Proof. Let

$$ \begin{align*}\{0\}=I_0\leq I=I_1\leq\ldots\leq I_n=B\end{align*} $$

be a chain of ideals of B such that, for all $0\leq i<n$ , either $(I_{i+1}/I_i,+)$ is infinite cyclic and $I_{i+1}/I_i\leq \operatorname {Soc}(B/I_i)$ , or $I_{i+1}/I_i$ has prime order. Then $X=(I,+)$ and $Y=(I,+)\rtimes _\lambda (I,{\boldsymbol {\cdot }})$ are normal subgroups of G such that both X and $Y/X$ are cyclic groups. Since $G/Y\simeq (B/I,+)\rtimes _\lambda (B/I,{\boldsymbol {\cdot }})$ , we are done by induction on n.

Proof. Since B is centrally nilpotent, it follows from [Reference Trombetti38], Theorem 3.7, that $(B,+)$ is finitely generated. Now, $(B,+)$ is finitely generated and nilpotent, so it satisfies the maximal condition on subgroups, which means that every section of $(B,+)$ is finitely generated. Let $Z=\zeta (B)$ . Then Z is a trivial brace whose additive subgroups are ideals of B. Since Z is additively finitely generated, its additive group is a direct sum of finitely many cyclic groups, so Z has a chain of ideals whose factors have cyclic additive groups. Similarly, every section of the upper central series of B can be refined to a finite chain of ideals whose factors have cyclic additive groups, and this completes the proof.

Proof. Let I be an ideal of B such that $|I|=p$ . Since $|B|=p^n$ for some nonnegative integer n, it follows that

$$ \begin{align*}(I,+)\leq C_{(B,+)}(B,+)\quad\text{and}\quad (I,{\boldsymbol{\cdot}})\leq C_{(B,{\boldsymbol{\cdot}})}(B,{\boldsymbol{\cdot}}).\end{align*} $$

Write $I=\langle b\rangle $ , and let $G=(B,+)\rtimes _\lambda (B,{\boldsymbol {\cdot }})$ ; in particular, $|G|=p^{2n}$ . Now, if $a\in (B,+)$ , then $a\in C_G(b)$ , so $a\ast b=0$ . On the other hand,

$$ \begin{align*}a\ast b=-a+ab-b=-a-b+ba-a+a=-a+b\ast a+a,\end{align*} $$

so $b\ast a=0$ and hence $b\in \operatorname {Ker}(\lambda )$ . Therefore, $I\leq \zeta (B)$ and B is centrally nilpotent by induction on the order of B.

Proof. Since B is, by assumption, supersoluble and of order $p^n$ for some positive integer n, there exists an ideal, say I, of order p. In particular, I is a minimal nonzero ideal and B is left-nilpotent. We claim that $B\ast I=0$ . Indeed, $B\ast I$ is a left-ideal of B contained in I. Hence, $B\ast I=\{0\}$ or $B\ast I=I$ . As B is a left-nilpotent, the latter is not possible. Thus, indeed, $B\ast I=\{0\}$ . So, $a\ast b=0$ for all $a\in B$ and $b\in I$ . Since $(I,{\boldsymbol {\cdot }})$ is normal in $(B,{\boldsymbol {\cdot }})$ , $(I,+)$ is normal in $(B,+)$ , and $(B,+)$ and $(B,{\boldsymbol {\cdot }})$ are nilpotent, we have that $I\leq Z(B)$ . Thus, $b{\boldsymbol {\cdot }} a=a{\boldsymbol {\cdot }} b$ . Then, $0=-a+a{\boldsymbol {\cdot }} b-b=-a+b{\boldsymbol {\cdot }} a-b=-a-b+b{\boldsymbol {\cdot }} a=-a-b+b{\boldsymbol {\cdot }} a-a+a=-a+b\ast a+a$ . So, $b\ast a=0$ and therefore $B\ast I=\{0\}$ . Hence, $I\leq \zeta (B)$ . The result now follows by induction on the order of B.

Proof. Since $(B,+)$ and $(B,{\boldsymbol {\cdot }})$ are finite groups whose Sylow subgroups are cyclic, we have that $(B,+)$ and $(B,{\boldsymbol {\cdot }})$ are supersoluble (see [Reference Robinson34], 10.1.10). If p is the largest prime dividing the order of B, then the set P of all p-elements of $(B,+)$ is a characteristic subgroup of $(B,+)$ by [Reference Robinson34], 5.4.8, which means that P is a strong left-ideal of B. Therefore, P is a subgroup of $(B,{\boldsymbol {\cdot }})$ , and order considerations show that P is the unique Sylow p-subgroup of $(B,{\boldsymbol {\cdot }})$ , so P is actually an ideal of B. Since $(P,+)$ and $(P,{\boldsymbol {\cdot }})$ are cyclic, they contain only one subgroup of order p; let I and J be the cyclic subgroup of order p of $(P,+)$ and $(P,{\boldsymbol {\cdot }})$ , respectively. Since I is characteristic in $(P,+)$ , it follows that $I=J$ is an ideal of B of prime order p. By induction, $B/I$ is supersoluble and hence B is supersoluble.

Proof. Given a finite chain of ideals of the type described in Definition 3.1, we show that it is essentially possible to swap adjacent factors that are not in the order prescribed by the statement (although in one case the swap could produce some new factors of order $2$ at the top). Then, in order to complete the proof, we only need to keep swapping the factors until there is no further possible swap. Let $H\leq K\leq L$ be ideals of B.

Suppose first $|K/H|=p$ and $|L/K|=q$ are prime numbers such that $p<q$ . Since $(L/H,+)$ and $(L/H,{\boldsymbol {\cdot }})$ are supersoluble, we have (see [Reference Robinson34], 5.4.8) that their Sylow q-subgroups are normal, so each of these groups is a direct product of a cyclic group of order p and a cyclic group of order q; in particular, both groups are nilpotent. Thus, the Sylow q-subgroup $Q/H$ of $(L/H,+)$ is an ideal of $B/H$ . Clearly, $|Q/H|=q$ and $|L/Q|=p$ . Hence, in this case, one can indeed swap ideals.

Suppose now that $K/H\leq \operatorname {Soc}(B/H)$ , $(K/H,+)$ is an infinite cyclic group and $|L/K|=p$ is an odd prime number. For the sake of clarity, we put $H=\{0\}$ . Since $L/K$ is an odd prime, we have that $(L,+)$ and $(L,{\boldsymbol {\cdot }})$ are abelian groups, and furthermore $(L,+)$ and $(L,{\boldsymbol {\cdot }})$ are either infinite cyclic or isomorphic to $C_p\times \mathbb {Z}$ .

Suppose $(L,+)$ is infinite cyclic and $(L,{\boldsymbol {\cdot }})\simeq C_p\times \mathbb {Z}$ , so there exists $0\neq x\in L$ such that $x^p=0$ . Choose a prime $q\neq p$ . Then $L/K^{q,{\boldsymbol {\cdot }}}$ is the direct product of two ideals, one of order p and the other of order q. Clearly, $xK^{q,{\boldsymbol {\cdot }}}$ is contained in the ideal of order p, so $px\in K^{q,{\boldsymbol {\cdot }}}$ . The arbitrariness of q and the fact that $\bigcap _{p\neq q\in \mathbb {P}}K^{q,{\boldsymbol {\cdot }}}=\{0\}$ show that $px=0$ . Because, by assumption, $(L, +)$ is infinite cyclic, this implies $x = 0$ . This contradiction shows that if $(L,+)$ is infinite cyclic, then also $(L,{\boldsymbol {\cdot }})$ is such. In this case, Corollary 4.3 of [Reference Stefanello and Trappeniers36] yields that L is abelian (as a brace). Let $(L,+)=\langle y\rangle _+$ ; in particular, $(K,+)=\langle y^p\rangle _+$ . Now, since $(L,+)$ is infinite cyclic and $(K,+)\leq Z(B,+)$ , we have that $(L,+)\leq Z(B,+)$ . Moreover, if b is any element of B, then (by Equations (†))

$$ \begin{align*}0 = y^2 \ast b =y\ast (y\ast b)+2(y\ast b)=2(y\ast b)\end{align*} $$

as $y \ast (y \ast b) \in L \ast L = 0$ because L is an ideal which is a trivial brace. More in general, by induction one proves

$$ \begin{align*}0=y^p\ast b=p(y\ast b),\end{align*} $$

which means that $y\ast b=0$ and hence that $L\leq \operatorname {Soc}(B)$ .

We may therefore assume that $(L,+)=L_1\times L_2$ , where $L_2\subseteq (\operatorname {Soc}(B),+)$ is infinite cyclic, and $L_1$ has order p. In this case, we have that L is a brace of abelian type such that $(L,{\boldsymbol {\cdot }})$ is not cyclic (otherwise, it would be a trivial brace by [Reference Stefanello and Trappeniers36], Corollary 4.3, with elements of order p, a contradiction). Thus, $(L,{\boldsymbol {\cdot }})\simeq C_p\times \mathbb {Z}$ so that $L_1$ is an ideal of B such that $L/L_1$ is contained in $\operatorname {Soc}(B/L_1)$ and is $1$ -generator.

Finally, suppose $|K/H|=2$ , $L/K\leq \operatorname {Soc}(B/K)$ and $(L/K,+)$ is infinite cyclic. Also in this case, both $(L/H,+)$ and $(L/H,{\boldsymbol {\cdot }})$ are abelian. It follows from Theorem 3.10 that there is an ideal $U/H$ of $B/H$ such that $U/H\leq \operatorname {Soc}(B/H)\cap L/H$ , $(U/H,+)$ is infinite cyclic, and $L/U$ is finite. Now, $L/U$ is the direct product of an even-order ideal $P/U$ of $B/U$ and an odd-order ideal $Q/U$ of $B/U$ . An application of the previous case to the chain $H\leq U\leq Q$ shows that $Q/H\leq \operatorname {Soc}(B/H)$ and that $(Q/H,+)$ is infinite cyclic. Since $|L/Q|$ is a power of $2$ , we can refine $L/Q$ with a finite chain of ideals of order $2$ . Thus, we have swapped the types of $K/H$ and $L/K$ with some possible addition of factors of order $2$ at the top. The statement is proved.

Proof. Let q be the largest odd prime in $\pi (B,+)$ ; of course, we may assume $p\leq q$ . It follows from Theorem 3.11 that B has an ideal I such that $|I|$ is a power of q and both $(B/I,+)$ and $(B/I,{\boldsymbol {\cdot }})$ have no q-elements. This means that $I=U_r^+(B)=U_r^{{\boldsymbol {\cdot }}}(B)$ is an ideal of B, where r is the largest prime that is strictly less than q. Now, the largest prime in $\pi (B/I,+)$ is strictly less than q, so induction yields that

$$ \begin{align*}U_p^+(B)/I=U_p^+(B/I)=U_p^{{\boldsymbol{\cdot}}}(B/I)=U_p^{{\boldsymbol{\cdot}}}(B)/I\end{align*} $$

is an ideal of $B/I$ . Thus, $U_p^+(B)=U_p^{{\boldsymbol {\cdot }}}(B)\trianglelefteq B$ .

Proof. (1) It follows from Theorem 3.10 that B has an ideal J such that $B/J$ is finite and $(J,+)$ is torsion-free. Now, $J\cap I=\{0\}$ , so if we can prove that $I+J/J\leq \operatorname {Soc}(B/J)$ , then

$$ \begin{align*} [a,b]_++J=a\ast b+J=J \end{align*} $$

for all $a\in I$ and $b\in B$ , so $[a,b]_+=a\ast b=0$ and consequently $I\leq \operatorname {Soc}(B)$ . Thus, without loss of generality we may replace B by $B/J$ and assume that B is finite. If $p=2$ , then obviously $I\leq \zeta (B)$ . Assume $p>2$ , and let q be the largest prime strictly less than p. Put $P=U_q^+(B)$ and $Q=U_p^+(B)$ ; in particular, $Q\leq P$ and $(P/Q,+)$ is a p-group. A further replacement of B by $B/Q$ allows us to assume that p is the biggest prime dividing the order of B. It follows from Theorem 3.7 that $I\leq \zeta (P)$ . Now, if b is an arbitrary element of B, we can write $b=c+d$ , where $c\in P$ and d is a $p'$ -element of $(B,+)$ of order m. If $I=\langle a\rangle $ , then $a\in Z(B,+)$ , so (by Equations (†) and the fact that I is an ideal)

$$ \begin{align*} a\ast (\ell d)=\ell(a\ast d) \end{align*} $$

for all $\ell \in \mathbb {N}$ . Thus,

$$ \begin{align*} m(a\ast d)=a\ast (md)=0 \end{align*} $$

and consequently $a\ast d=0$ since $a\ast d$ is a p-element of $(B,+)$ . Therefore (again by Equations (†) and the fact that I is an ideal),

$$ \begin{align*} a\ast b=a\ast(c+d)=a\ast c+a\ast d=0 \end{align*} $$

and hence $a\in \operatorname {Soc}(B)$ , as desired.

(2) It follows from Theorem 3.11 that $|I/\operatorname {Soc}(B)\cap I|=2^n$ for some $n\in \mathbb {N}$ . Since $(I,+)$ is infinite cyclic, and $\big (\operatorname {Soc}(B)\cap I,+\big )$ is central in $(B,+)$ , we have that $I\leq Z(B,+)$ (recall that the only automorphism of the infinite cyclic group is the inversion). Now, let $a\in I$ and $b\in B$ . Then (by Equations (†))

$$ \begin{align*}0=\big(a^{2^n}\big)\ast b=2^n(a\ast b),\end{align*} $$

so $a\ast b=0$ and $a\in \operatorname {Ker}(\lambda )$ . This proves that $a\in \operatorname {Soc}(B)$ and completes the proof.

Proof. Suppose B is weakly supersoluble, and let I be an abelian ideal of B such that either $I\leq Z(B,+)$ and $(I,+)$ is infinite cyclic, or I has prime order. In the latter case, $B/I$ is supersoluble by induction, and so B is supersoluble. Assume $(I,+)$ is infinite cyclic and $I\leq Z(B,+)$ . If p is any prime, then, by induction, $B/I^{p,+}$ is supersoluble, so Lemma 3.16 yields that $I/I^{p,+}\leq \operatorname {Soc}(B/I^{p,+})$ . Since $\bigcap _{p\in \mathbb {P}} I^{p,+}=\{0\}$ , we have that $I\leq \operatorname {Soc}(B)$ . The proof is complete.

Proof. It follows from [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], Theorem 3.10, that we may assume B is finite. Let I be any ideal of B whose order is a prime, p say. Then either $I\leq C$ , so an easy induction completes the proof, or $I+C=B$ and consequently

$$ \begin{align*}|B:C|=|(B,+):(C,+)|=p.\end{align*} $$

The statement is proved.

Proof. Let $n=|B:C|$ , and let D be the normal core of $(C,+)$ in $(B,+)$ . Then the index of D in B divides $n!$ . Thus, $U_p^+(B)\leq D\leq C$ , and the statement is proved.

Proof. By Lemma 3.21, $U_p^+(B)\leq C$ , so replacing B by $B/U_p^+(B)$ , we may assume that the order of B is a power of p. Then Theorem 3.7 shows that B is centrally nilpotent, and an application of [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti4], Theorem 4.6, yields that $C\trianglelefteq B$ .

Proof. It follows from [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], Theorem 3.10, that we may assume B is finite. Then Theorem 3.22 yields that C is an ideal of B.

Proof. Let

$$\begin{align*}(B,+) = \langle a \rangle \times \langle b \rangle \simeq \operatorname{C}_4\times \operatorname{C}_2 \quad \text{and}\quad (C,{\boldsymbol{\cdot}})\!=\! \langle x, y \mid x^4 \!= y^2 =\! 1,\, xy = yx^3\rangle \simeq \operatorname{D}_8\end{align*}$$

be groups of order $8$ . The multiplicative group acts on the additive by means of the map $\lambda $ defined by

$$\begin{align*}\begin{array}{lllll} \lambda_x(a) = 3a+b, & & \lambda_y(a) = 3a +b, \\ \lambda_x(b) = 2a+b, & & \lambda_y(b) = b. \\ \end{array} \end{align*}$$

Consider the semidirect product G of B and C with respect to this action (we use multiplicative notation in G). Then G turns out to be a trifactorised group as it possesses a subgroup $D=\langle ax, a^2y\rangle $ such that $D\cap C=D\cap B=\{1\}$ and $DC=BD=G$ . D leads to the bijective $1$ -cocycle $\delta \colon C\longrightarrow B$ with respect to $\lambda $ given by Table 1. This yields a product in B, and we get a brace of abelian type $(B,+,{\boldsymbol {\cdot }})$ of order $8$ . This brace corresponds to SmallBrace(8,18) in the Yang–Baxter library for GAP.

Table 1 Bijective $1$ -cocycle associated with Example 3.24.

From Table 1, we see that $S = \{1, x^2,y,x^2y\}$ is the unique subgroup of order $4$ in $(C,{\boldsymbol {\cdot }})$ such that $\delta (S) \leq (B,+)$ . Thus, $\delta (S)$ is a maximal subbrace of B. Moreover, $\delta (xy)$ and $\delta (x^3y)$ are not elements of order $2$ . Therefore, all subbraces of order $2$ of B are contained in $\delta (S)$ . Hence, $\delta (S)$ is the unique maximal subbrace of B and it is of prime index. Finally, we note that B is not supersoluble because no subbrace of order $2$ is $\lambda $ -invariant.

Proof. We use induction on the Hirsch length h of $(B,+)$ . If $h=0$ , then the order of B is a $\pi $ -number, so we are done. Assume $h>0$ , and put $U=U_2^+(B)$ . It follows from Theorem 3.11 that there is an ideal I of B such that $I/U\leq \operatorname {Soc}(B/U)$ , $(I/U,+)$ is infinite cyclic and $U_2^+(B/I)=\{0\}$ . By Theorem 3.10, there is an ideal J of B such that $J\leq I\cap \operatorname {Soc}(B)$ and $(J,+)$ is infinite cyclic, and by Theorem 3.11 we may even assume $U_p^+(I/J)=\{0\}$ . Thus, for each $n\in \mathbb {N}$ , the largest prime in $\pi \big (B/J^{2^n,{\boldsymbol {\cdot }}},+\big )$ is precisely p, so, by induction, $B/J^{2^n,{\boldsymbol {\cdot }}}$ is residually $\pi $ -finite. Since $\bigcap _{n\in \mathbb {N}}J^{2^n,{\boldsymbol {\cdot }}}=\{0\}$ , we have that B is residually $\pi $ -finite.

Proof. Clearly, both the additive and multiplicative groups are polycyclic-by-finite. Let $S=\big (\operatorname {Soc}(B),{\boldsymbol {\cdot }}\big )$ . Since $(B,{\boldsymbol {\cdot }})$ is supersoluble, we can find a nontrivial cyclic normal subgroup X of $(B,{\boldsymbol {\cdot }})$ in S. By Lemma 1.10 of [Reference Cedó, Smoktunowicz and Vendramin17], we have that X is an ideal of B, so an easy induction on the Hirsch length of $(B,+)$ yields that $B/X$ is supersoluble. Therefore, B is supersoluble and the statement is proved.

Proof. For the sake of clarity we prove (2). Since every $2$ -generator multiplicative subgroup is contained in a $2$ -generator subbrace, it follows from the above mentioned result that $(B,{\boldsymbol {\cdot }})$ is locally supersoluble. We claim by induction on n that B is supersoluble. If $n\leq 1$ , there is nothing to prove. Suppose $n>1$ . By induction, $B/\operatorname {Soc}(B)$ is supersoluble, so it is finitely presented. Therefore, Corollary 3.3 of [Reference Trombetti38] yields that $S=\operatorname {Soc}(B)$ is finitely generated as an ideal of B by a set E. Let F be a finite subset of B that generates both $(B,{\boldsymbol {\cdot }})$ and $(B,+)$ modulo $(S,{\boldsymbol {\cdot }})$ and $(S,+)$ , respectively.

Now, if X is the multiplicative subgroup generated by E and F, then X is supersoluble, so every subgroup of X is finitely generated. Let L be the smallest normal subgroup L of X containing E. Then $(L,{\boldsymbol {\cdot }})$ is normal in $(B,{\boldsymbol {\cdot }})$ , so L is an ideal of B, being contained in S. Since S is the smallest ideal of B containing E, we have that $L=S$ . This means that $(S,{\boldsymbol {\cdot }})$ is finitely generated, so also $(B,{\boldsymbol {\cdot }})$ is finitely generated. Therefore, $(B,{\boldsymbol {\cdot }})$ is supersoluble and B is supersoluble as well by Theorem 3.28.

Proof. Let

$$\begin{align*}\begin{array}{l} (B,+) = \langle a \rangle \times \langle b \rangle \times \langle c \rangle \times \langle d \rangle \times \langle e \rangle \simeq \operatorname{C}_2^5 \quad \text{and}\\[0.2cm] (C,{\boldsymbol{\cdot}})\!=\! \langle x, y, z \mid x^4 \!= y^4 \!= z^2 = 1,\, xy = yx,\, zxz = x^3y^2,\, zyz = x^2y\rangle\!\simeq\! \big(\operatorname{C}_4 \times \operatorname{C}_4\big)\!\rtimes\!\operatorname{C}_2. \end{array} \end{align*}$$

Of course, $(B,+)$ and $(C,{\boldsymbol {\cdot }})$ are groups of order $32$ . The multiplicative group acts on the additive by means of the map $\lambda $ defined as

$$\begin{align*}\begin{array}{lllll} \lambda_x(a) = c+d+e & & \lambda_y(a) = a & & \lambda_z(a) = a\\ \lambda_x(b) = c+e & & \lambda_y(b) = a+c+e & & \lambda_z(b) = b+c+d+e\\ \lambda_x(c) = a+c & & \lambda_y(c) = a+b+c+d & & \lambda_z(c) = a+b+c+d\\ \lambda_x(d) = a+b & & \lambda_y(d) = b & & \lambda_z(d) = c+e \\ \lambda_x(e) = a+b+c & & \lambda_y(e) = a+e & & \lambda_z(e) = a+b+e.\\ \end{array} \end{align*}$$

Consider the semidirect product G of B and C with respect to this action (we use multiplicative notation in G). Then G turns out to be a trifactorised group as it possesses a subgroup $D=\langle acx, acey, az\rangle $ such that $D\cap C=D\cap B=\{1\}$ , $DC=BD=G$ . Thus, there is a bijective $1$ -cocycle $\delta \colon C\longrightarrow B$ with respect to $\lambda $ given by Table 2. This yields a product in B, and we get a brace of abelian type $(B,+,{\boldsymbol {\cdot }})$ of order $32$ . This brace corresponds to SmallBrace(32, 25055) in the Yang–Baxter library for GAP.

Table 2 Bijective $1$ -cocycle associated with Example 3.30.

The ideals of B of order $16$ are given by the $\lambda $ -invariant subgroups $I = \langle a,c,b+d,b+e\rangle _+$ , $J = \langle a,b,d,c+e\rangle _+$ and $K = \langle a, b+c, b+d, e\rangle _+$ . In addition, B has only one more proper ideal, $L = \langle a, b+d, b+c+e\rangle _+$ of order $8$ . Thus, B is not supersoluble.

Note also that L, $L_2 = \langle a+b+d, b+c+e\rangle _+$ and $L_3 = \langle b+c+e\rangle _+$ are ideals of both I and J, so I and J are supersoluble braces.

Since B is of abelian type, $\partial (B) = B \ast B = L$ . Therefore, we also conclude that $\partial (B)$ is centrally nilpotent as it has a central series of ideals (see also Theorem 3.7).

Proof. (1) We claim that $\operatorname {Soc}(B)/I\cap \zeta (B/I)\neq \{0\}$ for every ideal I of B such that $I<\operatorname {Soc}(B)$ . For the sake of clarity, we prove the previous claim when $I=\{0\}$ . Thus, let $0\neq b\in \operatorname {Soc}(B)$ . Since B is left-nilpotent, there are elements $b_1,\ldots ,b_\ell $ of B such that

$$ \begin{align*}c=b_1\ast(b_2\ast(\ldots \ast (b_\ell\ast b)))\ldots)\neq0\end{align*} $$

and $a\ast c=0$ for all $a\in B$ . It follows from Lemma 1.10 of [Reference Cedó, Smoktunowicz and Vendramin17] that $c\in \zeta (B)$ , and the claim is proved.

The previous claim enables us to construct an ascending chain of ideals of B whose factors are central in B. This certainly implies that $\operatorname {Soc}(B)\leq \overline {\zeta }(B)$ .

Finally, if $\operatorname {Soc}(B)$ is finitely generated, then it satisfies the maximal condition on subbraces, so the previous ascending series stops after finitely many steps and $\operatorname {Soc}(B)$ is contained in some finite term of its upper central series.

(2) If $a\ast b\neq 0$ , then $I=\langle a\ast b\rangle $ , so $a\ast b=nb$ for some n which is prime to p. It follows that $a\ast (a\ast b)=n(a\ast b)\neq 0$ and hence that $I=\langle a\ast (a\ast b)\rangle $ . Continuing in this way, we contradict the left-nilpotency of B. Thus, $\lambda _a(b)=b$ for all $a\in B$ .

Proof. Let

$$ \begin{align*}\{0\}=I_0\leq I_1\leq\ldots\leq I_\ell=B\end{align*} $$

be a finite chain of ideals of B such that, for each $0\leq i<\ell $ , either $I_{i+1}/I_i\leq \operatorname {Soc}(B/I_i)$ and $\big (I_{i+1}/I_i,{\boldsymbol {\cdot }}\big )$ is infinite cyclic, or $I_{i+1}/I_i$ has prime order. By Lemma 3.31, $(B,{\boldsymbol {\cdot }})$ stabilizes the finite series

$$ \begin{align*}\{0\}=\big(I_0,+\big)\leq \big(I_1,+\big)\leq\ldots\leq \big(I_\ell,+\big)=(B,+)\end{align*} $$

in the natural semidirect product $(B,+)\rtimes _\lambda (B,{\boldsymbol {\cdot }})$ . Thus, $(B,{\boldsymbol {\cdot }})/\operatorname {Ker}(\lambda )$ is nilpotent by a well-known theorem of Philip Hall (see [Reference Hall28], Theorem 1, and [Reference Newell and Trombetti31], Theorem), and consequently $(B,{\boldsymbol {\cdot }})$ is nilpotent.

Proof. By Lemma 3.31, $\operatorname {Ker}(\lambda )$ is contained in some finite term of the upper central series of $(B,{\boldsymbol {\cdot }})$ , so we may apply Theorem 3.32.

Proof. By [Reference Jespers, Van Antwerpen and Vendramin30], Corollary 2.15, a brace of nilpotent type is centrally nilpotent if and only if it is left and right nilpotent. Then, the result follows from Corollary 3.34 and Theorem 3.27.

Proof. Let

$$ \begin{align*}\{0\}=J_0\leq J_1\leq\ldots\leq J_n=B\end{align*} $$

be a finite chain of ideals of B such that, for each $0\leq i<n$ , either $J_{i+1}/J_i\leq \operatorname {Soc}(B/I_i)$ and $(J_{i+1}/J_i,+)$ is an infinite cyclic group, or $J_{i+1}/J_i$ has prime order. Consider the induced chain

$$ \begin{align*}\{0\}=J_0\cap I=:H_0\leq H_1:=J_1\cap I\leq\ldots\leq H_n:=J_n\cap I=I\end{align*} $$

of ideals of B. In order to show that I is B-centrally nilpotent, we only need to prove that

$$ \begin{align*}H_{n-i+1}/H_{n-i}\leq\zeta(I/H_{n-i})\end{align*} $$

for all $1\leq i\leq n$ . Now, $H_{n-i+1}/H_{n-i}$ has either prime order or is contained in the socle of $I/H_{n-i}$ and its additive group is infinite cyclic. In the former case, the quotient $H_{n-i+1}/H_{n-i}$ is a minimal ideal of the centrally nilpotent brace $I/H_{n-i}$ , so $H_{n-i+1}/H_{n-i}\leq \zeta (I/H_{n-i})$ (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti4], Theorem 4.6). Suppose the latter condition holds. Since $(I/H_{n-i},{\boldsymbol {\cdot }})$ is nilpotent, and $(H_{n-i+1}/H_{n-i},{\boldsymbol {\cdot }})$ is an infinite cyclic normal subgroup, we have that

$$ \begin{align*}(H_{n-i+1}/H_{n-i},{\boldsymbol{\cdot}})\leq Z(I/H_{n-i},{\boldsymbol{\cdot}}).\end{align*} $$

Thus, again $H_{n-i+1}/H_{n-i}\leq \zeta (I/H_{n-i})$ and the theorem is proved.

Proof. This follows from Theorem 3.36 and the maximal condition on subbraces.

Proof. Let

$$\begin{align*}\begin{array}{l} (B,+) = \langle a \rangle \times \langle b \rangle \simeq \operatorname{C}_{12}\times \operatorname{C}_2 \quad \text{and}\\[0.2cm] (C,{\boldsymbol{\cdot}})\!=\! \big(\langle x\rangle\rtimes\langle y \rangle\big) \times \langle z\rangle \times \langle t\rangle \simeq \operatorname{Sym}(3) \times \operatorname{C}_2\times \operatorname{C}_2\hspace{-1.3pt}. \end{array} \end{align*}$$

Of course, $(B,+)$ and $(C,{\boldsymbol {\cdot }})$ are groups of order $24$ . The multiplicative group acts on the additive by means of the map $\lambda $ defined by

$$\begin{align*}\begin{array}{llllll l} \lambda_x(a) = a, & & \lambda_y(a) = 5a, & & \lambda_z(a) = 7a, & & \lambda_t(a) = 7a, \\ \lambda_x(b) = b, & & \lambda_y(b) = b, & & \lambda_z(b) = b, & & \lambda_t(b) = 6a + b.\\ \end{array} \end{align*}$$

Consider the semidirect product G of B and C with respect to this action (we use multiplicative notation in G). Then G turns out to be a trifactorised group as it possesses a subgroup $D=\langle a^8x, a^6y, bz, a^3t\rangle $ such that $D\cap C=D\cap B=\{1\}$ , $DC=BD=G$ . Thus, there is a bijective $1$ -cocycle $\delta \colon C\longrightarrow B$ with respect to $\lambda $ given by Table 3. This yields a product in B, and we get a brace of abelian type $(B,+,{\boldsymbol {\cdot }})$ and of order $24$ . This brace corresponds to SmallBrace(24, 629) in the Yang–Baxter library for GAP.

Table 3 Bijective $1$ -cocycle associated with Example 3.39.

From the $\lambda $ -action described, it follows that $\operatorname {Soc}(B) = \langle 4a\rangle _+$ , $\operatorname {Soc}_2(B) = \langle 2a \rangle _+$ and $\operatorname {Soc}_3(B) = B$ .

Now, take $I:= \langle 2a, b\rangle _+$ , which is an ideal of B. Observe that the multiplicative group of I is $\delta ^{-1}(I) = \langle x, y, z\rangle \simeq \operatorname {D}_{12}$ , which is not nilpotent. Thus, I is not centrally nilpotent. Moreover, $\langle 4a, b\rangle _+$ is an ideal of I which is centrally nilpotent as it is a trivial brace of order $6$ . Therefore, $\operatorname {Fit}(I) = \langle 4a, b \rangle _+$ . But, $\operatorname {Fit}(I)$ is not an ideal of B, as it is not $\lambda $ -invariant.

Proof. It follows from Lemma 3.5 that the semidirect product $G=(B,+)\rtimes _\lambda (B,{\boldsymbol {\cdot }})$ is supersoluble. Thus, $[G,G]$ is nilpotent. On the other hand, $[G,G]\cap (B,+)$ contains $[B,B]_+$ and $B\ast B$ , so the additive group of $\partial (B)$ is nilpotent. By Theorem 3.27, $\partial (B)$ has finite multipermutational level, and the statement is proved.

Proof. Let $(B,+) = \langle a \rangle \simeq \operatorname {C}_{12}$ and $(C,{\boldsymbol {\cdot }}) = \langle \sigma , \tau \mid \sigma ^6 = \tau ^2 = 1,\, \sigma \tau = \tau \sigma ^{-1}\rangle \simeq \operatorname {D}_{12}$ be groups of order $12$ . There exists an action $c \in (C,{\boldsymbol {\cdot }}) \mapsto \lambda _c \in \operatorname {Aut}(B,+)$ given by

$$\begin{align*}\lambda_{\sigma}(a) = 7a\quad\text{and}\quad \lambda_{\tau}(a) = 5a.\end{align*}$$

If we consider the semidirect product G of B and C with respect to this action, then G turns out to be a trifactorised group, as it possesses a subgroup $D=\langle a^5\sigma , a^6\tau \rangle $ such that $D\cap C=D\cap B=\{1\}$ , $DC=BD=G$ . Thus, there is a bijective $1$ -cocycle $\delta \colon C\longrightarrow B$ with respect to $\lambda $ given by Table 4. This yields a product in B, and we get a brace of abelian type $(B,+,{\boldsymbol {\cdot }})$ and of order $12$ , corresponding to SmallSkewbrace(12, 15) in the Yang–Baxter library for GAP.

Table 4 Bijective $1$ -cocycle associated with Example 3.41.

Since $\operatorname {Ker} \lambda = \langle \sigma ^2\rangle $ , it holds that $\langle 4a\rangle _+ = \operatorname {Soc}(B)$ . Then, $\overline {B}:= B/\operatorname {Soc}(B)$ is a brace of order $4$ . Moreover, if we consider $\overline {\lambda }\colon \big (C/\langle \sigma ^2\rangle ,{\boldsymbol {\cdot }}\big ) \rightarrow \operatorname {Aut}(B/\langle 4a \rangle , +)$ , it follows that $\operatorname {Ker}\overline {\lambda } = \big \langle \tau + \langle \sigma ^2\rangle \big \rangle $ . Therefore, $\operatorname {Soc}_2(B) = \langle 2a \rangle _+$ and $B/\operatorname {Soc}_2(B)$ is a (trivial) brace of order $2$ . Hence, $\operatorname {Soc}_3(B) = B$ .

On the other hand, $B \ast B = \langle 2a \rangle _+$ and $\delta ^{-1}\big (\langle 2a \rangle _+\big ) = \langle \sigma ^2, \tau \rangle \simeq \operatorname {Sym}(3)$ , which is not nilpotent. Hence, $\partial (B) = B \ast B$ is not centrally nilpotent.

Proof. By hypothesis, B has a chain of ideals

$$ \begin{align*}\{0\}=I_0\leq I_1\leq\ldots\leq I_n=B\end{align*} $$

such that, for each $0\leq i<n$ , either $I_{i+1}/I_i$ is contained in $\operatorname {Soc}(B/I_i)$ and is $1$ -generator, or $I_{i+1}/I_i$ has prime order. In particular, for each $0\leq i<n$ , we have that $C_{(B,{\boldsymbol {\cdot }})}(I_{i+1}/I_i,{\boldsymbol {\cdot }})$ and $C_{(B,+)}(I_{i+1}/I_i,+)$ have finite index in $(B,{\boldsymbol {\cdot }})$ and $(B,+)$ , respectively. By Theorem 3.10 of [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], there is a finite-index ideal J of B that is contained in all these centralizers. Intersecting the chain of the $I_i$ ’s with J, we see that J is centrally nilpotent (one also needs to use Lemma 3.16). Thus, $J\leq \operatorname {Fit}(B)$ by Corollary 3.37 and the statement is proved.

Proof. By Theorem 3.11, there is an ideal J of B contained in I such that $J\leq \operatorname {Soc}(B)$ and $(J,+)$ is infinite cyclic. Since $(I,{\boldsymbol {\cdot }})$ is nilpotent and $(J,{\boldsymbol {\cdot }})$ is a normal infinite cyclic subgroup of $(B,{\boldsymbol {\cdot }})$ , we have that $J\leq Z(I,{\boldsymbol {\cdot }})$ , so $J\leq \zeta (I)$ .

In order to use induction (consequently completing the proof), we should find such an ideal J with an additional property: $(I/J,+)$ must be torsion-free. Since I is centrally nilpotent, the set $T/J$ of all periodic elements of $(I/J,+)$ is an ideal of $I/J$ and coincides with the set of all periodic elements of $(I/J,{\boldsymbol {\cdot }})$ (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti4], Theorem 4.12). Thus, $T/J$ is also an ideal of $B/J$ . Now, since $(T,+)$ is torsion-free, central-by-finite and has Hirsch length $1$ , we have that $(T,+)$ is infinite cyclic. Moreover, $T\leq \zeta (I)$ by [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti4], Corollary 4.17, so, in particular, T is abelian. Also, since $(T,+)$ is infinite cyclic and normal in $(B,+)$ , we have $T\leq Z(B,+)$ . Let $x\in T$ and $b\in B$ . Then

$$ \begin{align*} m(x\ast b)=(x^m)\ast b=0, \end{align*} $$

where $m=|T/J|$ , and hence $x\ast b=0$ . Therefore, $x\in \operatorname {Soc}(B)$ and the statement is proved.

Proof. Since every ideal of order $2$ of a brace is always contained in the centre, it follows from Theorem 3.11 that we may assume $(I,+)$ is torsion-free. Now, by Lemma 3.43, there is a chain

$$ \begin{align*}\{0\}=I_0\leq I_1\leq\ldots\leq I_n=I\end{align*} $$

of ideals of B such that $I_{i+1}/I_i\leq \operatorname {Soc}(B/I_i)\cap \zeta (I/I_i)$ and $(I_{i+1}/I_i,+)$ is infinite cyclic. By induction on n, we only need to show that $I_1\leq \zeta (U)$ .

Now, let $x\in U$ . Then there are $y\in U$ and $z\in I$ such that $x=y^2z$ . Since $(I_1,{\boldsymbol {\cdot }})$ is infinite cyclic, we have $[y^2,I_1]_{{\boldsymbol {\cdot }}}=\{0\}$ , so also $[x,I_1]_{{\boldsymbol {\cdot }}}=\{0\}$ . This implies $I_1\leq Z(U,{\boldsymbol {\cdot }})$ and consequently $I_1\leq \zeta (U)$ .

Proof. Let $F=\operatorname {Fit}(B)$ . It follows from Corollary 3.37 and Theorem 3.42 that F is a centrally nilpotent ideal of B and $B/F$ is finite. Moreover, Lemma 3.44 yields that $U_2^+(B/F)=\{0\}$ , so $|B/F|$ is a power of $2$ .

Proof. Let T be the largest finite ideal of B (it exists because B satisfies the maximal condition on subbraces). Replacing B by $B/T$ , we may assume that $T=\{0\}$ . In particular, the additive group of $\zeta (B)$ is torsion-free, and consequently, the additive group of every factor of the upper central series of B is torsion-free (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti4], Theorem 4.16). Then, also using Theorem 3.11, we see that B has a finite chain of ideals

$$ \begin{align*}\{0\}=I_0\leq I_1\leq \ldots\leq I_\ell=J_0\leq J_1\leq\ldots\leq J_m=U\leq B,\end{align*} $$

where

1. (a) $I_{i+1}/I_i\leq \zeta (B/I_i)$ and $(I_{i+1}/I_i,+)$ is infinite cyclic for all $0\leq i<\ell $ ,

2. (b) $|J_{i+1}/J_i|$ is an odd prime for all $0\leq i<m$ , and

3. (c) $|B/U|$ is a power of $2$ .

We now claim that there is a finite chain of ideals of B contained in $J_1$ whose factors satisfy (a). Write $p=|J_1/J_0|$ . It follows from Theorem 3.11 that only one of the following two possibilities holds:

1. (1) $J_1/I_{\ell -1}\leq \operatorname {Soc}(B/I_{\ell -1})$ and $\big (J_1/I_{\ell -1},+\big )$ is infinite cyclic.

2. (2) $J_1/I_{\ell -1}=I_\ell /I_{\ell -1}\times P/I_{\ell -1}$ , where $P/I_{\ell -1}$ is an ideal of $B/I_{\ell -1}$ having order p.

In case (2), the claim follows by induction on $\ell $ (note that the base case $\ell =0$ implies that $B=\{0\}$ , so we may assume $\ell \geq 1$ ). In case (1), $(J_1/I_{\ell -1},{\boldsymbol {\cdot }})$ is infinite cyclic, and $(I_\ell /I_{\ell -1},{\boldsymbol {\cdot }})\leq Z(B/I_{\ell -1},{\boldsymbol {\cdot }})$ , so also $(J_1/I_{\ell -1},{\boldsymbol {\cdot }})\leq Z(B/I_{\ell -1},{\boldsymbol {\cdot }})$ and hence $J_1/I_{\ell -1}\leq \zeta (B/I_{\ell -1})$ . The claim is proved.

Now, repeated applications of the previous claim allow us to assume $m=0$ (compare with the proof of Theorem 3.48); in particular, $|B/\zeta _\ell (B)|$ is a power of $2$ . It follows from Theorem 3.7 that $B/\zeta _\ell (B)$ is centrally nilpotent, so B is centrally nilpotent. Since the additive groups of the factors of the upper central series of B are torsion-free and $B/\zeta _n(B)$ is finite by hypothesis, we have that $B=\zeta _n(B)$ , so $\Gamma _{n+1}(B)$ is finite.

Proof. By Lemma 3.48, $\Gamma _n(B)$ contains a finite-index subbrace I of B such that $I\leq \zeta (B)$ . By induction, $U/I=\zeta _{n-1}(B/I)$ has finite-index in B, so also $\zeta _n(B)\geq U$ has finite index in B.

Proof. If B is supersoluble, then obviously such are also its (finite) homomorphic images. Conversely, suppose B is infinite and that all its finite homomorphic images are supersoluble. Consider the natural semidirect product $G=N\rtimes _\lambda X$ , where $N=(B,+)$ and $X=(B,{\boldsymbol {\cdot }})$ . Then, G is polycyclic-by-finite since from the definition of almost polycyclic brace it follows that both $(B,+)$ and $(B,{\boldsymbol {\cdot }})$ are polycyclic-by-finite, and this class is closed with respect to forming extensions. If H is any finite-index normal subgroup of G, then there is a finite-index ideal I of B such that $M=(I,+)\leq H\cap N$ and $Y=(I,{\boldsymbol {\cdot }})\leq H\cap X$ (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], Theorem 3.10). Since $G/MY\simeq (B/I,+)\rtimes _\lambda (B/I,{\boldsymbol {\cdot }})$ , we have that $G/H$ is supersoluble by Lemma 3.5. The arbitrariness of H shows that G is supersoluble (see [Reference Baer10], Satz 3.1).

Now, let I be a nontrivial normal cyclic subgroup of G contained in $(\operatorname {Soc}(B),+)\leq N$ (this is possible by Theorem 3.10). Then Lemma 1.10 of [Reference Cedó, Smoktunowicz and Vendramin17] yields that $I\trianglelefteq B$ . By induction, $B/I$ is supersoluble, and consequently B is supersoluble.

Proof. Two recursive procedures are set in motion. The first constructs all finite braces in increasing orders and finds finitely many homomorphisms from B into each of these braces. Then it tests if the image of B is supersoluble. By Theorem 3.51, this procedure stops if B is not supersoluble.

In order to complete the proof of the statement, we need a procedure stopping if B is supersoluble. By Theorem 4.7 of [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], we find a set $\{b_1,\ldots ,b_n\}$ of additive and multiplicative generators of B, and, by Theorem 4.13 of [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], we find $n\in \mathbb {N}$ and $I\trianglelefteq B$ such that $I\leq \operatorname {Soc}_n(B)$ and $B/I$ is finite; in particular, we find a set $\{a_1,\ldots ,a_m\}$ of additive and multiplicative generators of I. Suppose $I\neq \{0\}$ (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], Theorem 4.4). Then we recursively enumerate all b-words with symbols $x_1,\ldots ,x_m$ , and for each of these b-words, say $\theta (x_1,\ldots ,x_m)$ , we check if $[a,b_i]_+=a\ast b_i=a\ast a=0\neq a$ for all $1\leq i\leq n$ , where $a=\theta (a_1,\ldots ,a_m)$ (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], Theorem 4.1). If the result is negative, we move to the next b-word. If the result is positive, it means that $\langle a\rangle _+\leq \operatorname {Soc}(B)$ ; then we also check if $\langle a\rangle _+$ is an ideal of B by using Theorems 4.4 and 4.9 of [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5]. If so, we check if $I=\langle a\rangle _+$ (see [Reference Ballester-Bolinches, Esteban-Romero, Ferrara, Pérez-Calabuig and Trombetti5], Theorem 4.4); otherwise, we move to the next b-word. If $I=\langle a\rangle _+$ , then $B/\langle a\rangle _+$ is finite; if not, we move to $B/\langle a\rangle _+$ and we repeat the previous procedure until we find, if possible, a finite chain of ideals of B

$$ \begin{align*}\{0\}=I_0\leq I_1\leq\ldots\leq I_\ell=I\end{align*} $$

such that, for each $0\leq i<\ell $ , $I_{i+1}/I_i\leq \operatorname {Soc}(B/I_i)$ and $(I_{i+1}/I_i,+)$ is cyclic. Then we move to $B/I$ , which is finite, so without loss of generality we may assume B is finite. As before, we recursively enumerate all elements of B, and for each of them, say x, we check if $\langle x\rangle _+$ is an ideal of B such that $\langle x\rangle _{{\boldsymbol {\cdot }}}=\langle x\rangle _+$ . If x is such an element, then we check if $B=\langle x\rangle _+$ . If so, the algorithm stops. If not, we repeat the same procedure to the quotient $B/\langle x\rangle _+$ , until we find, if possible, a finite chain of ideals of B

$$ \begin{align*}\{0\}=J_0\leq J_1\leq\ldots\leq J_r=I\end{align*} $$

such that, for each $0\leq i<r$ , $(I_{i+1}/I_i,+)$ and $(I_{i+1}/I_i,{\boldsymbol {\cdot }})$ are cyclic groups (see also Remark 3.4). It is clear that if B is supersoluble, then the algorithm stops after finitely many steps.

Proof. We only prove the result for a hypercyclic brace B. By definition, B has a nonzero ideal I such that either $I\leq \operatorname {Soc}(B)$ and $(I,+)$ is infinite cyclic, or $|I|$ is a prime. Then $M=(I,+)$ and $N=(I,+)\rtimes _\lambda (I,{\boldsymbol {\cdot }})$ are normal subgroups of G such that M and $N/M$ are cyclic groups, and $G/N\simeq (B/I,+)\rtimes _\lambda (B/I,{\boldsymbol {\cdot }})$ . Iterating this procedure by transfinite recursion, we may define an ascending series of normal subgroups of G with cyclic factors that must reach G at some point.

Proof. If F is any finitely generated subbrace of B containing I, then $I\leq \operatorname {Soc}(B)$ by Lemma 3.16. The arbitrariness of F yields $I\leq \operatorname {Soc}(B)$ and completes the proof.

Proof. It follows from Corollary 3.3 of [Reference Trombetti38] that B is locally weakly supersoluble, so also locally supersoluble by Theorem 3.17. Now, let $J\leq I$ be ideals of B such that $I/J$ is abelian and $(I/J,+)$ is a central cyclic subgroup of $(B/J,+)$ . If $I/J$ is infinite, then, since $B/J$ is locally supersoluble, it follows from Lemma 4.2 that $I/J\leq \operatorname {Soc}(B/J)$ . Suppose $I/J$ is finite. Since $(I/J,+)$ and $(I/J,{\boldsymbol {\cdot }})$ are finite cyclic groups, there is a finite chain of ideals of B

$$ \begin{align*}I=J_0\leq J_1\leq\ldots\leq J_\ell=J\end{align*} $$

whose factors have prime order (see the proof of Theorem 3.8 for more details). Therefore, B is hypercyclic.

Proof. Of course, we only need to show that if all countable subbraces of B are hypercyclic, then B is hypercyclic. Moreover, by Theorem 4.4 (see also the proof of Theorem 3.8), we only need to find a nonzero $1$ -generator subgroup I of $(B,+)$ that has the following property with respect to B:

• (⋆) I is an abelian ideal of B, and either $|I|$ is finite or $I\leq \operatorname {Soc}(B)$ .

Suppose there is no such a subgroup. Let $0\neq x\in B$ , and put $X_0=\langle x\rangle $ ; in particular, $X_0$ is a countable subbrace of B. Assume now we have defined a countable subbrace $X_n$ of B for some nonnegative integer n. For each $y\in X_n\setminus \{0\}$ , we have that $\langle y\rangle _+$ does not satisfy ( $\star $ ), so either $y\ast y\neq 0$ (in this case, we let $b_y=0$ ), or there is an element $b_y\in B$ such that one of the following conditions holds:

• ○ $\langle y\rangle _+$ is not an ideal of $\langle X_n,b_y\rangle $ .

• ○ $\langle y\rangle _+$ is an infinite ideal of $\langle X_n,b_y\rangle $ and $y\ast b_y\neq 0$ .

Define $X_{n+1}=\langle X_n,b_y\,:\, y\in X_n\setminus \{0\}\rangle $ . Finally, put

$$ \begin{align*}X=\bigcup_{n\geq0}X_n.\end{align*} $$

Since $X_i\leq X_{i+1}$ for all i, we have that X is a countable subbrace of B. Thus, $(X,+)$ contains a nonzero $1$ -generator subgroup $U=\langle a\rangle _+$ satisfying ( $\star $ ) with respect to X. Therefore, there is $\ell \geq 0$ such that $a\in X_\ell $ , and this means that $U=\langle a\rangle _+$ does not satisfy property ( $\star $ ) with respect to $\langle X_{\ell },b_a\rangle \leq X_{\ell +1}\leq X$ , an obvious contradiction.

Proof. Let $x\in U_p^{+}(B)$ , $b\in B$ , and put $X=\langle x,b\rangle $ . Then

$$ \begin{align*}\langle x\rangle^X\leq U_p^+(X)=U_p^{{\boldsymbol{\cdot}}}(X)\leq U_p^{{\boldsymbol{\cdot}}}(B).\end{align*} $$

A symmetric argument shows that $\langle y\rangle ^Y\leq U_p^+(B)$ , where $y\in U_p^{{\boldsymbol {\cdot }}}$ and $Y=\langle y,b\rangle $ . This is clearly enough to prove the statement.

Proof. Suppose I is nonabelian. Then there exist elements $a,b\in I$ such that $S=\{[a,b]_+,a\ast b\}\neq \{0\}$ . Let $c\in S\setminus \{0\}$ . The minimality of I in B implies that $I=\langle c\rangle ^B$ , so there is a finite subset E of B such that $c\in E$ and $\langle a,b\rangle \leq H=\langle c\rangle ^{K}$ , where $K=\langle E\rangle $ . Since $H\leq K\cap I$ , we have that H is K-hypoabelian and hence there is an ideal L of K which is strictly contained in H and such that $H/L$ is abelian. Thus, $c\in S\subseteq L$ and consequently $H=\langle c\rangle ^K\leq L<H$ , a contradiction. Therefore, I is abelian.

Now, let p be any prime. Clearly, $I[p]=\{x\in I\,:\, x^p=0\}$ is an ideal of B, so either $I[p]=I$ or $I[p]=\{0\}$ . Similarly, $I^{p,+}=I^{p,{\boldsymbol {\cdot }}}$ is an ideal of B. This implies that either $(I,+)$ is an elementary abelian q-group for some prime q, or it is torsion-free and divisible, which means it is a restricted direct product of copies of the additive group of rational numbers.

Proof. Since local solubility is preserved with respect to forming quotients, we only need to prove the statement for a minimal ideal I of B. Now, let F be any finitely generated subbrace of B. Then F is soluble, so $F\cap I$ is F-soluble and hence F-hypoabelian. An application of Theorem 4.8 completes the proof.

Proof. By Theorem 4.8, I is abelian. Suppose by contradiction that $(I,+)$ is torsion-free, and let $0\neq x\in I$ . Put $y=x^2$ . The minimality of I shows that $I=\langle y\rangle ^B$ . Thus, there is a finite subset E of B such that $y\in E$ and $x\in \langle y\rangle ^K$ , where $K=\langle E\rangle $ . By assumption, $K\cap I$ is finitely generated and abelian, so $(K\cap I,+)$ is finitely generated, and hence, it is a direct product of finitely many infinite cyclic groups. Put $X=\langle x\rangle ^K$ ; in particular, $X\leq K\cap I$ . Now, $y=x^2\in X^{2,{\boldsymbol {\cdot }}}=X^{2,+}\trianglelefteq K$ , so

$$ \begin{align*}x\in \langle y\rangle^K\leq X^{2,{\boldsymbol{\cdot}}}\trianglelefteq K\end{align*} $$

and hence $X=X^{2,{\boldsymbol {\cdot }}}$ , a contradiction. Thus, $(I,+)$ is an elementary abelian p-group for some prime p and we are done.

Proof. Since the property of being locally polycyclic is preserved with respect to forming quotients, we only need to prove the statement for a minimal ideal I of B. Now, let F be any finitely generated subbrace of B. Then F is polycyclic, so $F\cap I$ is F-hypoabelian. Moreover, since every subbraces of a polycyclic brace is finitely generated, we have that $F\cap I$ is finitely generated. An application of Theorem 4.10 completes the proof.

Proof. It is clearly enough to prove the statement for a minimal ideal I of B. Let F be a nonzero, finitely generated subbrace of B, and let J be any nonzero ideal of F. Since F satisfies the maximal condition on subbraces, there is an ideal L of F that is maximal with respect to the property of being strictly contained in J. Thus, $J/L$ is a chief factor of F and is consequently abelian. The arbitrariness of J in F shows that $F\cap I$ is F-hypoabelian. Since $F\cap I$ is finitely generated, it follows from Theorem 4.10 that I is abelian and that $(I,+)$ is an elementary abelian p-group for some prime p. Suppose the rank of I is at least $n+1$ , and let $x_1,\ldots ,x_{n+1}$ be additively independent elements of I. There are elements $b_1,\ldots ,b_\ell $ of B such that if $c\in \langle x_1,\ldots ,x_{n+1}\rangle =V$ , then $\langle c\rangle ^U\geq V$ , where $U=\langle x_1,\ldots ,x_{n+1},b_1,\ldots ,b_\ell \rangle $ . Let M be an ideal of U which is maximal with respect to the property of not containing $x_1$ . Then $V^UM/M$ is a nonzero chief factor of U whose additive group is an elementary abelian p-group of rank at least $n+1$ . This contradiction completes the proof.

Proof. Let X be any subbrace of B such that $|(B,+):(X,+)|$ is finite and is equal to n. Let $b_1,\ldots ,b_n$ be a transversal to $(X,+)$ in $(B,+)$ , and put $C=\langle b_1,\ldots ,b_n\rangle $ . Then $|(C,+):(C\cap X,+)|=n$ . On the other hand, since C is finitely generated, we have that

$$ \begin{align*}|(C,+):(C\cap X,+)|=|(C,{\boldsymbol{\cdot}}):(C\cap X,{\boldsymbol{\cdot}})|,\end{align*} $$

so $|(B,{\boldsymbol {\cdot }}):(X,{\boldsymbol {\cdot }})|\geq n$ . If $|(B,{\boldsymbol {\cdot }}):(X,{\boldsymbol {\cdot }})|>n$ , then we can find $c_1,\ldots ,c_{n+1}\in B$ such that $c_iX\neq c_jX$ for all $1\leq i,j\leq n+1$ with $i\neq j$ . However, this means that

$$ \begin{align*}|(D,+):(D\cap X,+)|=n<|(D,{\boldsymbol{\cdot}}):(D\cap X,{\boldsymbol{\cdot}})|,\end{align*} $$

where $D=\langle c_1,\ldots ,c_{n+1}\rangle $ , a contradiction. Thus, $|(B,+):(X,+)|=|(B,{\boldsymbol {\cdot }}):(X,{\boldsymbol {\cdot }})|$ and B has the index-property. We argue in a similar way in case $|(B,{\boldsymbol {\cdot }}):(X,{\boldsymbol {\cdot }})|$ is finite.

Proof. Suppose C is not an ideal of B, and choose $b\in B$ and $c\in C$ such that one of the elements $\lambda _b(c)$ , $c^{b,{\boldsymbol {\cdot }}}$ , $c^{b,+}$ is not contained in C. Now, the subbrace $D=\langle b,c\rangle $ is supersoluble and $C\cap D$ has index $2$ in D, so $C\cap D\trianglelefteq D$ by Corollary 3.23. However, $c\in C\cap D$ and $b\in D$ , although one of the elements $\lambda _b(c)$ , $c^{b,{\boldsymbol {\cdot }}}$ , $c^{b,+}$ does not belong to $C\cap D$ . This contradiction completes the proof.

Proof. Let F be any finitely generated subbrace of B; in particular, F is supersoluble and all the subbraces of F are finitely generated. Being a finitely generated subbrace of the locally centrally-nilpotent ideal I, we have that $I\cap F$ is centrally nilpotent. Then $I\cap F$ is an F-centrally nilpotent ideal of F by Theorem 3.36. The arbitrariness of F proves that I is locally B-nilpotent.

Proof. Suppose $I\cap M$ is not an ideal of B; in particular, $I\not \leq M$ . Since M is maximal in B, we have that $B=IM$ , so there are elements $x\in I$ and $a\in I\cap M$ such that $\mathcal {A}:=\big \{a^{x,{\boldsymbol {\cdot }}},a^{x,+},\lambda _x(a)\big \}\not \subseteq M$ . Let $b\in \mathcal {A}\setminus M$ , and let F be a finite subset of M such that $x\in \langle b,F\rangle $ (this subset exists because M is maximal). Put $C=\langle b,a,F\rangle $ , and let D be a subbrace of C that is maximal with respect to containing $\langle a,F\rangle $ but not b; in particular, D is a maximal subbrace of C. Since C satisfies the maximal condition on subbraces, we have that $I\cap C$ is a C-centrally nilpotent ideal of C. Moreover, $a\in (I\cap C)\cap D$ and $x\in C$ but $b\not \in (I\cap C)\cap D$ , so without any loss of generality we may assume that I is B-centrally nilpotent.

Now, let i be the largest nonnegative integer such that $\zeta _i(I)_B\leq I\cap M$ . Then $Z:=\zeta _{i+1}(I)_B\not \leq I\cap M$ , so $B=ZM$ . Since $I\cap M\trianglelefteq M$ and

$$ \begin{align*}[z,I\cap M]_{{\boldsymbol{\cdot}}}\cup [z,I\cap M]_+\cup\lambda_z(I\cap M)\subseteq I\cap M\end{align*} $$

for all $z\in Z$ , we have that $I\cap M\trianglelefteq B$ , thus completing the proof of the first part of the statement.

Finally, suppose that $\partial _B(I)\not \leq I\cap M$ . Since $I\cap M\trianglelefteq B$ , we have $\partial (I)\not \leq I\cap M$ . Let $a\in \partial (I)\setminus (I\cap M)$ ; in particular, a does not belong to M. Then there are finitely many elements $x_1,\ldots ,x_n$ of I such that $a\in \partial \big (\langle x_1,\ldots ,x_n\rangle \big )$ . For each $i\in \{1,\ldots ,n\}$ , let $F_i$ be a finite subset of M such that $x_i\in \langle a,F_i\rangle $ . Put $C=\langle a,F_1,\ldots ,F_n\rangle $ . Since $x_i\in C$ for all $1\leq i\leq n$ , we also have that $a\in \partial (I\cap C)$ . Since C satisfies the maximal condition on subbraces, we have that $I\cap C$ is a C-centrally nilpotent ideal of C. Let D be a subbrace of C that is maximal with respect to containing $F_1,\ldots ,F_n$ but not a. It follows that D is actually a maximal subbraces of C and that $\partial (I\cap C)\not \leq (I\cap C)\cap D$ . Thus, without loss of generality we may assume I is B-centrally nilpotent. Now, $B = \partial _B(I)M$ , and hence