Proof of Proposition 1·1. (i) $ \Rightarrow $ (ii). Suppose that $z\!\in\! A^{(2)}$ and $a\!\in\! A$ . Then $z*a=A^{(3)}=\{0\}$ .

(ii) $ \Rightarrow $ (iii). Let $z\in A^{(2)}$ and $a\in A$ . Since $\lambda_z(a)-a=z*a=0$ , we have that $\lambda_z(a)=a$ . This implies that $\lambda_z=\mathrm{id}$ and so $z\in \mathrm{Ker}\ \lambda$ .

(iii) $ \Rightarrow $ (iv). Since $ab=a*b+a+b=(a*b)\lambda_{a*b}^{-1}(a+b)$ , we have that $\lambda_{a}\circ\lambda_b=\lambda_{ab}=\lambda_{(a*b)\lambda_{a*b}^{-1}(a+b)}=\lambda_{a*b}\circ\lambda_{\lambda_{a*b}^{-1}(a+b)}=\lambda_{a+b}$ and so $\lambda\colon (A, {+})\longrightarrow \mathrm{{Aut}}(A, {+})$ is a group homomorphism.

(iv) $ \Rightarrow $ (v). We have that $\lambda_a\circ\lambda_b=\lambda_{ab}=\lambda_{a+\lambda_a(b)}=\lambda_a\circ\lambda_{\lambda_a(b)}$ . Therefore, $\lambda_b=\lambda_{\lambda_a(b)}$ .

(v) $ \Rightarrow $ (iii). Let a, $b\in A$ . We have that

\begin{align*} \lambda_{a*b}&=\lambda_{\lambda_a(b)-b}=\lambda_{\lambda_a(b)\lambda^{-1}_{\lambda_a(b)}({-}b)}=\lambda_{\lambda_a(b)\lambda_b^{-1}({-}b)}=\lambda_{\lambda_a(b)}\circ\lambda_{\lambda_b^{-1}({-}b)}\\ &=\lambda_b\circ\lambda_{\lambda_b^{-1}}=\lambda_{b\lambda_b^{-1}({-}b)}=\lambda_{b-b}=\lambda_0=\mathrm{id}. \end{align*}

Therefore, $a*b\in \mathrm{Ker}\ \lambda$ . Since $\mathrm{Ker}\ \lambda=\mathrm{Soc}\ A$ is an ideal of A, we have that $A^{(2)}\subseteq \mathrm{Ker}\ \lambda$ .

(iii) $ \Rightarrow $ (i). Suppose that $A^{(2)}\subseteq \mathrm{Ker}\ \lambda$ . It is enough to show that all generators of $A^{(3)}$ , that have the form $d*e$ for $d\in A^{(2)}$ and $e\in A$ , are zero. Since $A^{(2)}\subseteq \mathrm{Ker}\ \lambda$ , we have that $d*e=\lambda_d(e)-e=e-e=0$ if $d\in A^{(2)}$ and $e\in A$ . Therefore, $A^{(3)}=\{0\}$ .

Proof. (i) and (ii) imply (iii). Since $\lambda_a\circ\lambda_b=\lambda_{ab}=\lambda_{a+\lambda_a(b)}=\lambda_a\circ\lambda_{\lambda_a(b)}$ , we conclude that $\lambda_b=\lambda_{\lambda_a(b)}$ .

(ii) and (iii) imply (i). We have that $\lambda_{a+b}=\lambda_{a\lambda_a^{-1}(b)}=\lambda_a\circ\lambda_{\lambda_a^{-1}(b)}=\lambda_a\circ\lambda_{\lambda_a(\lambda_a^{-1}(b))}=\lambda_a\circ\lambda_b$ .

(iii) and (i) imply (ii). We have that $\lambda_{ab}=\lambda_{a+\lambda_a(b)}=\lambda_a\circ\lambda_{\lambda_a(b)}=\lambda_a\circ\lambda_b$ .

Proof. The first equality is just Lemma 2·3 (ii). The second equality follows from the fact that $\lambda_{x+y}=\lambda_x\lambda_y$ by the equivalence between 1 and 4 in Proposition 1·1 and so $(x+z)*y=\lambda_{x+z}(y)-y=\lambda_x(\lambda_z(y))-y=\lambda_{xz}(y)-y=(xz)*y$ .

Proof. We have that $b*y=0$ by Proposition 1·1. By Proposition 2·4, we have that

\[(x+b)*y=x*(b*y)+x*y+b*y=x*0+x*y+0=x*y.\]

Proof. We prove this proposition by induction on n. For $n=1$ , the result is clearly true. Suppose that

\[a_1^n=\sum_{k=1}^n\binom{n}{k}a_k.\]

Then, by using Lemma 2·3, we have:

\begin{align*} a^{n+1}&=a+a*a^n+a^n\\ &=a+a*\left(\sum_{k=1}^n\binom{n}{k}a_k\right)+\sum_{k=1}^n\binom{n}{k}a_k\\ &=a+\sum_{k=1}^n\binom{n}{k}(a*a_k)+\sum_{k=1}^n\binom{n}{k}a_k\\ &=a+\sum_{k=1}^n\binom{n}{k}a_{k+1}+\sum_{k=1}^n\binom{n}{k}a_k\\ &=a+\sum_{k=2}^{n+1}\binom{n}{k-1}a_k+\sum_{k=1}^n\binom{n}{k}a_k\\ &=\sum_{k=1}^{n+1}\binom{n}{k-1}a_k+\sum_{k=1}^n\binom{n}{k}a_k\\ &=\sum_{k=1}^{n+1}\binom{n+1}{k}a_k. \end{align*}

We conclude that the result is true for all positive integers n.

Proof. By Proposition 2·6, we have that $na=a^n+c$ , where

\[c=-\left(\binom{n}{2}a_2+\binom{n}{3}a_3+\dots+\binom{n}{n-1}a_{n-1}+a_n\right)\in A^2.\]

By Corollary 2·5, we have that $(na)*x=a^n*x$ for every element $x\in A$ . We argue by induction on n. For $n=1$ , the result is clear. Suppose that

\[(a^n)*a_j=\sum_{k=1}^n\binom{n}{k}a_{k+j}.\]

Then, by Lemma 2·3,

\begin{align*} (a^{n+1})*a_j&=(aa^n)*a_j\\ &=a*(a^n*a_j)+a^n*a_j+a*a_j\\ &=a*\left(\sum_{k=1}^n\binom{n}{k}a_{k+j}\right)+\sum_{k=1}^n\binom{n}{k}a_{k+j}+a_{j+1}\\ &=a_{j+1}+\sum_{k=1}^n\binom{n}{k}(a*a_{k+j})+\sum_{k=1}^n\binom{n}{k}a_{k+j}\\ &=a_{j+1} +\sum_{k=1}^n\binom{n}{k}a_{k+j+1}+\sum_{k=1}^n\binom{n}{k}a_{k+j}\\ &=a_{j+1} +\sum_{k=2}^{n+1}\binom{n}{k-1}a_{k+j}+\sum_{k=1}^n\binom{n}{k}a_{k+j}\\ &=\sum_{k=1}^{n+1}\binom{n+1}{k}a_{k+j}. \end{align*}

The result follows by induction.

Proof of Theorem A. We show first that $(B, {\cdot})$ is a group. Clearly, $\cdot$ is an internal operation on B. In addition, given x, y, $z\in B$ , we have that

\begin{align*} (xy)z&=xy+\lambda_{xy}(z)=(x+\lambda_x(y))+\lambda_{x+\lambda_x(y)}(z)= x+\lambda_x(y) +\lambda_x(\lambda_{\lambda_x(y)}(z))\\ &=x+\lambda_x(y)+\lambda_x(\lambda_y(z))=x+\lambda_x(y+\lambda_y(z))=x+\lambda_x(yz)=x(yz).\end{align*}

It follows that $\cdot$ is associative. Furthermore, $0x=0+\lambda_0(x)=0+x=x$ and $x0=x+\lambda_x(0)=x+0=x$ for each $x\in B$ . Consequently, 0 is the neutral element for $\cdot$ . Finally, if $x\in B$ , we have that $x({-}\lambda_x^{-1}(x))=x+\lambda_x({-}\lambda_x^{-1}(x))=x-\lambda_x(\lambda^{-1}_x(x))=x-x=0$ , while $({-}\lambda_x^{-1}(x))x=-\lambda_x^{-1}(x)+\lambda_{-\lambda_x^{-1}(x)}(x)=-\lambda_x^{-1}(x)+\lambda_{\lambda_x^{-1}(x)}^{-1}(x)=-\lambda_x^{-1}(x)+\lambda_x^{-1}(x)=0$ , and so $x^{-1}=-\lambda_x^{-1}(x)$ is the inverse of x under $\cdot$ . Therefore $(B, {\cdot})$ is a group. Furthermore, if x, y, $z\in B$ , we have that $xy-x+xz=x+\lambda_x(y)-x+x+\lambda_x(z)=x+\lambda_x(y)+\lambda_x(z)=x+\lambda_x(y+z)=x(y+z)$ . Therefore, $(B, {+}, {\cdot})$ is a left brace.

Let a, $b\in B$ . We have that $a*b=\lambda_a(b)-b$ and so $\lambda_{a*b}=\lambda_{\lambda_a(b)-b}=\lambda_{\lambda_a(b)}\circ\lambda_b^{-1}=\lambda_b\circ\lambda_b^{-1}=\operatorname{id_B}$ . It follows that $a*b\in\mathrm{Ker}\ \lambda$ . Therefore $B^{(2)}\subseteq\mathrm{Ker}\ \lambda$ , as $\mathrm{Ker}\ \lambda=\mathrm{Soc}(B)$ and $B^{(2)}$ are ideals of B. Let $u\in B^{(2)}$ and $c\in B$ , then $u*c=\lambda_u(c)-c=c-c=0$ , which implies that $B^{(3)}=\{0\}$ .

Proof of Theorem B. By Proposition 1·1, the map $\lambda\colon (A, {+})\longrightarrow \mathrm{{Aut}}(A, {+})$ is a group homomorphism and $A^2\subseteq \mathrm{Ker}(\lambda)$ . In particular, for $i\in{\mathbb{Z}}$ ,

\[\lambda_{a_i}=\lambda_{-a^i+a^{i+1}}=\lambda_{a^i}^{-1}\lambda_{a^{i+1}}=\lambda_{(a^i)^{-1}a^{i+1}}=\lambda_a.\]

Clearly, B is contained in the left subbrace generated by a. Furthermore, B is closed under the addition and by taking additive opposites. If $x=\sum_{i\in{\mathbb{Z}}}x_ia_i\in B$ , then

\[\lambda_{\sum_{i\in{\mathbb{Z}}}x_ia_i}=\prod_{i\in{\mathbb{Z}}}\lambda_{a_i}^{x_i}=\prod_{i\in{\mathbb{Z}}}\lambda_a^{x_i}=\lambda_{a}^t, \qquad\text{where $t={\sum_{i\in{\mathbb{Z}}}x_i}$.}\]

Here the product symbol corresponds to the iterated composition of group homomorphisms. It follows that, for $y=\sum_{i\in{\mathbb{Z}}}y_ia_i\in B$ ,

\[\lambda_{\sum_{i\in{\mathbb{Z}}}x_ia_i}\Biggl(\sum_{j\in{\mathbb{Z}}}y_ja_j\Biggr)=\sum_{j\in{\mathbb{Z}}}y_ja_{j+t}=\sum_{j\in{\mathbb{Z}}}y_{j-t}a_j,\qquad t=\sum_{i\in{\mathbb{Z}}}x_i.\]

Therefore, $xy=x+\lambda_x(y)\in B$ .

Furthermore, if $x\;=\;\sum_{i\in {\mathbb{Z}}}x_ia_i$ , then $y\;=\;\sum_{i\in{\mathbb{Z}}}({-}x_{i-\sum_{j\in{\mathbb{Z}}}x_j})a_i\;\in\; B$ satisfies that $xy=yx=0$ . Consequently $y=x^{-1}$ is the multiplicative inverse of x. We conclude that B is a subbrace, and so B is the subbrace generated by a.

We note that if $x=\sum_{i\in {\mathbb{Z}}}x_ia_i\in B$ and $y=\sum_{i\in I}y_ia_i$ , then $x*y=\lambda_x(y)-y=\sum_{i\in {\mathbb{Z}}}(y_{i-\sum_{j\in {\mathbb{Z}}}x_i}-y_i)a_i$ satisfies that $\sum_{i\in{\mathbb{Z}}}(y_{i-\sum_{j\in {\mathbb{Z}}}x_i}-y_i)=\sum_{i\in{\mathbb{Z}}}y_i-\sum_{i\in{\mathbb{Z}}}y_i=0$ . It follows that $B*B\subseteq D$ . Furthermore, if $x=\sum_{i\in{\mathbb{Z}}}x_ia_i\in D$ , then $x=\sum_{i\in{\mathbb{Z}}}x_ia_i-\sum_{i\in {\mathbb{Z}}}({-}x_ia)=\sum_{i\in{\mathbb{Z}}}x_i(a_i-a)= \sum_{i\in{\mathbb{Z}}}x_i(\lambda_{a^i}(a)-a)=\sum_{i\in{\mathbb{Z}}}x_i(a^i*a)\in C*C$ . It follows that $D=C*C$ .

Proof of Theorem C. As $\sum_{i\in{\mathbb{Z}}}y_{i-t}=\sum_{i\in {\mathbb{Z}}}y_i$ for each $t\in{\mathbb{Z}}$ , we obtain that $\lambda_{\lambda_x(y)}=\lambda_y$ for all x, $y\in C$ and so we conclude that $(C, {+}, {\cdot})$ is a left brace by Theorem A. In particular, for each a, $b\in C$ , $-a+ab=-a+a+\lambda_a(b)$ and the map $\lambda\colon (C, {\cdot})\longrightarrow \mathrm{{Aut}}(C, {+})$ is also a group homomorphism.

If $i=0$ , $a_0=\lambda_{a^0}(a)=\lambda_{0}(a)=a=c_0$ . Assume that $i \gt 0$ . Then $a_i=\lambda_{a^i}(a)=\lambda_a\circ\overset{(i)}{\cdots}\circ \lambda_a(a)=\lambda_{ia}(c_0)=c_i$ and $a_{-i}=\lambda_{a^{-i}}(a)=\lambda_a^{-1}\circ\overset{(i)}{\cdots}\circ \lambda_a^{-1}=\lambda_{({-}i)a}(c_0)=c_{-i}$ . Consequently, for every $i\in{\mathbb{Z}}$ , $\lambda_{a^i}(a)=c_i$ and so C coincides with the subbrace generated by $a=c_0$ by Theorem B.

Let $s_1=a$ , $s_j=a*s_{j-1}$ , $j\ge 2$ . Let us show by induction that $s_j=\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j-k-1}$ for $j\ge 2$ . For $j=2$ , we have that $s_2=a*a=\lambda_a(a)-a=c_1-c_0=-c_0+c_1$ . Suppose that $s_j=\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j-k-1}$ . Then

\begin{align*} s_{j+1}&=a*s_j-s_j=a*\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j-k-1}-\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j-k-1}\\ &=\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}(a*c_{j-k-1})-\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j-k-1}\\ &=\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j+1-k-1}-\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j-k-1}\\ &=\sum_{k=0}^{j-1}({-}1)^k\binom{j-1}{k}c_{j+1-k-1}+\sum_{k=1}^{j}({-}1)^k\binom{j-1}{k-1}c_{j+1-k-1}\\ &=\sum_{k=0}^{j}({-}1)^k\binom{j}{k}c_{j+1-k-1}. \end{align*}

We conclude that $0\ne s_j\in C^j$ for all $j\in{\mathbb{N}}$ and so $C^j\ne \{0\}$ for all $j\in{\mathbb{N}}$ .

The description of the subbrace generated by one element given in Theorem B shows that, with the notation of that theorem, the assignment

\[\sum_{i\in {\mathbb{Z}}}x_ic_i\longmapsto \sum_{i\in {\mathbb{Z}}}x_ia_i\]

is a left brace epimorphism between C and the subbrace of A generated by a and maps $c_0$ to a. Furthermore it is clear that it is the unique left brace epimorphism satisfying this condition.

Proof. We argue by induction on k. Note that, for all k, we have that

(2) \begin{equation} 0=(({-}a)+a)*a_{m-k}=({-}a)*a_{m-k+1}+({-}a)*a_{m-k}+a_{m-k+1} \end{equation}

by Proposition 2·4.

Proof. The result is obvious for $n=0$ and known for positive integers n by Equation (1). Hence it is enough to prove it for negative integers. We argue by induction on $-n$ . When $n=-1$ , we observe that $\binom{-1}{j}=({-}1)^j\binom{j}{j}=({-}1)^j$ . The result follows by Proposition 6·1.

Assume that the result is true for $-n=k$ , that is,

\[(na)*a_j=\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}.\]

We want to prove that it is also true for $-n=k+1$ . By Proposition 2·4, Lemma 2·3, Corollary 2·5 and Proposition 6·2, we have that

\begin{align*} (({-}1&+n)a)*a_j=({-}a)*((na)*a_j)+({-}a)*a_j+(na)*a_j\\ &=({-}a)*\left(\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}\right)+\sum_{k=1}^{m-j}\binom{-1}{k}a_{k+j}+\sum_{k=1}^{m-j}\binom{n}{k}a_{k+j}\\ &=\sum_{k=1}^{m-j}\binom{n}{k}(({-}a)*a_{k+j})+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{j+k}\\ &=\sum_{k=1}^{m-j}\binom{n}{k}\sum_{t=1}^{m-k-j}\binom{-1}{t}a_{t+k+j} + \sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{k+j}\\ &=\sum_{r=2}^{m-j}\biggl(\sum_{\substack{1\le k\le m-j\\1\le t\le m-k-j\\t+k=r}}\binom{n}{k}\binom{-1}{t}\biggr)a_{r+j}+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{k+j}\\ &=\sum_{r=2}^{m-j}\biggl(\sum_{t=1}^{r-1}\binom{n}{r-t}\binom{-1}{t}\biggr)a_{r+j}+\sum_{k=1}^{m-j}\left(\binom{-1}{k}+\binom{n}{k}\right)a_{k+j}\\ &=\sum_{r=2}^{m-j}\biggl(\sum_{t=0}^r\binom{n}{r-t}\binom{-1}{t}\biggr)a_{r+j}+\left(\binom{-1}{1}+\binom{n}{1}\right)a_{1+j}\\ &=\sum_{r=2}^{m-j}\binom{n-1}{r}a_{r+j}+({-}1+n)a_{1+j}\\ &=\sum_{r=1}^{m-j}\binom{n-1}{r}a_{r+j}=\sum_{k=1}^{m-j}\binom{n-1}{k}a_{k-j}. \end{align*}

By induction, the result is valid for all negative integers n.

We are now in a position to give a description of the subbrace generated by an element a in a left brace with $A^{(3)}=A^{m+1}=\{0\}$ .

Proof of Theorem D. It is clear that if T is a subbrace of A containing a, then $S\subseteq T$ . Observe also that $a_k=0$ if $k \gt m$ . It is enough to check that S is a subbrace of A, that is, S is closed by taking sums, products, additive opposites, and multiplicative inverses. It is clear that S is closed by taking sums and additive opposites.

Since $xy=x+x*y+y$ for every x, $y\in A$ and S is closed by taking sums, it will be closed under taking products if it is closed under taking the star operation $*$ . Consider

\[t=\sum_{k=1}^m t_ka_k\in S,\quad u=\sum_{k=1}^mu_ka_k\in S.\]

Since $\sum_{k=2}^mt_ka_k\in A^2$ , by Corollary 2·5 and Lemma 2·3 we see that

\[t*u=(t_1a_1)*u=\sum_{k=1}^mu_k (t_1a)*a_k.\]

By Proposition 6·2, this is a linear combination with integer coefficients of elements of S, and so $t*u\in S$ .

To conclude the proof, we must show that S is closed under taking inverses. Take $t=\sum_{i=1}^mt_ia_i$ , we will show that

\[t^{-1}=\sum_{j=1}^m\left({-}\sum_{k=0}^{j-1}\binom{-t_1}{k}t_{j-k}\right)a_j.\]

Call w the right-hand side of the previous equality. Since $wt=w+w*t+t$ , we compute

\begin{align*} w*t&=\left({-}\binom{-t_1}{0}t_1a_1\right)*\sum_{i=1}^mt_ia_i\\ &=({-}t_1a_1)*\sum_{i=1}^mt_ia_i\\ &=\sum_{i=1}^mt_i(({-}t_1a_1)*a_i)\\ &=\sum_{i=1}^mt_i\sum_{k=1}^{m-i}\binom{-t_1}{k}a_{i+k}\\ &=\sum_{r=1}^m\biggl(\sum_{\substack{1\le i\le m\\1\le k\le m-i\\i+k=r}}t_i\binom{-t_1}{k}\biggr)a_r\\ &=\sum_{r=1}^m\biggl(\sum_{k=1}^{r-1}t_{r-k}\binom{-t_1}{k}\biggr)a_r.\end{align*}

by Lemma 2·3, Corollary 2·5 and Proposition 6·2. It follows that $wt=w+w*t+t=0$ , the neutral element of $\cdot$ , and so w is the multiplicative inverse of t in A. As $w\in S$ and the inverse is unique, S is closed under taking inverses. This completes the proof.

Proof of Theorem E. Consider an integer $m\ge 2$ . We define in

\[B_m=\overbrace{\mathbb{Z}\times\dots\times\mathbb{Z}}^{(m)}\]

the addition in the usual form

\[(n_1,\dots, n_m)+(t_1,\dots, t_m)=(n_1+t_1,\dots,n_m+t_m).\]

Given $\mathbf{n}=(n_1,\dots, n_m)\in B_m$ and $\mathbf{t}=(t_1,\dots, t_m)\in B_m$ , we define $\lambda_{\mathbf{n}}(\mathbf{t})=(z_1,\dots, z_m)$ where, for $1\le i\le m$ ,

\[z_i=\sum_{k=0}^{i-1}\binom{n_1}{k}t_{i-k}.\]

Let us denote $z_i$ by $(\lambda_{\mathbf{n}}(\mathbf{t}))_i$ , $1\le i\le m$ . It is clear that $\lambda_{\mathbf{n}}$ is an endomorphism of $(B, {+})$ for all $\mathbf{n}\in B_m$ . We prove that $\lambda$ is a homomorphism from $(B, {+})$ to $\operatorname{End}(B, {+})$ . Consider the elements of $B_m$ $\mathbf{n}=(n_1,\dots, n_m)$ , $\mathbf{t}=(t_1,\dots, t_m)$ , and $\mathbf{w}=(w_1,\dots, w_m)$ . Then

\[(\lambda_{\mathbf{w}}(\mathbf{t}))_1=\sum_{k=0}^0\binom{w_1}{0}t_{1-k}=t_1\]

and so, for $1\le i\le m$ ,

\begin{align*} (\lambda_{\mathbf{n}}(\lambda_{\mathbf{w}}(\mathbf{t})))_i&= \sum_{k=0}^{i-1}\binom{n_1}{k}(\lambda_{\mathbf{w}}(\mathbf{t}))_{i-k}\\ &=\sum_{k=0}^{i-1}\binom{n_1}{k}\sum_{l=0}^{i-k-1}\binom{w_1}{l}t_{i-k-l}\\ &=\sum_{k=0}^{i-1}\sum_{l=0}^{i-k-1}\binom{n_1}{k}\binom{w_1}{l}t_{i-k-l}\\ &=\sum_{k=0}^{i-1}\sum_{r=k}^{i-1}\binom{n_1}{k}\binom{w_1}{r-k}t_{i-r}\\ &=\sum_{k=0}^{r-1}\binom{n_1+w_1}{r}t_{i-r}=(\lambda_{\mathbf{n}+\mathbf{w}}(\mathbf{t}))_i.\end{align*}

We also note that if $\mathbf{0}=(0,\dots, 0)\in B_m$ , then

\[(\lambda_{\mathbf{0}}(\mathbf{t}))_i=\sum_{k=0}^{i-1}\binom{0}{k}t_{i-k}=t_i,\]

therefore $\lambda_{\mathbf{0}}=\mathrm{id}$ . Consequently, if $\mathbf{n}\in B_m$ , we have that $\lambda_{\mathbf{n}}$ has an inverse $\lambda_{-\mathbf{n}}$ and so the image of $\lambda$ is contained in the automorphism group of $(B, {+})$ . Furthermore, for $\mathbf{n}$ , $\mathbf{t}$ , $\mathbf{w}\in B_m$ and $1\le i\le m$ ,

\[(\lambda_{\lambda_{\mathbf{w}}(\mathbf{n})}(\mathbf{t}))_i=\sum_{k=0}^{i-1}\binom{(\lambda_{\mathbf{w}}(\mathbf{n}))_1}{k}t_{i-k}=\sum_{k=0}^{i-1}\binom{n_1}{k}t_{i-k}=(\lambda_{\mathbf{n}}(\mathbf{t}))_i.\]

We conclude that $\lambda_{\lambda_{\mathbf{w}}(\mathbf{n})}=\lambda_{\mathbf{n}}$ .

By Theorem A, we have that $B_m$ is a left brace with the product $\cdot$ given by $\mathbf{n}\mathbf{t}=\mathbf{n}+\lambda_{\mathbf{n}}(\mathbf{t})$ for $\mathbf{n}$ , $\mathbf{t}\in B_m$ and $B^{(3)}=\{0\}$ .

Let $\mathbf{b}=(1,0,\dots, 0)$ . If $\mathbf{b}_1=b$ and $\mathbf{b}_{k+1}=\mathbf{b}*\mathbf{b}_k$ for $k\ge 1$ , we prove that ${(\mathbf{b}_i)}_j=\delta_{i,j}$ for $1\le j\le m$ by induction on i. The result is clear for $i=1$ . If ${(\mathbf{b}_i)}_j=\delta_{i,j}$ for $1\le j\le m$ and some i, then

\begin{align*} {(\mathbf{b}_{i+1})}_j&={(\mathbf{b}*\mathbf{b}_{i})}_j={(\lambda_{\mathbf{b}}(\mathbf{b}_i)-\mathbf{b}_i)}_j\\ &=\sum_{k=0}^{j-1}\binom{1}{k}{(\mathbf{b}_i)}_{j-k}-\delta_{i,j}=\binom{1}{0}\delta_{i,j}+\binom{1}{1}\delta_{i,j-1}-\delta_{i,j}\\ &=\delta_{i,j-1}=\delta_{i+1,j}.\end{align*}

By Theorem D, we have that the subbrace of $B_m$ generated by $\mathbf{b}$ coincides with $B_m$ . By Remark 6·3, we obtain that

\[B_m^r=\{(t_1,\dots, t_m)\in B_m\mid \text{$t_i=0$ for $1\le i \lt m$}\}\]

if $r\le m$ , and $B_m^r=\{\mathbf{0}\}$ if $r \gt m$ .

It is clear that, given a left brace A with $A^{(3)}=A^{m+1}=\{0\}$ for a natural number $m\ge 2$ , with the same notation as in Theorem D, the map $(n_1,\dots, n_m)\longmapsto \sum_{k=1}^mn_ka_k$ defines a left brace epimorphism $\varphi$ from $B_m$ to the subbrace of A generated by a such that $\varphi(\mathbf{b})=a$ , and this is the unique possible homomorphism with this condition.