Proof For all $a, b, c \in B$

$$ \begin{align*} \lambda_{c}(a \circ b) = \lambda_{c}(a + \lambda_{a}(b)) = \lambda_{c}(a) + \lambda_{c} \lambda_{a}(b), \end{align*} $$

and

$$ \begin{align*} \lambda_{c}(a) \circ \lambda_{c}(b) = \lambda_{c}(a) + \lambda_{\lambda_{c}(a)}\lambda_{c}(b). \end{align*} $$

Thus, the two expressions coincide precisely when for all $a, c \in B,$ one has

$$ \begin{align*} \lambda_{\lambda_{c}(a)} = \lambda_{c} \lambda_{a} \lambda_{c}^{-1}, \end{align*} $$

which is, once one translates the notation, condition 4 of [Reference Caranti9, Theorem 3.1].

The fact that (1) implies (2) also follows from the fact that if $\lambda $ is the lambda function of the bi-skew brace $(B, +, \circ )$ , then the lambda function of $(B, \circ , +)$ is given by

$$ \begin{align*}c\in (B,+) \mapsto \lambda^{-1}_c= \lambda_{c^{-1}}\in \mathrm{Aut}(B,\circ)\end{align*} $$

(see [Reference Caranti9, Subsection 3.1]).

Proof The fact that $C$ is a sub-skew brace of $B^{\mathrm {opp}}$ is obvious because the (normal) subgroups of $(B,+)$ are also (normal) subgroups of $(B,+^{\mathrm {opp}})$ . Suppose $C$ is a (strong left) ideal of $B$ . We only need to prove that $C$ is $\lambda $ -invariant. Thus, let $a\in B$ and $b\in C$ . We have that

$$ \begin{align*}\lambda_a(b)^{\mathrm{opp}}=-a+^{\mathrm{opp}}a\circ b=a\circ b-a=a-a+a\circ b-a=a+\lambda_a(b)-a\end{align*} $$

belongs to $C$ because $C$ is a strong left ideal of $B$ .

Proof We only prove the case for a prime $p$ , the proof for the case of a set of primes $\pi $ being totally analogous (the only difference is the use of the theorems of Hall instead of those of Sylow). Since the sub-skew braces of $B$ and its opposite are the same by Lemma 6, it is enough to prove the case when $(B,\circ )/\mathrm {Ker}(\lambda )$ is a $p$ -group. Let $G=(B,+)\rtimes _\lambda (B,\circ )$ . Now, since $P$ is a $p$ -sub-skew brace of $B$ , $S=(P,+)\rtimes _\lambda (P,\circ )$ is a $p$ -subgroup of $G$ . By the hypotheses, $S$ can be extended to a $p$ -subgroup of the form $S_1=(P,+)\rtimes _\lambda (P_1,\circ )$ , where $(P_1,\circ )$ is a Sylow $p$ -subgroup of $(B,\circ )$ containing $(P,\circ )$ , and subsequently to a Sylow $p$ -subgroup of $G$ , which is necessarily of the form ${S_2=(P_2,+)\rtimes _\lambda (P_1,\circ )}$ , where $(P_2,+)$ contains $(P,+)$ . Consequently, $\lambda _a(P_2)=P_2$ for every $a\in P_1$ . But $(B,\circ )=P_1\circ \mathrm {Ker}(\lambda )$ , and so $P_2$ is $\lambda $ -invariant. Therefore, $P_2$ is the required Sylow $p$ -sub-skew brace.

Proof (1) In this case, $\lambda $ is an antihomomorphism of $(B,+)$ (see [Reference Caranti9, Theorem 3.1(3)]), so that the kernel $K = \mathrm {Ker}(\lambda )$ is an ideal of $B$ . If $K = \{ 0 \}$ , then $B$ is an almost trivial skew brace (see [Reference Caranti9, Theorem 3.1(7)]), for which the results hold true trivially. Let $B$ be a minimal counterexample, so that $K \ne \{ 0 \}$ . By the minimality of $B,$ we have that $B/K$ is a $\pi $ -skew brace, and thus the results follow from Lemma 7.

(2) We use induction on the right nilpotency class $n$ of $B$ . Clearly, if $n\leq 1$ , then the statement is trivial. Suppose $n>1$ , and let $I$ be an ideal of $B$ such that $I\ast B=\{0\}$ and $B/I$ is right nilpotent of right nilpotency class $n-1$ . By induction, $B/I$ has a Hall $\pi $ -sub-skew brace, so, without loss of generality, we may assume that $|B/I|$ is a $\pi $ -number. Since $I\subseteq \mathrm {Ker}(\lambda )$ , we may apply Lemma 7 to conclude the proof.

(3) This follows from [Reference Stefanello and Trappeniers15, Theorem 3.13] and (2).

Proof We only need to prove that $J$ is multiplicatively normal. Let $B$ be a minimal counterexample to the statement. Since $B$ is left nilpotent, $I+(J\ast J)$ is a proper ideal of $B$ , so $J\ast J$ is an ideal of $I+(J\ast J)$ , and hence also an ideal of $B$ . By minimality of $B$ , $J\ast J=\{0\}$ . Now, since $B$ is left nilpotent, if $B=B^1\supseteq B^2\supseteq \cdots \supseteq B^m=\{0\}$ denotes the left series of $B$ , then the subgroups $I_r=I\cap B^r$ give a series of $(I,+)$ . Moreover, for every $j\in J$ and every $x\in I_r$ , one has $j\ast x\in I\cap B^{r+1}=I_{r+1}$ , because $I$ is an ideal and $J\ast B^r\subseteq B^{r+1}$ . Thus, $\lambda _j$ acts trivially on each factor $I_r/I_{r+1}$ ; in this sense, $(J,\circ )$ stabilizes this series of $(I,+)$ .

Consider now the image of $(J,\circ )$ in $\mathrm {Aut}(I,+)$ . It is contained in the group of automorphisms stabilizing the above series, and this group has order divisible only by primes dividing $|I|$ . On the other hand, this image is a quotient of $(J,\circ )$ , hence its order divides $|J|$ . Since $|I|$ and $|J|$ are coprime, the image is trivial. Therefore, $\lambda _j(i)=i$ for all $j\in J$ and $i\in I$ , and so $J\ast I=\{0\}$ . Thus, $J\ast B=\{0\}$ . Hence, $J\leq \mathrm {Ker}(\lambda )$ and hence $[J,I]_\circ =\{0\}$ , which means that $J$ is multiplicatively normal, a contradiction.

Proof Let $B$ be a minimal counterexample to the statement, and let $I$ be a minimal ideal of $B$ contained in $B\ast B$ . By minimality of $B$ , we have that $|B/I|$ is a $\pi $ -number. Moreover, since $B$ is soluble, $I$ is an abelian $q$ -group for some prime $q\not \in \pi $ . Let $P$ be any Hall $\pi $ -subgroup of $(B,\circ )$ , and extend it to a Hall $\pi $ -subgroup $P_1\rtimes _\lambda P$ of $G=(B,+)\rtimes _\lambda (B,\circ )$ . By hypothesis, $B$ is left nilpotent. If $B=B^1\supseteq B^2\supseteq \cdots \supseteq B^m=\{0\}$ is the left series of $B$ , then the subgroups $I_r=I\cap B^r$ form a finite series of $(I,+)$ . Moreover, for every $p\in P$ and every $x\in I_r$ , one has $p\ast x\in I\cap B^{r+1}=I_{r+1}$ , because $I$ is an ideal and $P\ast B^r\subseteq B^{r+1}$ . Hence, $\lambda _p$ acts trivially on each factor $I_r/I_{r+1}$ . Thus, the image of $P$ in $\mathrm {Aut}(I,+)$ is contained in the group of automorphisms acting trivially on this series. Since $(I,+)$ is an abelian $q$ -group, this latter group is a $q$ -group, and in particular, it is nilpotent. On the other hand, the image of $P$ is a quotient of the $\pi $ -group $P$ . Since $q\notin \pi $ , this image is trivial. Therefore, $P\ast I=\{0\}$ .

Let $a\in I$ , $b\in P_1$ , and $c\in P$ . Then $b+a-b=a_1$ for some $a_1\in I$ . Since $I\leq B\ast B$ , $P\ast I=\{0\}$ , so conjugating the previous equality by $c$ in $G$ yields $\lambda _c(b)+a-\lambda _c(b)=a_1$ , so $-b+\lambda _c(b)=[-b,c]_\lambda $ additively centralizes $a$ . The arbitrariness of $a$ , $b,$ and $c$ shows that $P\ast P_1$ additively centralizes $I$ .

It follows that $(B\ast B,+)$ is the direct sum of $(I,+)$ and $(P\ast P_1,+)$ . By Lemma 10, $P\ast P_1$ is an ideal of $B\ast B$ , and so also an ideal of $B$ . Thus, the minimality of $B$ yields that $P\ast P_1=\{0\}$ . But then $P\ast B=P\ast (P_1+I)=\{0\}$ , so $P\ast P=\{0\}$ , and hence $P$ is a Hall $\pi $ -sub-skew brace of $B$ .

Proof Suppose the statement is false, and let $B$ be a minimal counterexample to the statement; in particular, $B\ast B\neq \{0\}$ otherwise $B$ would be a group and the statement would trivially holds. Let $I$ be a minimal ideal of $B$ contained in $B\ast B$ . Since $B$ is soluble, $I$ is a trivial skew brace of abelian type, and so its characteristic subgroups are ideals of $B$ . This implies that $(I,+)$ is an elementary abelian $q$ -group for some prime $q$ . By minimality of $B$ , we have that $|B/I|$ is a $\pi $ -number and (obviously) that $q\not \in \pi $ . By hypothesis, $(B\ast B,+)$ contains a characteristic Hall $\pi $ -subgroup $P_1$ . But then $P_1$ is a sub-skew brace (and in particular a subgroup of $(B,\circ )$ ), which, by order considerations, coincides with the only Hall $\pi $ -subgroup of $(B\ast B,\circ )$ . Thus, $P_1$ is an ideal of $B$ . Now, $B\ast B$ is the direct product (as a skew brace) of $I$ and $P_1$ , so if we factor $P_1$ out, then we may assume that $I=B\ast B$ .

Let $P$ be the unique Hall $\pi $ -subgroup of $(B,\circ )$ . In the natural semi direct product $G=(B,+)\rtimes _\lambda (B,\circ )$ , extend $P$ to a Hall $\pi $ -subgroup $R$ of $G$ . Since $(B,+)$ is normal in $G$ , there is a Hall $\pi $ -subgroup $P_2$ of $(B,+)$ such that $R=(P_2,+)\rtimes _\lambda (P,\circ )$ (use Schur Zassenhaus). Thus, if $a\in P_2$ and $x\in P$ , then $x\ast a\in P_2$ ; but $B/I$ is a trivial skew brace, so $x\ast a\in I$ , and hence $x\ast a\in P_2\cap I=\{0\}$ . Therefore, $P\ast P_2=\{0\}$ .

If $b\in (B,\circ )$ , then the action by conjugation in $G$ on $R$ gives

$$ \begin{align*}^bR=\lambda_b(P_2)\rtimes_\lambda P,\end{align*} $$

which shows that $P\ast \lambda _b(P_2)=\{0\}$ . Let $U$ be the subgroup of $(B,+)$ generated by the $\lambda _b(P_2)$ ’s with $b\in B$ . Then $P\ast U=\{0\}$ . By its very definition, $U$ (and consequently $U\cap I$ ) is $\lambda $ -invariant, so $U$ is a sub-skew brace. Since $P_2$ is an additive subgroup of $U$ , the order of $U$ is divisible by $|P_2|=|P|$ , and this means that the multiplicative Hall $\pi $ -subgroup of $U$ must coincide with $P$ (the latter being the unique Hall $\pi $ -subgroup of $(B,\circ )$ ). Consequently, $P\ast P=\{0\}$ and so $P$ is a Hall $\pi $ -sub-skew brace of $B$ .

Proof Suppose the statement is false, and let $B$ be a minimal counterexample to the statement. Let $I$ be a minimal ideal of $B$ . Since $B$ is soluble, $I$ is a trivial skew brace of abelian type. Hence, every characteristic subgroup of $(I,+)$ is an ideal of $B$ . By the minimality of $I$ , the group $(I,+)$ has no proper nonzero characteristic subgroups, and therefore, it is an elementary abelian $q$ -group for some prime $q$ . Moreover, $q\neq p$ , because otherwise, since $I$ a trivial elementary abelian $p$ -brace, any subgroup of $(I,+)$ of order $p$ would be a sub-skew brace of $B$ , contrary to our assumption.

We claim that $|B/I|=p$ . Since $q\neq p$ and $p$ divides $|B|$ , the prime $p$ divides $|B/I|$ . The image of $P$ in $(B/I,\circ )$ is therefore a normal Sylow $p$ -subgroup. Hence, by the minimality of $B$ , the quotient $B/I$ has a sub-skew brace $\overline K$ of order $p$ . Let $K$ be the inverse image of $\overline K$ in $B$ . If $|B/I|>p$ , then $K$ is a proper sub-skew brace of $B$ . Since $|K/I|=p$ and $|I|$ is a $q$ -number with $q\neq p$ , the subgroup $K\cap P$ is a normal Sylow $p$ -subgroup of $(K,\circ )$ . Thus, $K$ satisfies the hypotheses of the corollary and, by the minimality of $B$ , it contains a sub-skew brace of order $p$ , again a contradiction. Therefore, $|B/I|=p$ .

Since every skew brace of prime order is trivial, we get $B\ast B\leq I$ . In particular, $(B\ast B,+)$ is a $q$ -group, and so its Hall $p$ -subgroup is trivial and normal. Theorem 12, applied with $\pi =\{p\}$ , implies that $P$ is a Hall $p$ -sub-skew brace of $B$ . But $|B/I|=p$ and $q\neq p$ , so $|P|=p$ . Hence, $P$ itself is a sub-skew brace of order $p$ , the final contradiction.

Proof We prove the result for a set $\pi $ of primes, the case of a single prime $p$ being easier. Since the multiplicative inner automorphisms of $B$ are skew brace automorphisms (see [Reference Trappeniers17, Proposition 2.3]) and $(B,+)$ is soluble, we only need to prove the existence of the Hall $\pi $ -sub-skew braces (because of the well-known theorems of Philip Hall). Thus, let $B$ be a minimal counterexample to the existence, and set $H=(B\ast B)\cap (B\ast ^{\mathrm {opp}} B)$ . Then $H$ is an ideal of $B$ and $(H,+)$ is abelian by Theorem 4.3 of [Reference Trappeniers17].

Suppose first that $H=\{0\}$ . Clearly, neither $B\ast B$ nor $B\ast ^{\mathrm {opp}}B$ can be $\{0\}$ (because otherwise $B$ has Hall $\pi $ -sub-skew braces). By minimality of $B$ , both the orders of $B/B\ast B$ and $B/B\ast ^{\mathrm {opp}}B$ are $\pi $ -numbers. Thus, $|B|$ is a $\pi $ -number itself, and we have a contradiction.

Thus, $H\neq \{0\}$ . Let $I$ be a minimal ideal of $B$ contained in $H$ . Then $(I,+)$ is an elementary abelian $q$ -group for some prime $q$ . By minimality, the statement holds for $B/I$ , so we have that the order of $B/I$ is a $\pi $ -number and $q\not \in \pi $ . Also $(H,+)$ is abelian, so it contains a unique Hall $\pi $ -subgroup $R$ , which is then an ideal of $H$ (see [Reference Trappeniers17, Corollary 2.4]). But this means that $H=I\times R$ , so $R$ is actually both additively and multiplicatively characteristic, implying that $R$ is an ideal of $B$ . By minimality of $B$ , $R=\{0\}$ . Thus, $H=I$ is contained in $Z(B\ast B,+)$ by [Reference Trappeniers17, Theorem 4.3]. Let $J/I$ be a minimal ideal of $B/I$ contained in $B\ast B/I$ , if any. Then $J$ is an ideal of $B$ of nilpotent type. Again, if $S$ is a Hall $\pi $ -subgroup of $(J,+)$ , then $S$ is an ideal of $B$ which must then be $\{0\}$ by minimality of $B$ . Therefore, $I=H=B\ast B$ .

Let $P$ be a Hall $\pi $ -subgroup of $(B,\circ )$ . Then there is a Hall $\pi $ -subgroup $P_1$ of $(B,+)$ such that $P_1\rtimes _\lambda P$ is a Hall $\pi $ -subgroup of $G=(B,+)\rtimes _\lambda (B,\circ )$ . Clearly, $P\ast P_1\subseteq I\cap P_1=\{0\}$ . Now, let $U$ be the largest subgroup of $(B,+)$ such that $P\ast U=\{0\}$ . Clearly, $P_1\leq U$ . We claim that $P_1<U$ . Suppose not, so $P_1=U$ . Now, since $B$ is two-sided, so $a\ast y=0$ implies $a\ast y^{-1}=0$ for every $a,y\in B$ . Thus, $P_1^{-1}\subseteq U=P_1$ . Similarly, if ${a\ast b=a\ast c=0}$ for some $a,b,c\in B$ , then $a\ast (b\circ c)=0$ . Consequently, $P_1\circ P_1\subseteq U=P_1$ . This implies that $P_1$ is a multiplicative subgroup and so a Hall $\pi $ -sub-skew brace of $B$ , a contradiction.

This last contradiction shows that $U>P_1$ . Then $V=U\cap I$ is not $\{0\}$ and is additively normal because $U$ contains an additive complement to $I$ . Now, $P\ast V=\{0\}$ , and so also $B\ast V=\{0\}$ because $(B,\circ )=I\circ P$ and $I\ast V=\{0\}$ .

Since the multiplicative inner automorphisms are skew brace automorphisms, we have that $B\ast \big (x\circ V\circ x^{-1}\big )=\{0\}$ for all $x\in (B,\circ )$ , so $x\circ V\circ x^{-1}\subseteq U$ and hence ${V=x\circ V\circ x^{-1}}$ . Therefore, $V$ is multiplicatively normal. Since $B\ast V=\{0\}$ , so $V$ is an ideal of $B$ , which must coincide with the minimal ideal $I$ . Thence, $B\ast I=\{0\}$ .

Finally, if $a,b$ are multiplicative $\pi $ -elements of $B$ , then $\lambda _a(b)=b+u$ for some ${u\in I}$ (because $I=B\ast B$ ). But $\lambda _a(u)=u$ and $u$ has additive order dividing $q$ , so $u=0$ (because $q\not \in \pi $ ), and consequently $a\ast b=0$ . This shows that every Hall $\pi $ -subgroup of $(B,\circ )$ is a (Sylow $\pi $ -)sub-skew brace of $B$ and gives the last contradiction.

Proof Let $B$ be a minimal counterexample to the statement, and let $I$ be a minimal ideal of $B$ . Then $(I,+)$ is an elementary abelian $q$ -group for some prime $q$ . By minimality of $B$ , we have that $B/I$ has order $p^n$ for some $p\neq q$ .

Let $P$ be a Sylow $p$ -subgroup of $(B,\circ )$ . Then there is a Sylow $p$ -subgroup $P_1$ of $(B,+)$ such that $P_1\rtimes _\lambda P$ is a Sylow $p$ -subgroup of $G=(B,+)\rtimes _\lambda (B,\circ )$ . Thus, $\lambda _x(P_1)=P_1$ for every $x\in P$ . Since $\lambda _x$ is a skew brace isomorphism for every $x\in P$ , we have that $\lambda _x(P_1)=P_1$ for every $x\in \lambda _b(P)$ with $b\in P$ .

Let $U$ be the normalizer of $(P_1,+)$ in $(B,\circ )$ . Clearly, $U$ contains every $\lambda _x(U)$ with $x\in P$ because $\lambda _x$ is a skew brace isomorphism, and we may write $U=P\circ (I\cap U)$ . Moreover, since $P$ is not a sub-skew brace, $I\cap U\neq \{0\}$ . If $c\in I\cap U$ , then $\lambda _c(P_1)=P_1$ . On the other hand, $(I,+)\rtimes _\lambda (I,\circ )$ is a normal Sylow $q$ -subgroup of $G$ , so $\lambda _c(u)=u$ for every $u\in P_1$ . Therefore, $c\ast P_1=\{0\}$ . But obviously, $c\ast I=\{0\}$ , and hence $c\in K=\mathrm {Ker}(\lambda )$ . It follows that $I\cap U\leq K$ , which means that $I\cap U=K\cap I$ because $K\leq U$ .

Now, if $a\in P$ , $b\in I\cap U\leq K,$ and $u\in U$ , then $\lambda _{a\circ b}(u)=\lambda _a(u)\in U$ . Thus, $\lambda _u(v)\in U$ for every $u,v\in U$ , and consequently, $u+v=u\circ \lambda _{u^{-1}}(v)\in U$ for every $u,v\in U$ . By choosing $w=\lambda _u(u^{-1})$ , we see that $u+w=0$ . Therefore, $U$ is a sub-skew brace of $B$ . Since $|P|$ divides $|U|$ , we have by minimality that $U=B$ , which implies that $I\subseteq K$ . Lemma 7 provides the final contradiction.

Proof Let $B$ be a counterexample of minimal order, and $M$ a minimal ideal of $B$ . By minimality, $|B/M|$ is a $\pi $ -number, and $M$ has prime order $q\not \in \pi $ . In particular, $(B,+) = (P,+)+(M,+)$ , where $P$ is a Hall $\pi $ -subgroup of $(B,+)$ . If $P$ is normal in $(B,+)$ , then $P$ is characteristic in $(B,+)$ , and so it is a Hall $\pi $ -sub-skew brace of $B$ , a contradiction. Thus, $P$ is not additively normalized by $M$ .

For $m\in M$ , write

$$ \begin{align*} \lambda_m = \begin{bmatrix} a_m & b_m\\ 0 & d_m \end{bmatrix}, \end{align*} $$

where $a_m, b_m, d_m$ are defined as in Theorem 17. Since $M$ is additively characteristic, it is also a subgroup of $(B,\circ )$ , so $\lambda (M)$ is a finite $q$ -group, and each $\lambda (m)$ is a $q$ -element, for $m \in M$ . In particular, $a_m$ is an automorphism of order a power of $q$ of the group $M$ of order $q$ , and thus $a_m = 1$ for all $m \in M$ . We also have $d_m = 1$ for $m \in M$ , because $M$ is an ideal.

In order to obtain the final contradiction, we prove that there exists a function $\tau $ on $M$ such that $b_m(x) = [m^{\tau },x]$ , for $x \in P$ . The existence of such a function obviously implies that $\lambda _m = \iota (m^{\tau })$ , for $m \in M$ , where $\iota $ denotes additive conjugation (on the left). Writing $\tau = -\sigma $ , Proposition 18 now applies to $G = (B,+)$ , so either $\sigma =0$ and $M\leq \mathrm {Ker}(\lambda )$ , or $\sigma =1$ and $M$ is easily seen to lie in $\mathrm {Ker}(\lambda ^{\mathrm {opp}})$ . In any case, we may apply Lemma 7 and obtain a contradiction.

It remains to prove the existence of such a function $\tau $ . We provide two proofs of this fact, one using cohomology results, and the other cohomology-free. As for the former, $b_m:P\rightarrow M$ is a derivation for the action of $P$ on $M$ by conjugation in $(B,+)$ . Since $P$ and $M$ have coprime orders, we have $H^{1}(P, M) = 0$ (see [Reference Brown6, III.10.2]), that is, all such derivations are inner, so $b_m(x)=[a,x]_+=a+x-a-x$ for some $a\in M$ and for all $x\in P$ . This obviously proves our claim.

The latter proof is as follows. Let $c_{1}, c_{2} \in C=C_{(P,+)}(M,+)$ and $m\in M$ . Then

$$ \begin{align*} b_m(c_{1} + c_{2}) = b_m(c_{1})+ {}^{+,c_1}b_m(c_{2}) = b_m(c_{1})+ b_m(c_{2}). \end{align*} $$

Therefore, ${b_m}_{\restriction C} : (C,+) \longrightarrow (M,+)$ is a group homomorphism, and since $\gcd (|P|, |M|) = 1$ , it is trivial. Let $t \in P$ be such that $(P / C,+) = \langle t +C\rangle _+$ . Then every element of $P$ is of the form $\ell t+c$ , for some nonnegative integer $\ell $ and some $c \in C$ , so that

$$ \begin{align*} b_m(\ell t+c) = b_m(\ell t) = {}^{1+t+\cdots +t^{\ell-1}}(b_m(t)). \end{align*} $$

It follows that $b_m$ is completely determined by its value $b_m(t)\in M$ , so there are at most $q$ possibilities for $b_m$ . But for $m \in M$ , the map $\beta _m: x\in P\mapsto [m,x]\in M$ satisfies

$$ \begin{align*} \beta_m(x+y) = [m,x +y]_+ = [m, x]_+ + {}^{x}[m, y]_+ = \beta_m(x)+{}^x\beta_m(y), \end{align*} $$

and they are $q$ distinct maps, as for $m, n \in M$ and $x \in P,$ we have

$$ \begin{align*} [m - n,x]_+ = [m, x]_+ - [n,x]_+, \end{align*} $$

as $M$ is abelian, and $[m-n,x]_+ = 0$ for all $x \in P$ if and only if $m-n=0$ , as $P$ acts nontrivially on $M$ . This again proves the existence of the map $\tau $ .