $M = |G/G'|$ is the largest orbit size of $G$ on $V$, and among orbits of $G$ on $V$ there are exactly two orbits of size $M$. Does this imply that $G$ is dihedral of order 8, and $|V| = 9$?

Let $G$ be a finite $p$-group of maximal class such that all its irreducible characters are induced from linear characters of normal subgroups. Let $S$ be the set of the derived subgroups of the members of the central series of $G$. Is there a logarithmic bound for the derived length of $G$ in terms of $|S|$?

Every finite group $G$ can be embedded in a group $H$ in such a way that every element of $G$ is a square of an element of $H$. The overgroup $H$ can be chosen such that $|H| \leqslant 2|G|^2$. Is this estimate sharp?

It is known that the best possible estimate cannot be better than $|H| \leqslant |G|^2$ (D. V. Baranov, Ant. A. Klyachko, Siber. Math. J., 53, no. 2 (2012), 201–206).

Is it algorithmically decidable whether a group generated by three given class transpositions (for the definition, see 17.57)
$\qquad$ a) has only finite orbits on $\mathbb{Z}$?
$\qquad$ b) acts transitively on the set of nonnegative integers in its support?
A difficult case is the group $\langle \tau_{1(2), 4(6)}, \tau_{1(3), 2(6)}, \tau_{2(3), 4(6)} \rangle$, which acts transitively on $\mathbb{N} \setminus 0(6)$ if and only if Collatz’ $3n + 1$ conjecture is true.

Is it true that there are only finitely many integers which occur as orders of products of two class transpositions? (For the definition, see 17.57.)

Let $n \in \mathbb{N}$. Is it true that for any $a, b, c \in \mathbb{N}$ satisfying $1 < a, b, c \leqslant n - 2$ the symmetric group $S_n$ has elements of order $a$ and $b$ whose product has order $c$?

Let $n \in \mathbb{N}$. Is it true that for every $k \in \{1, \dots, n!\}$ there is some group $G$ and pairwise distinct elements $g_1, \dots, g_n \in G$ such that the set $\{g_{\sigma(1)} \dots g_{\sigma(n)} \mid \sigma \in S_n\}$ of all products of the $g_i$ obtained by permuting the factors has cardinality $k$?

Given a prime $p$ and $n \in \mathbb{N}$, let $f_p(n)$ be the smallest number such that there is a group of order $p^{f_p(n)}$ into which every group of order $p^n$ embeds. Is it true that $f_p(n)$ grows faster than polynomially but slower than exponentially when $n$ tends to infinity?

Is every finite simple group generated by two elements of prime-power orders $m, n$? (Here numbers $m, n$ may depend on the group.) The work of many authors shows that it remains to verify this property for a small number of finite simple groups.

Is there a non-linear simple locally finite group in which every centralizer of a non-trivial element is almost soluble, that is, has a soluble subgroup of finite index?

Can a group be equal to the union of conjugates of a proper finite nonabelian simple subgroup?

a) Can a locally finite $p$-group $G$ of finite exponent be the union of conjugates of an abelian proper subgroup?
b) Can this happen when $G$ is of exponent $p$?

Let $G$ be a finite 2-group, of order greater than 2, such that $|H/H_G| \leqslant 2$ for all $H \leqslant G$, where $H_G$ denotes the largest normal subgroup of $G$ contained in $H$. Must $G$ have an abelian subgroup of index 4?

Let $G$ be a finite 2-group generated by involutions in which $[x, u, u] = 1$ for every $x \in G$ and every involution $u \in G$. Is the derived length of $G$ bounded?

Let $G$ be a group generated by finite number $n$ of involutions in which $(uv)^4 = 1$ for all involutions $u, v \in G$. Is it true that $G$ is finite? is a 2-group? This is true for $n \leqslant 3$.

Does there exist a periodic group $G$ such that $G$ contains an involution, all involutions in $G$ are conjugate, and the centralizer of every involution $i$ is isomorphic to $\langle i \rangle \times L_2(P)$, where $P$ is some infinite locally finite field of characteristic 2?

Let $V$ be an infinite countable elementary abelian additive 2-group. Does $\text{Aut}\,V$ contain a subgroup $G$ such that
$\qquad$ a) $G$ is transitive on the set of non-zero elements of $V$, and
$\qquad$ b) if $H$ is the stabilizer in $G$ of a non-zero element $v \in V$, then $V = \langle v \rangle \oplus V_v$, where $V_v$ is $H$-invariant, $H$ is isomorphic to the multiplicative group $P^*$ of a locally finite field $P$ of characteristic 2, and the action $H$ on $V_v$ is similar to the action of $P^*$ on $P$ by multiplication?

Conjecture: such a group $G$ does not exist. If so, then the group $G$ in 18.59 does not exist too.

Is a 2-group nilpotent if all its finite subgroups are nilpotent of class at most 3? This is true if 3 is replaced by 2.

The spectrum of a finite group is the set of orders of its elements. Let $\omega$ be a finite set of positive integers. A group $G$ is said to be $\omega$-critical if the spectrum of $G$ coincides with $\omega$, but the spectrum of every proper section of $G$ is not equal to $\omega$.
$\qquad$ a) Does there exist a number $n$ such that for every finite simple group $G$ the number of $\omega(G)$-critical groups is less than $n$?
$\qquad$ b) For every finite simple group $G$, find all $\omega(G)$-critical groups.

Let $G$ be a periodic group generated by two fixed-point-free automorphisms of order 5 of an abelian group. Is $G$ finite?

(K. Harada). Conjecture: Let $G$ be a finite group, $p$ a prime, and $B$ a $p$-block of $G$. If $J$ is a non-empty subset of $\text{Irr}(B)$ such that $\sum_{\chi \in J} \chi(1)\chi(g) = 0$ for every $p$-singular element $g \in G$, then $J = \text{Irr}(B)$.

(R. Guralnick, G. Malle). Conjecture: Let $p$ be a prime different from 5, and $C$ a class of conjugate $p$-elements in a finite group $G$. If $[c, d]$ is a $p$-element for any $c, d \in C$, then $C \subseteq O_p(G)$.

Suppose that a finite group $G$ admits a Frobenius group of automorphisms $FH$ with kernel $F$ and complement $H$ such that $G^F$ is also a Frobenius group with kernel $G$ and complement $F$. Is the derived length of $G$ bounded in terms of $|H|$ and the derived length of $C_G(H)$?

Suppose that a finite group $G$ admits a Frobenius group of automorphisms $FH$ with kernel $F$ and complement $H$ such that $C_G(F) = 1$. Is the exponent of $G$ bounded in terms of $|F|$ and the exponent of $C_G(H)$?

What are the nonabelian composition factors of a finite nonsoluble group all of whose maximal subgroups have complements?
Note that finite simple groups with all maximal subgroups having complements are, up to isomorphism, $L_2(7)$, $L_2(11)$, and $L_5(2)$ (V. M. Levchuk, A. G. Likharev, Siberian Math. J., 47, no. 4 (2006), 659–668). The same groups exhaust finite simple groups with Hall maximal subgroups and composition factors of groups with Hall maximal subgroups (N. V. Maslova, Siberian Math. J., 53, no. 5 (2012), 853–861). It is also proved that in a finite group with Hall maximal subgroups all maximal subgroups have complements (N. V. Maslova, D. O. Revin, Siberian Adv. Math., 23, no. 3 (2013), 196–209).

Does there exist a relatively free group $G$ containing a free subsemigroup and having $[G, G]$ finitely generated?

Note that $G$ cannot be locally graded (Publ. Math. Debrecen, 81, no. 3-4 (2012), 415–420.)

Is every finitely generated Coxeter group conjugacy separable?

(N. Aronszajn). Suppose that $W(x, y) = 1$ in an open subset of $G \times G$, where $G$ is a connected topological group. Must $W(x, y) = 1$ for all $(x, y) \in G \times G$?

Is it true that an existentially closed subgroup of a nonabelian free group of finite rank is a nonabelian free factor of this group?

a) Does every finitely generated solvable group of derived length $l \geqslant 2$ embed into a 2-generated solvable group of length $l + 1$?
b) Does every finitely generated solvable group of derived length $l \geqslant 2$ embed into some $k$-generated $(l + 1)$-solvable group, where $k = k(l)$?

Let $G$ be a finitely generated elementary amenable group which is not virtually nilpotent. Is there a finitely generated metabelian non-virtually-nilpotent section in $G$?

Does every finite solvable group $G$ have the following property: there is a number $d = d(G)$ such that $G$ is a homomorphic image of every group with $d$ generators and one relation?

This property holds for finite nilpotent groups and does not hold for every non-solvable finite group; see (S. A. Zaĭtsev, Moscow Univ. Math. Bull., 52, no. 4 (1997), 42–44).

Let $A$ be a division ring, $G$ a subgroup of the multiplicative group of $A$, and $E$ an extension of the additive group of $A$ by $G$ such that $G$ acts by multiplication in $A$. Is it true that $E$ splits? This is true if $G$ is finite.

Let $G$ be a finite $p$-group and let $p^e$ be the largest degree of an irreducible complex representation of $G$. If $p > e$, is it necessarily true that $\bigcap \text{ker}\,\Theta = 1$, where the intersection runs over all irreducible complex representations $\Theta$ of $G$ of degree $p^e$?

Let $K^t G$ be a twisted group algebra of the finite group $G$ over the field $K$. If $K^t G$ is a simple $K$-algebra, is $G$ necessarily solvable? This is known to be true if $K^t G$ is central simple.

Let $K[G]$ be the group algebra of the finitely generated group $G$ over the field $K$. Is the Jacobson radical $\mathcal{J} K[G]$ equal to the join of all nilpotent ideals of the ring? This is known to be true if $G$ is solvable or linear.

(I. Kaplansky). For $G \neq 1$, show that the augmentation ideal of the group algebra $K[G]$ is equal to the Jacobson radical of the ring if and only if $\text{char}\,K = p > 0$ and $G$ is a locally finite $p$-group.

Let $G$ be a finitely generated $p$-group that is residually finite. Are all maximal subgroups of $G$ necessarily normal?

Is there a function $f : \mathbb{N} \to \mathbb{N}$ such that for any prime $p$, if $p^{f(n)}$ divides the order of a finite group $G$, then $p^n$ divides the order of $\text{Aut } G$?

A generating system $X$ of a group $G$ is fast if there is an integer $n$ such that every element of $G$ can be expressed as a product of at most $n$ elements of $X$ or their inverses. If not, we say that it is slow. For instance, in $(\mathbb{Z}, +)$, the squares are fast, but the powers of 2 are slow.
Do there exist countable infinite groups without an infinite slow generating set? Uncountable ones do exist.

Let $\pi$ be a set of primes. We say that a finite group is a $\text{BS}_\pi$-group if every conjugacy class in this group any two elements of which generate a $\pi$-subgroup itself generates a $\pi$-subgroup. Is every normal subgroup of a $\text{BS}_\pi$-group a $\text{BS}_\pi$-group?

A subset of a group is said to be rational if it can be obtained from finite subsets by finitely many rational operations, that is, taking union, product, and the submonoid generated by a set.

Conjecture: every finitely generated solvable group in which all rational subsets form a Boolean algebra is virtually abelian.

Is the group $G = \langle a, b \mid [[a, b], b] = 1 \rangle$, which is isomorphic to the group of all unitriangular automorphisms of the free group of rank 3, linear?

A system of equations with coefficients in a group $G$ is said to be independent if the matrix composed of the sums of exponents of the unknowns has rank equal to the number of equations.
$\qquad$ a) The Kervaire–Laudenbach Conjecture (KLC): every independent system of equations with coefficients in an arbitrary group $G$ has a solution in some over group $\overline{G}$. This is true for every locally residually finite group $G$ (M. Gerstenhaber and O. S. Rothaus).
$\qquad$ b) KLC — nilpotent version: every independent system of equations with coefficients in an arbitrary nilpotent group $G$ has a solution in some nilpotent overgroup $\overline{G}$.
$\qquad$ c) KLC — solvable version: every independent system of equations with coefficients in an arbitrary solvable group $G$ has a solution in some solvable overgroup $\overline{G}$.

Can a finitely generated infinite group of finite exponent be the quotient of a residually finite group by a locally finite normal subgroup? If not, then there exists a hyperbolic group that is not residually finite.

Consider the set of balanced presentations $\langle x_1, \dots, x_n \mid r_1, \dots, r_n \rangle$ with fixed $n$ generators $x_1, \dots, x_n$. By definition, the group $\text{AC}_n$ of Andrews–Curtis moves on this set of balanced presentations is generated by the Nielsen transformations together with conjugations of relators. Is $\text{AC}_n$ finitely presented?

Prove that the group $Y(m, n) = \langle a_1, a_2, \dots, a_m \mid a_k^n = 1 \, (1 \leqslant k \leqslant m), (a_k^i a_l^i)^2 = e \, (1 \leqslant k < l \leqslant m, 1 \leqslant i \leqslant \frac{n}{2}) \rangle$ is finite for every pair $(m, n)$.

Some known cases are (where $C_n$ denotes a cyclic group of order $n$): $Y(m, 2) = C_2^m$; $Y(m, 3) = A_{m+2}$ (presentation by Carmichael); $Y(m, 4)$ has order $2^{m(m+3)/2}$, nilpotency class 3 and exponent 4; $Y(2, n)$ is the natural extension of the augmentation ideal $\omega(C_n)$ of $\text{GF}(2)[C_n]$ by $C_n$ (presentation by Coxeter); $Y(3, n) \cong \text{SL}(2, \omega(C_n))$.

These groups have connections with classical groups in characteristic 2. When $n$ is odd, the group $Y(m, n)$ has the following presentation, where $\tau_{ij}$ are transpositions:
$$y(m, n) = \langle a, S_m \mid a^n = e, [\tau_{12}^{a^i}, \tau_{12}] = e \, (1 \leqslant i \leqslant n/2), \tau_{12}^{1+a+\dots+a^{n-1}} = e, \tau_{i, i+1} a \tau_{i, i+1} = a^{-1} \, (2 \leqslant i \leqslant m-1) \rangle,$$ and it is known that, for example, $y(3, 5) \cong \text{SL}(2, 16) \cong \Omega^-(4, 4)$, $y(4, 5) \cong \text{Sp}(2, 16) \cong \Omega(5, 4)$, $y(5, 5) \cong \text{SU}(4, 16) \cong \Omega^-(6, 4)$, $y(6, 5) \cong 4^6 \Omega^-(6, 4)$, $y(3, 7) \cong \text{SL}(2, 8)^2 \cong \Omega^+(4, 8)$, $y(4, 7) \cong \text{Sp}(4, 8) \cong \Omega(5, 8)$. Extensive computations by Felsch, Neubüser and O'Brien confirm this trend. Different types reveal Bott periodicity and a connection with Clifford algebras.

A subgroup $H$ of a group $G$ is said to be propermutable in $G$ if there is a subgroup $B \leqslant G$ such that $G = N_G(H)B$ and $H$ permutes with every subgroup of $B$.
$\qquad$ a) Is there a finite group $G$ with subgroups $A \leqslant B \leqslant G$ such that $A$ is propermutable in $G$ but $A$ is not propermutable in $B$?
$\qquad$ b) Let $P$ be a non-abelian Sylow 2-subgroup of a finite group $G$ with $|P| = 2^n$. Suppose that there is an integer $k$ such that $1 < k < n$ and every subgroup of $P$ of order $2^k$ is propermutable in $G$, and also, in the case of $k = 1$, every cyclic subgroup of $P$ of order 4 is propermutable in $G$. Is it true that then $G$ is 2-nilpotent?

A non-empty set $\theta$ of formations is called a complete lattice of formations if the intersection of any set of formations in $\theta$ belongs to $\theta$ and $\theta$ has the largest element (with respect to inclusion). If $L$ is a complete lattice, then an element $a \in L$ is said to be compact if $a \leqslant \bigvee X$ for any $X \subseteq L$ implies that $a \leqslant \bigvee X_1$ for some finite $X_1 \subset X$. A complete lattice is called algebraic if every element is the join of a (possibly infinite) set of compact elements.
$\qquad$ a) Is there a non-algebraic complete lattice of formations of finite groups?
$\qquad$ b) Is there a non-modular complete lattice of formations of finite groups?

Let $\mathfrak{M}$ be a one-generated saturated formation, that is, the intersection of all saturated formations containing some fixed finite group. Let $\mathfrak{F}$ be a subformation of $\mathfrak{M}$ such that $\mathfrak{F} \neq \mathfrak{F}\mathfrak{F}$.
$\qquad$ a) Is it true that then $\mathfrak{F}$ can be written in the form $\mathfrak{F} = \mathfrak{F}_1 \dots \mathfrak{F}_t$, where $\mathfrak{F}_i$ is a non-decomposable formation for every $i = 1, \dots, t$?
$\qquad$ b) Suppose that $\mathfrak{F} = \mathfrak{F}_1 \dots \mathfrak{F}_t$, where $\mathfrak{F}_i$ is a non-decomposable formation for every $i = 1, \dots, t$. Is it true that then all factors $\mathfrak{F}_i$ are uniquely determined?

Let $G$ be a group without involutions, $a$ an element of it that is not a square of any element of $G$, and $n$ an odd positive integer. Is it true that the quotient $G/\langle (a^n)^G \rangle$ does not contain involutions?

Suppose that a group $G = AB$ is a product of an abelian subgroup $A$ and a locally quaternion group $B$ (that is, $B$ is a union of an increasing chain of finite generalized quaternion groups). Is $G$ soluble?

Suppose that a periodic group $G$ contains involutions and the centralizer of each involution is locally finite. Is it true that $G$ has a nontrivial locally finite normal subgroup?

Let $G$ be a periodic Zassenhaus group, that is, a two-transitive permutation group with trivial stabilizer of every three points. Suppose that the stabilizer of a point is a Frobenius group with locally finite kernel $U$ containing an involution. Is it true that $U$ is a 2-group?

The work of many authors shows that most finite simple groups are generated by two elements of orders 2 and 3; for example, see the survey. Which finite simple groups cannot be generated by two elements of orders 2 and 3? In particular, is it true that, among classical simple groups of Lie type, such exceptions, apart from $\text{PSU}(3, 5^2)$, arise only when the characteristic is 2 or 3?

The question of which finite simple groups are (2,3)-generated remains open only for the orthogonal groups of even dimension $2m > 8$ (M. A. Pellegrini, M. C. Tamburini Bellani, J. Austral. Math. Soc., 117 (2024), 130–148).

Let $\text{cd}(G)$ be the set of all complex irreducible character degrees of a finite group $G$. Huppert’s Conjecture: if $H$ is a nonabelian simple group and $G$ is any finite group such that $\text{cd}(G) = \text{cd}(H)$, then $G \cong H \times A$, where $A$ is an abelian group.

For a given class of groups $\mathcal{X}$, let $(\mathcal{X}, \infty)^*$ denote the class of groups in which every infinite subset contains two distinct elements $x$ and $y$ that satisfy $\langle x, x^y \rangle \in \mathcal{X}$. Let $G$ be a finitely generated soluble-by-finite group in the class $(\mathcal{X}, \infty)^*$, let $m$ be a positive integer, and let $\mathcal{F}, \mathcal{E}_m, \mathcal{E}, \mathcal{N}$, and $\mathcal{P}$ denote the classes of finite groups, groups of exponent dividing $m$, groups of finite exponent, nilpotent groups, and polycyclic groups, respectively.
$\qquad$ a) If $\mathcal{X} = \mathcal{E}\mathcal{N}$, then is $G$ in $\mathcal{E}\mathcal{N}$?
$\qquad$ b) If $\mathcal{X} = \mathcal{E}_m\mathcal{N}$, then is $G$ in $\mathcal{E}_m(\mathcal{F}\mathcal{N})$?
$\qquad$ c) If $\mathcal{X} = \mathcal{N}(\mathcal{P}\mathcal{F})$, then is $G$ in $\mathcal{N}(\mathcal{P}\mathcal{F})$?

(E. C. Dade). Let $C$ be a Carter subgroup of a finite solvable group $G$, and let $\ell(C)$ be the number of primes dividing $|C|$ counting multiplicities. It was proved in (E. C. Dade, Illinois J. Math., 13 (1969), 449–514) that there is an exponential function $f$ such that the nilpotent length of $G$ is at most $f(\ell(C))$. Is there a linear (or at least a polynomial) function $f$ with this property?

A group is said to be minimax if it has a finite subnormal series each of whose factors satisfies either the minimum or the maximum condition on subgroups. Is it true that in the class of nilpotent minimax groups only finitely generated groups may have faithful irreducible primitive representations over a finitely generated field of characteristic zero?

Construct an example of a pro-$p$ group $G$ and a proper abstract normal subgroup $K$ such that $G/K$ is perfect.

Let $R$ be the formal power series algebra in commuting indeterminates $x_1, \dots, x_n, \dots$ over the field with $p$ elements, with $p$ a prime. (Thus its elements are (in general infinite) linear combinations of monomials $x_1^{a_1} \dots x_n^{a_n}$ with $n, a_1, \dots, a_n$ non-negative integers.) Let $I$ be the maximal ideal of $R$ and $J$ the (abstract) ideal generated by all products of two elements of $I$. Is there an ideal $U$ of $R$ such that $U < I$ and $U + J = I$?

If so, then $\text{SL}_3(R)$ and the kernel of the map to $\text{SL}_3(R/U)$ provide an answer to 18.104.

Let $R$ be a group that acts coprimely on the finite group $G$. Let $p$ be a prime, let $P$ be the unique maximal $R C_G(R)$-invariant $p$-subgroup of $G$ and assume that $C_G(O_p(G)) \leqslant O_p(G)$. If $p > 3$ then $P$ contains a nontrivial characteristic subgroup that is normal in $G$ (P. Flavell, J. Algebra, 257 (2002), 249–264). Does the same result hold for the primes 2 and 3?

Is there a finitely generated infinite residually finite $p$-group such that every subgroup of infinite index is cyclic?

The answer is known to be negative for $p = 2$. Note that in (M. Ershov, A. Jaikin Zapirain, J. Reine Angew. Math., 677 (2013), 71–134) it is shown that for every prime $p$ there is a finitely generated infinite residually finite $p$-group such that every finitely generated subgroup of infinite index is finite.

A group $G$ is said to have property $(\tau)$ if its trivial representation is an isolated point in the subspace of irreducible representations with finite images (in the topological unitary dual space of $G$). Is it true that there exists a finitely generated group satisfying property $(\tau)$ that homomorphically maps onto every finite simple group?

The motivation is the theorem in (E. Breuillard, B. Green, M. Kassabov, A. Lubotzky, N. Nikolov, T. Tao) saying that there are $\epsilon > 0$ and $k$ such that every finite simple group $G$ contains a generating set $S$ with at most $k$ elements such that the Cayley graph of $G$ with respect to $S$ is an $\epsilon$-expander. In (M. Ershov, A. Jaikin Zapirain, M. Kassabov, Mem. Amer. Math. Soc., 1186, 2017) it is proved that there exists a group with Kazhdan’s property (T) that maps onto every simple group of Lie type of rank $\geqslant 2$. (A group $G$ is said to have Kazhdan’s property (T) if its trivial representation is an isolated point in the unitary dual space of $G$.)

Is it true that a group satisfying Kazhdan’s property (T) cannot homomorphically map onto infinitely many simple groups of Lie type of rank 1?

It is known that a group which maps onto $\text{PSL}_2(q)$ for infinitely many $q$ does not have Kazhdan’s property (T) (see 18.108).

The non-$p$-soluble length of a finite group $G$ is the number of non-$p$-soluble factors in a shortest normal series each of whose factors either is $p$-soluble or is a direct product of non-abelian simple groups of order divisible by $p$. For a given prime $p$ and a given proper group variety $\mathfrak{V}$, is there a bound for the non-$p$-soluble length of finite groups whose Sylow $p$-subgroups belong to $\mathfrak{V}$?

Let $G$ be a discrete countable group, given as a central extension $0 \to \mathbb{Z} \to G \to Q \to 0$. Assume that either $G$ is quasi-isometric to $\mathbb{Z} \times Q$, or that $\mathbb{Z} \to G$ is a quasi-isometric embedding. Does that imply that $G$ comes from a bounded cocycle on $Q$?

Is it true that the orders of all elements of a finite group $G$ are powers of primes if, for every divisor $d$ ($d > 1$) of $|G|$ and for every subgroup $H$ of $G$ of order coprime to $d$, the order $|H|$ divides the number of elements of $G$ of order $d$?

The converse is true (W. J. Shi, Math. Forum (Vladikavkaz), 6 (2012), 152–154).

Let $\mathcal{M}$ be a finite set of finite simple nonabelian groups. Is it true that a periodic group saturated with groups from $\mathcal{M}$ (see 14.101) is isomorphic to one of groups in $\mathcal{M}$? The case where $\mathcal{M}$ is one-element is of special interest.

Does there exist an irreducible $5'$-subgroup $G$ of $\text{GL}(V)$ for some finite $\mathbb{F}_5$-space $V$ such that the number of conjugacy classes of the semidirect product $V \rtimes G$ is equal to $|V|$ but $G$ is not cyclic?

Such a subgroup exists if 5 is replaced by $p = 2, 3$ but does not exist for primes $p > 5$ (J. Group Theory, 14 (2011), 175–199).

Let $G$ be a finite simple group, and let $X, Y$ be isomorphic simple maximal subgroups of $G$. Are $X$ and $Y$ conjugate in $\text{Aut } G$?

Given a finite 2-group $G$ of order $2^n$, does there exist a finite set $S$ of rational primes such that $|S| \leqslant n$ and $G$ is a quotient group of the absolute Galois group of the maximal 2-extension of $\mathbb{Q}$ unramified outside $S \cup \{\infty\}$? This is true for $p$-groups, $p$ odd, as noted by Serre.

Is every element of a nonabelian finite simple group a commutator of two elements of coprime orders?

The answer is known to be “yes” for the alternating groups (P. Shumyatsky, Forum Math., 27, no. 1 (2015), 575–583) and for $\text{PSL}(2, q)$ (M. A. Pellegrini, P. Shumyatsky, Arch. Math., 99 (2012), 501–507). Without the coprimeness condition, this is Ore’s conjecture proved in (M. W. Liebeck, E. A. O’Brien, A. Shalev, Pham Huu Tiep, J. Eur. Math. Soc., 12, no. 4 (2010), 939–1008).

Let $G$ be a profinite group such that $G/Z(G)$ is periodic. Is $[G, G]$ necessarily periodic?

A multilinear commutator word is any commutator of weight $n$ in $n$ distinct variables. Let $w$ be a multilinear commutator word and let $G$ be a finite group. Is it true that every Sylow $p$-subgroup of the verbal subgroup $w(G)$ is generated by $w$-values?

Let $P = AB$ be a finite $p$-group factorized by an abelian subgroup $A$ and a class-two subgroup $B$. Suppose, if necessary, $A \cap B = 1$. Then is it true that $\langle A, [B, B] \rangle = AB_0$, where $B_0$ is an abelian subgroup of $B$?