Proof. Only the second part of the statement is in doubt.

Clearly, $1\in \operatorname {Ker}\lambda$. Then, for every $a,b\in \operatorname {Ker} \lambda$, $a+b = a\lambda _{a^{-1}}(b) = ab \in \operatorname {Ker} \lambda$. Moreover, if $b \in \operatorname {Ker}\lambda$, then $-b = b(b^{-1} + b^{-1}) \in \operatorname {Ker} \lambda$.

Proof. It is clear that only the second statement is in doubt.

By proposition 2.2, we have that $IJ = I+J$. Therefore, $IJ = I+J$ is a normal subgroup of $(B,+)$ and a normal subgroup of $(B,\cdot )$.

Let $b \in B$, $x \in I$ and $y \in J$. Then, $\lambda _b(x+y) = \lambda _b(x) + \lambda _b(y) \in I+J$. Thus, $\lambda _b(I+J) \subseteq I+J$. Consequently, $IJ$ is an ideal of $B$.

Proof. By lemma 2.1, we have that $\operatorname {Ker}\lambda$ is a subgroup of $(B,+)$ and then $\operatorname {Soc}(B) = \operatorname {Ker}\lambda \cap \operatorname {Z}(B,+)$ is a normal subgroup of $(B,+)$.

Since $\operatorname {Z}(B,+)$ is a characteristic subgroup of $(B,+)$, it is clear that $\lambda _b(\operatorname {Soc}(B)) \subseteq \operatorname {Z}(B,+)$, for every $b\in B$. Therefore, for every $a \in \operatorname {Soc}(B)$ and $b\in B$

\[ \lambda_b(a) = b + \lambda_b(a) - b = ba - b = b(a+b^{{-}1}) = bab^{{-}1}. \]

Thus, $\lambda _b(a) \in \operatorname {Ker} \lambda$, as $\operatorname {Ker} \lambda$ is normal in $(B,\cdot )$.

Hence, $\operatorname {Soc}(B)$ is a left ideal of $B$ and, by definition, $\operatorname {Soc}(B) \ast B = 1 \subseteq \operatorname {Soc}(B)$. Therefore, $\operatorname {Soc}(B)$ is an ideal of $B$.

Proof. Let $x\in I$. Then,

\begin{align*} xJ \in \operatorname{Fix}(B/J) & \Leftrightarrow \lambda_b(x)J = xJ, \ \text{for every}\ b\in B \\ & \Leftrightarrow bJ \ast xJ = J, \ \text{for every}\ b\in B\\ & \Leftrightarrow b \ast x \in J, \ \text{for every}\ b\in B. \end{align*}

Then, $I/J$ is an $f$-factor of $B$ if, and only if, $b \ast x \in J$, for every $b\in B$ and every $x \in I$, and the first statement holds.

Let $x\in I$. Then, $xJ \in \operatorname {Soc}(B/J)$ implies that $xJ \ast bJ = J$, for every $b\in B$. Thus, if $I/J$ is an $s$-factor, it follows that $I \ast B \subseteq J$. Moreover, since $I/J \subseteq Z(B/J,+)$, we have that $[I,B]_+ \subseteq J$.

Conversely, suppose that $I \ast B \subseteq J$ and $[I,B]_+ \subseteq J$. Given $x \in I$ and $b\in B$, it follows that $x \ast b = \lambda _x(b) - b \in J$, i.e. $\lambda _x(b)J = bJ$. Moreover, $(x+b)J = (b+x)J$. Thus, $I/J \subseteq \operatorname {Soc}(B/J)$.

Proof. We only prove (3.5) as (3.5) is analogous. We can assume without loss of generality that $I \cap J = 1$.

Suppose that $I \subseteq \operatorname {Soc}(B)$. Let $zJ \in L/J$ and $bJ \in B/J$. Then, $zJ = xJ$ for some element $x \in I$. Since $\lambda _x(b)= b$ and $x + b = b+ x$, it follows that $\lambda _{zJ}(bJ) = \lambda _{xJ}(bJ) = \lambda _x(b)J= bJ$ and $zJ + bJ = (x+b)J = (b + x)J = bJ + zJ$. Consequently, $L/J \subseteq \operatorname {Soc}(B/J)$.

Conversely, suppose that $L/J \subseteq \operatorname {Soc}(B/J)$. We shall prove that $I \subseteq \operatorname {Soc}(B)$. Let $x \in I$ and $y \in B$. Since $L/J \subseteq \operatorname {Soc}(B/J)$, then $\lambda _{xJ}(yJ) = \lambda _x(y)J = yJ$. By proposition 2.2, $\lambda _x(y)J = \lambda _x(y) + J = y + J = yJ$. Thus, $\lambda _x(y) - y = x\ast y \in J \cap I = 1$. Therefore, $\lambda _x(y) = y$.

Furthermore, $(x+y) + J = (y+x) + J$. Note that $I$ is a normal subgroup of $(B, +)$. Hence, $y -x +y \in I$ so that $x+y -x -y \in I \cap J = 1$ and $x+y = y+x$. Therefore, $I \subseteq \operatorname {Soc}(B)$.

Suppose that $I\subseteq \operatorname {Fix}(B)$. Let $aJ \in B/J$ and $bJ\in B/J$. Then, $\lambda _{bJ}(aJ) = \lambda _b(a)J = aJ$. This yields $L/J \subseteq \operatorname {Fix}(B/J)$. Conversely, if $L/J \in \operatorname {Fix}(B/J)$, then for every $a \in I$ and $b \in B$, $aJ = \lambda _{bJ}(aJ) = \lambda _b(a)J$. By proposition 2.2, $a +J = \lambda _b(a) + J$. Thus, $\lambda _b(a) - a = b\ast a \in J \cap I = 1$. Therefore, $\lambda _b(a) = a$ and so $I\subseteq \operatorname {Fix}(B)$.

Proof. Note that $UI$ and $VI$ are ideals of $B$ by corollary 2.3. The isomorphism theorem implies that

\[ UI/VI=U(VI)/(VI)\cong U/(U\cap VI). \]

Since $U\cap VI$ is an ideal of $B$ by corollary 2.3, $V\subseteq U\cap VI\subseteq U$ and $U/V$ is simple, it follows that either $V=U\cap VI$ or $U=U\cap VI$. If $V=U\cap VI$, then $U\ne U\cap VI$. Hence, $UI\ne VI$ and $UI/VI$ is a chief factor of $B$ isomorphic to $U/V$. If $U=U\cap VI$, then $U=V(U\cap I)$ and, by the isomorphism theorem, we have

\[ U/V = (U\cap I)V/V\cong (U\cap I)/(V\cap I). \]

Consequently, $U\cap I\ne V\cap I$ and $(U\cap I)/(V\cap I)$ is a chief factor of $B$ isomorphic to $U/V$.

Proof. Let $1=J_0\subseteq J_1\subseteq \dots \subseteq J_n=B$ be the given chief series of $B$ and write $U_i=J_i\cap I$, so that $1=U_0\subseteq U_1\subseteq \dots \subseteq U_n=I$ is a series of ideals of $B$. By lemma 3.6, each of the factors $U_i/U_{i-1}$ is either trivial or a chief factor of $B$. It follows that if we delete $U_i$ from the above series whenever $U_i=U_{i-1}$, what remains is part of a chief series of $B$. In particular, it is a chief series of $I$.

For the second statement, we denote $V_i=J_iI$ for $0\le i\le n$. Then, $I=V_0\le V_1\subseteq \dots \subseteq V_n=B$ is a series of ideals of $B$. By lemma 3.6, either $V_i=V_{i-1}$ or $V_i/V_{i-1}$ is a chief factor of $B$. It follows that by deleting $V_i$ if $V_i=V_{i-1}$ we obtain part of a chief series of $B$.

It is clear that $1=U_0\subseteq \dots \subseteq U_n=I=V_0\subseteq \dots \subseteq V_n=B$ is a chief series of $B$ passing through $I$. Furthermore, $I = V_0/I \subseteq \dots \subseteq V_n/I=B/I$ is a chief series of $B/I$ because, by the isomorphism theorem,

\[ (V_i/I)/(V_{i-1}/I) = V_i/V_{i-1},\quad i \in \{1, \dots ,n\}.\]

□

Proof. We use induction on the order of $B$. If $J = 1$, the result is obviously true. Hence, we suppose $J\neq 1$ and let $L$ be a minimal ideal of $B$ contained in $J$. Since the hypothesis carry over to $B/L$, we conclude by induction that $I/L \subseteq \operatorname {Rad}(B/L)J/L$.

If $L \subseteq \operatorname {Rad}(B)$ then $\operatorname {Rad}(B/L) = \operatorname {Rad}(B)/L$ and thus, $I \subseteq \operatorname {Rad}(B)J$, as desired.

Therefore, suppose that $L$ is not contained in $\operatorname {Rad}(B)$. Let $M$ be a maximal ideal of $B$ such that $L \nsubseteq M$. Then, $B = L+M = LM$ and $L\cap M = 1$. Hence, $I = L(I\cap M)$.

The isomorphism $x \mapsto xL$ from $M$ onto $B/L$ yields $\operatorname {Rad}(M)L/L = \operatorname {Rad}(B/L)$ and therefore, $I \subseteq \operatorname {Rad}(M)LJ = \operatorname {Rad}(M)J$. By [Reference Jespers, Kubat, Van Antwerpen and Vendramin11, proposition 3.9], $\operatorname {Rad}(M) \subseteq \operatorname {Rad}(B)$. Consequently, we obtain $I \subseteq \operatorname {Rad}(B)J$.

Proof. Put $L = IJ$, and assume that $I \subseteq \operatorname {Rad}(B)$. Then, $L/J \subseteq \operatorname {Rad}(B)J/J \subseteq \operatorname {Rad}(B/J)$.

If $L/I\subseteq \operatorname {Rad}(B/I)$ then $L \subseteq \operatorname {Rad}(B)$, because $\operatorname {Rad}(B/I) = \operatorname {Rad}(B)/I$. In this case, the map $f(I) = L/J$ and $f(L/I) = J$ satisfies the requirements. If $L/I\nsubseteq \operatorname {Rad}(B/I) = \operatorname {Rad}(B)/I$, then $J$ is not an $r$-chief factor of $B$ and the same choice of $f$ will suffice. It only remains to consider the case where $I\cap \operatorname {Rad}(B) = J \cap \operatorname {Rad}(B) = 1$ and $L/J \subseteq \operatorname {Rad}(B/J)$.

Let $M$ be a maximal ideal of $B$ such that $B = I+M = IM$ and $I \cap M = 1$. Write $K = L \cap M$. Then, $K$ is an ideal of $B$ and $KI = L$. Hence, $K \cong L/I\cong J$.

If $K = J$, then $L \subseteq M$, which is a contradiction to the fact $I \cap M = 1$. Therefore, $K \neq J$, $L = KJ$ and $I \cong K \cong J$. By lemma 3.8, $L \subseteq \operatorname {Rad}(B)J$. Hence, $L = J(L \cap \operatorname {Rad}(B))$.

Note that $J \cap (L\cap \operatorname {Rad}(B)) = 1$ and $L \cap \operatorname {Rad}(B)\subseteq K$. Thus, $K = L \cap \operatorname {Rad}(B)$ and $K$ is an $r$-chief factor of $B$ which is isomorphic to $L/J$. Since $L/I = KI/I \subseteq \operatorname {Rad}(B)I/I = \operatorname {Rad}(B)/I$, it follows that $f(I) = J$ and $f(L/I) = L/J$ satisfies the requirements of the lemma.

Proof. We call two chief series $\mathcal {X}$ and $\mathcal {Y}$ of $B$ equivalent if $\mathcal {X}$ and $\mathcal {Y}$ have the same length and isomorphic factors (up to rearrangement) such that the $\tau$-factors of $\mathcal {I}$ correspond to the $\tau$-factors of $\mathcal {J}$. We write in this case $\mathcal {X}\sim \mathcal {Y}$. It is clear that $\sim$ is an equivalence relation on the set of all chief series of $B$.

We use induction on the length $n$ of $\mathcal {I}$ to show that $\mathcal {I} \sim \mathcal {J}$.

Assume that $n=1$. Then, $B$ is simple, and in that case $\mathcal {I} =\mathcal {J}$. Let $n > 1$ and suppose that the theorem is true for all skew left braces with chief series of length $\leq n-1$. Write $I=I_{1}$ and $J=J_{1}$. If $I = J$, we can form the following chief series of $B/I$

\begin{align*} \mathcal{I}/I\colon\quad 1& =I_1/I\subseteq I_2/I\subseteq \dots\subseteq I_n/I = B/ I,\\ \mathcal{J}/I\colon\quad 1& =J_1/I\subseteq J_2/I\subseteq \dots\subseteq J_m/I = B/I. \end{align*}

Since the length of $\mathcal {I}/I$ is $n-1$, the induction hypothesis applied to $B/I$ yields $\mathcal {I}/I\sim \mathcal {J}/I$. Hence, $\mathcal {I} \sim \mathcal {J}$.

Now assume that $I\ne J$ and set $U=IJ$. By corollary 2.3, $U$ is an ideal of $B$. In this case $U/I$ and $U/J$ are chief factors of $B$ and, by proposition 3.7, there exist chief series $\mathcal {V}_1$ and $\mathcal {V}_2$ of $B$ of the following form:

\begin{align*} \mathcal{V}_1\colon\quad 1& = V_0\subseteq V_1 = I \subseteq V_2 = U \subseteq V_3 \dots \subseteq V_r = B,\\ \mathcal{V}_2\colon\quad 1& = W_0\subseteq W_1 = J \subseteq W_2 = U \subseteq V_3 \dots \subseteq V_r = B. \end{align*}

Since $\mathcal {I}$ and $\mathcal {V}_1$ have the minimal ideal $I$ in common, the induction yields $\mathcal {I} \sim \mathcal {V}_1$. Similarly, $\mathcal {J} \sim \mathcal {V}_2$. Furthermore, as the chief series $\mathcal {V}_1$ and $\mathcal {V}_2$ coincide above $U$ and $I \cap J = 1$, we can apply lemma 3.5 to conclude that $\mathcal {V}_1 \sim \mathcal {V}_2$. Consequently, $\mathcal {I} \sim \mathcal {J}$.

Assume now that $B$ is finite. Then, arguing as above, it follows that the bijection stated in the theorem also respects $r$-chief factors by lemma 3.9.

Proof. Assume that $B \neq 1$ has finite chief length. Let $I_1\subseteq I_2 \dots$ be any ascending chain of ideals of $B$. We may assume that $I = I_1 \neq 1$. Then, $I_1/I\subseteq I_2/I \dots$ is an ascending chain of ideals of $B/I$. By proposition 3.11, $\operatorname {fcl} (B/I) < \operatorname {fcl} (B)$. If we argue by induction on the chief length of $G$, we have that $I_n = I_{n +1} = \dots$, and hence $B$ is noetherian. Assume that $I_1\supseteq I_2 \dots$ is any descending chain of ideals of $B$. Let $L$ be a minimal ideal of $B$. If $L$ is contained in $I_k$ for all $k \geq 1$, then we can argue as above. Otherwise, there exists a $t \geq 1$ such that $I_t \cap L = 1$. Since $(I_1 +L)/L \supseteq (I_2 +L)/L \dots$ is a descending chain of ideals of $B/L$, we can apply induction to conclude that $I_m + L = I_{m +e} + L$ for some $m \geq t$ and all $e \geq 0$. Since $I_m \cap I = I_{m +e} \cap I = 0$, we conclude that $I_m = I_{m +e}$ for all $e \geq 0$, and $B$ is artinian.

Now assume that $B$ is artinian and noetherian but $B$ has no finite length. Apply the noetherian condition to the set $\mathcal {T}$ of proper ideals $I$ of $B$ that have a chain of ideals of $B$: $1 = I_0\subseteq I_1 \cdots \subseteq I_n = I$ such that $I_t/I_{t-1}$ is a chief factor of $B$ for all $1 \leq t \leq n$ (note that $B \neq 1$), and select a maximal member $A$. Since $B$ has no finite length, $A$ is proper in $B$. Apply now the artinian condition to the set of all ideals of $B$ that properly contain $A$ and select a minimal element $C \leq B$. It is clear that $C/A$ is a chief factor of $B$, and so we can append $C$ to the end of part of a chief series of $B$: $1 = V_0\subseteq V_1 \cdots \subseteq V_n =A$ to conclude that $C \in \mathcal {T}$. This contradicts the maximality of $A$ in $\mathcal {T}$.

Proof. It is straightforward from the above definition.

Proof. If $B$ is left nilpotent, then left series of $B$ clearly is an $f$-series of left ideals.

Conversely, suppose that $B$ admits an $f$-series of left ideals

\[ 1 = L_0 \subseteq L_1 \subseteq \ldots \subseteq L_n = B. \]

We shall prove by induction that $B^{i} \subseteq L_{n-i+1}$, for every $1 \leq i \leq n+1$. It is clear that $B^1 = B = L_n$. Suppose that $B^{i}\subseteq L_{n-i+1}$, for some $1\leq i < n+1$. Then, by definition of $f$-series of left ideals

\[ B^{i+1} = B \ast B^{i} \subseteq B \ast L_{n-i+1} \subseteq L_{n-i+1}=L_{n-i}. \]

Hence, $B^{n+1} = 1$ and then, $B$ is left nilpotent.

Proof. Suppose that every chief factor of $B$ is an $s$-factor. Let

\[ 1 = I_0 \subseteq I_1 \subseteq \ldots \subseteq I_n = B. \]

be a chief series of $B$, $n = \operatorname {fcl} (B)$. Then,

\[ I_{i}/I_{i-1} \subseteq \operatorname{Soc}(B/I_{i-1}) \subseteq Z(B/I_{i-1}, +), \]

for every $1 \leq i \leq n$. Therefore, $B$ is of nilpotent type. Moreover, we shall see by induction that $B^{(i)} \subseteq I_{n-i+1}$, for every $1 \leq i \leq n+1$. The case $i = 1$ is obvious; suppose that $B^{(i)}\subseteq I_{n-i+1}$, for some $1 \leq i < n$. Since $I_{n-i+1}/I_{n-i} \subseteq \operatorname {Soc}(B/I_{n-i})$ we can apply Lemma 3.3 to conclude that $B^{(i+1)} = B^{(i)}\ast B \subseteq I_{n-i+1}\ast B \subseteq I_{n-i}$. Hence, $B^{(n)} = 1$ and, therefore, $B$ is right nilpotent. Furthermore, $\operatorname {nil\,class}_r(B) \leq n$.

Conversely, suppose that $B$ is of nilpotent type and right nilpotent. Assume that there exists a chief factor $I/J$ of $B$ such that $I/J \nsubseteq \operatorname {Soc}(B/J)$. Since $I/J$ is a minimal ideal of $B/J$ and $\operatorname {Soc}(B/J)$ is an ideal of $B/J$, it follows that $I/J \cap \operatorname {Soc}(B/J) = 1$. Furthermore, as $(B/J,+)$ is nilpotent, there exists $1\neq b \in I/J \cap Z(B/J,+)$.

By the minimality of $I/J$, we have that either $I/J \ast B/J = 1$ or $I/J \ast B/J = I/J$. Assume that $I/J \ast B/J = I/J$. Then, $R_t(I/J,B/J) = I/J \subseteq R_t(B/J,B/J) = (B/J)^{(t)}$ for every $t\in \mathbb {N}$ by lemma 4.1. But $B/J$ is also right nilpotent and so $(B/J)^{(m)} = 1$ for some $m \in \mathbb {N}$. In particular, $I/J = 1$. This contradiction yields $I/J \ast B/J = 1$. Then, $1 \neq b\in I/J\cap \operatorname {Soc}(B/J)$, which is a contradiction. Consequently, $I/J \subseteq \operatorname {Soc}(B/J)$.

Proof. We argue by induction on $n$. Assume that $n = 1$. We prove

\[ b \in \operatorname{Soc}(B) \Leftrightarrow \lambda_b = \rho_b = \operatorname{id}_B. \]

If $b\in \operatorname {Soc}(B)$ then $b \in \operatorname {Ker} \lambda$ and so $\lambda _b = \operatorname {id}_B$. Moreover, $b\in Z(B,+)$ and then, $\rho _b(a) = (a^{-1} + b)^{-1}b = (b+a^{-1})^{-1}b= (ba^{-1})^{-1} b = a$, for every $a\in B$. Thus, $\rho _b = \operatorname {id}_B$.

Conversely, suppose that $\lambda _b = \rho _b = \operatorname {id}_B$, for some $b\in B$. It remains to prove that $b \in Z(B,+)$. For every $a\in B$, it holds

\begin{align*} \rho_b(a) = a & \Leftrightarrow (a^{{-}1} +b)^{{-}1}b = a \Leftrightarrow (a^{{-}1} +b)^{{-}1} = ab^{{-}1} \Leftrightarrow a^{{-}1} + b = ba^{{-}1} \\ & \Leftrightarrow a^{{-}1} + b = b + \lambda_b(a^{{-}1}) = b + a^{{-}1}, \end{align*}

as desired.

Then, for the retraction relation on the solution $r_B$, it occurs that $a \sim b$ if, and only if, $ab^{-1}\in \operatorname {Soc}(B)$, for every $a,b\in B$. Therefore, the map $\varphi \colon B/\!\sim \ \rightarrow B/\operatorname {Soc}(B)$, defined as $\varphi ([b]) = b\operatorname {Soc}(B)$, turns out to be an isomorphism between the solutions $\operatorname {Ret}(B,r_B)$ and $(B/\operatorname {Soc}(B), r_{B/\operatorname {Soc}(B)})$, as $([\lambda _a(b)], [\rho _b(a)]) = (\lambda _a(b)\operatorname {Soc}(B), \rho _b(a)\operatorname {Soc}(B))$, for every $a,b\in B$.

Now, suppose that $\operatorname {Ret}^n(B,r_B)$ is isomorphic to the solution associated with $B/\operatorname {Soc}_n(B)$, for some $n\geq 1$. Recall that

\begin{align*} & \operatorname{Ret}^{n+1}(B,r_B) = \operatorname{Ret}(\operatorname{Ret}^n(B,r_B)), \\ & B/\operatorname{Soc}_{n+1}(B) \cong (B/\operatorname{Soc}_n(B))/\operatorname{Soc}(B/\operatorname{Soc}_n(B)). \end{align*}

A similar argument to the case $n = 1$ shows that the solutions $\operatorname {Ret}^{n+1}(B,r_B)$ and $B/\operatorname {Soc}_{n+1}(B)$ are isomorphic.

Proof. Assume that $B$ has finite multipermutational level $m$. We show that every chief factor of $B$ is an $s$-chief factor by induction on $m$. If $m = 1$, then $B = \operatorname {Soc}(B)$ and the result follows. Suppose that $m \geq 1$ and the statement holds for every skew left brace with chief series of multipermutational level $m- 1$. f

Note that $\operatorname {Soc}(B)$ is a non-trivial ideal of $B$. By lemma 3.7, $B/\operatorname {Soc}(B)$ has chief series. Furthermore, $B/\operatorname {Soc}(B)$ has finite multipermutational level $m-1$. Thus, by induction hypothesis, every chief factor of $B/\operatorname {Soc}(B)$ is an $s$-factor and $m-1$ is less or equal than the chief length of $B/\operatorname {Soc}(B)$. According to lemma 3.7, there exists a chief series of $B$ passing through $\operatorname {Soc}(B)$. Moreover every chief factor of this series is an $s$-chief factor of $B$. Applying theorem 3.10, it follows that every chief factor of $B$ is an $s$-factor. Furthermore, $m$ is less or equal than the chief length of $B$.

Conversely, suppose that every chief factor of $B$ is an $s$-factor. If $B$ has not finite multipermutational level, then there exists $n_0\in \mathbb {N}$ such that $\operatorname {Soc}_n(B) = \operatorname {Soc}_{n_0}(B) < B$, for every $n \geq n_0$ since $B$ is noetherian by theorem 3.12. Now, we consider a chief series of $B$ passing through $\operatorname {Soc}_{n_0}(B)$,

\[ 1 = I_1 < \ldots < I_i = \operatorname{Soc}_{n_0}(B) < I_{i+1} < \ldots < I_m = B. \]

Since every chief factor is an $s$-factor, it follows that $1 \neq I_{i+1}/I_i \subseteq \operatorname {Soc}(B/I_i) = \operatorname {Soc}(B/\operatorname {Soc}_{n_0}) = \operatorname {Soc}_{n_0+1}(B)/\operatorname {Soc}_{n_0}(B)$. Therefore, $\operatorname {Soc}_{n_0}(B) \lneq \operatorname {Soc}_{n_0+1}(B)$, which is a contradiction.

Proof. It remains to prove that if $B$ has finite multipermutational level $m$, then $\operatorname {nil\,class}_r(B) \leq m+1$. We show that $B^{(k)} \subseteq \operatorname {Soc}_{m-k+1}(B)$, for every $1 \leq k \leq m$, by induction on $k$. Clearly $B^{(1)} = B = \operatorname {Soc}_m(B)$. Suppose that $B^{(k)} \subseteq \operatorname {Soc}_{m-k+1}(B)$, for some $1 \leq k < m$. Since

\[ \operatorname{Soc}_{m-k+1}(B)/\operatorname{Soc}_{m-k}(B) = \operatorname{Soc}\left( B/\operatorname{Soc}_{m-k}(B)\right). \]

The induction hypothesis and lemma 3.3 yield

\[ B^{(k+1)} = B^{(k)} \ast B \subseteq \operatorname{Soc}_{m-k+1}(B) \ast B \subseteq \operatorname{Soc}_{m-k}(B), \]

as desired.

Then, $B^{(m)} \subseteq \operatorname {Soc}(B)$ and, thus, $B^{(m+1)} = 0$. Hence, $\operatorname {nil\,class}_r(B) \leq m+1$.