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Notice that adjoint groups of braces are the multiplicative groups of braces under the operation ◦. This provides us with easy-to- use formulas for one-to-one correspondence between right nilpotent Fp-braces and right k nilpotent pre-Lie algebras over Fp of nilpotency index k for p> 2 . As an example application, we answer a question asked by Vendramin in [38] on braces of cardinality p4 for p> 64 for right nilpotent braces. Many authors have developed methods to describe finite braces of a given cardinality. In particular, all braces and skew braces of cardinality p3 for all prime numbers p were described by Bachiller [4] and Zenouz [39] (for some related results, see also [15]). A research problem posed in [38] suggests to describe all braces of cardinality pn for n> 3, and at the end of paper [38] it is suggested to consider this question for F-braces and one-generator braces. We answer this question for n = 4 and p ≥ 17 for right nilpotent braces. In chapter 4, the connection between braces and pre-Lie algebras is used to generalise some results from the context of pre-Lie algebras to braces. For example, it is shown that the sum of a finite number of left nilpotent ideals in a left brace is a left nilpotent ideal. This is an analogon of the pre-Lie algebra result obtained in [13]. Therefore every finite brace contains the largest left nilpotent ideal.

1 Background information

Recall that a pre-Lie algebra A is a vector space with a binary operation (x, y) → xy satisfying (xy)z − x(yz)=(yx)z − y(xz), for every x, y, z ∈ A. We say that a pre-Lie algebra A is nilpotent if, for some n ∈ N, all products of n elements in A are zero. Pre-Lie algebras were introduced by Gerstenhaber, and independently by Vinberg. Recall that a set A with binary operations + and ∗ is a left brace if (A, +) is an abelian group and the following version of distributivity combined with associativity holds:

(a + b + a ∗ b) ∗ c = a ∗ c + b ∗ c + a ∗ (b ∗ c), a ∗ (b + c)= a ∗ b + a ∗ c, for all a, b, c ∈ A, moreover (A, ◦) is a group, where we define a ◦ b = a + b + a ∗ b. In what follows, we will use the definition in terms of operation ‘◦’ presented in [11] ( see [31] for the original definition): a set A with binary operations of addition +, and multiplication ◦ is a brace if (A, +) is an abelian group, (A, ◦) is a group and for every a, b, c ∈ A a ◦ (b + c)+ a = a ◦ b + a ◦ c. Braces have found application in several research areas, some of which we now men- tion. They form an important concept in Hopf-Galois extensions – see [3, 14, 18, 22, 39, 37] for related results. They have been shown to be equivalent to several concepts in group theory, such as groups with bijective 1-cocycles, regular subgroups of the holomorph, matched pairs of groups and Garside Groups [32, 3, 10, 16, 21, 37]. There is a connec- tion between braces and grupoids [24]. It is known that two-sided braces are exactly the Jacobson radical rings [31], [11]. In [19], applications of braces in quantum integrable

2 systems were investigated, and in [34] R-matrices constructed from braces were studied. Solutions of the pentagon equation related to braces have been investigated by several authors [17]. In [7], Brzezi´nski showed that braces are related to trusses. Simple braces were investigated in [3, 5], and cohomology of braces was investigated in [29]. An anal- ogon of the Artin- Wedderburn theorem for braces was obtained in [26]. Circle algebras related to braces were introduced by Catino and Rizzo in [10]. We now recall Definition 2 from [32], which we state for left braces, as it was originally stated for right braces. Notice that F-braces are related to circle algebras.

Definition 1. Let F be a field. We say that a left brace A is an F-brace if its additive group is an F-vector space such that a ∗ (αb) = α(a ∗ b) for all a, b ∈ A, α ∈ F. Here a ∗ b = a ◦ b − a − b.

In [31], Rump introduced left nilpotent and right nilpotent braces and radical chains Ai+1 = A ∗ Ai and A(i+1) = A(i) ∗ A for a left brace A, where A = A1 = A(1). Recall that a left brace A is left nilpotent if there is a number n such that An = 0, where inductively Ai consists of sums of elements a∗b with a ∈ A, b ∈ Ai−1. A left brace A is right nilpotent if there is a number n such that A(n) = 0, where A(i) consists of sums of elements a ∗ b with a ∈ A(i−1), b ∈ A. Strongly nilpotent braces and the chain of ideals A[i] of a brace [1] [i+1] i [j] [i+1−j] A were defined in [35]. Define A = A and A = Pj=1 A ∗ A . A left brace A is strongly nilpotent if there is a number n such that A[n] = 0, where A[i] consists of sums of elements a ∗ b with a ∈ A[j], b ∈ A[i−j] for all 0 ring theory and group theory, see [5, 28, 27, 26]. In this paper we introduce left nilpotent radical for finite braces, in analogy with pre-Lie algebras.

2 From finite braces to Pre-Lie algebras

Let p > 0 be a prime number. Let Fp denote the field of cardinality p. Let B be a left brace with operations + and ◦, and operation ∗ where a ∗ b = a ◦ b − a − b. p Let B be a finite left brace whose additive group is Zp ×···× Zp where (Z/pZ) , then B is an Fp-brace. On the other hand, the additive group of an Fp -brace is abelian, and n every element has order p, so it is Zp ×···× Zp, and hence B has cardinality p for some n. By a result of Rump [31], every brace of order pn is left nilpotent. Assume that B is also right nilpotent, then by a result from [35] it is strongly nilpotent. In other words there is k such that the product of any k element, in any order, in R is zero (where all products are under the operation ∗). If B[k] = 0 and B[k−1] =6 0, then we will say that B is strongly nilpotent of degree k. We recall Lemma 15 from [35]: Lemma 2. Let s be a natural number and let (A, +, ◦) be a left brace such that As =0 for some s. Let a, b ∈ A, and as usual define a ∗ b = a ◦ b − a − b. Define inductively elements ′ ′ ′ ′ di = di(a, b),di = di(a, b) as follows: d0 = a, d0 = b, and for i ≤ 1 define di+1 = di + di ′ ′ and di+1 = didi. Then for every c ∈ A we have

2s i+1 ′ ′ (a + b) ∗ c = a ∗ c + b ∗ c + X(−1) ((di ∗ di) ∗ c − di ∗ (di ∗ c)). i=0

3 Notation 1. Let A be a strongly nilpotent brace with operations +, ◦, ∗ defined as usual, so x◦y = x+y+x∗y for x, y, z ∈ A, and let E(x, y, z) ⊆ A denote the set consisting of any product of elements x and y and one element z at the end of each product under the operation ∗, in any order, with any distribution of brackets, each product consisting of at least 2 elements from the set {x, y}, each product having x and y appear at least once, and having element z at the end. Notice that E(x, y, z) is finite, provided that A is a strongly nilpotent brace. Let Vx,y,z be a vector obtained from products of elements x, y, z arranged in a such way that shorter products of elements are situated before longer products.

Below we associate to every such brace a pre-Lie algebra which is also strongly nilpo- tent and which has the same additive group.

Proposition 3. Let A be an Fp-brace which is strongly nilpotent of degree k. Let p be a prime number such that 2k < p. As usual, the operations on A are +, ◦ and ∗ where a ∗ b = a ◦ b − a − b. Define the binary operation · on A as follows

p−2 1 a · b = ((2ia) ∗ b), X 2i i=0 for a, b ∈ A, where 2ia denotes the sum of 2i copies of element a, and 2−i denotes the i inverse of 2 in Fp. Then (a + b) · c = a · c + b · c for every a, b, c ∈ A. Moreover a · (b + c)= a · b + a · c for every a, b, c ∈ A.

Proof. By the definition of a left Fp-brace, we immediately get that a·(b+c)= a·b+a·c. We will show that (a + b) · c = a · c + b · c for a, b, c ∈ A. Observe that

p−2 1 (a + b) · c = ((2ia +2ib) ∗ c). X 2i i=0 Lemma 2 applied several times yields 1 1 1 1 (2na +2nb) ∗ c = (2na) ∗ c + (2nb) ∗ c + C(n), 2n 2n 2n 2n

1 1 where C(n) is a sum of some products of elements 2n a and 2n b and an element c at the end (because A is a strongly nilpotent brace). Moreover, each product has at last one occurrence of element 2na and also at last one occurrence of element 2nb and an element c at the end. To show that (a + b) · c = a · c + b · c it suffices to prove that

p−2 1 C(n)=0. X 2n i=0

We may consider a vector V 1 1 obtained as in Notation 1 from products of elements 2n a, 2n b,c 1 1 2n a, 2n b, c. By Lemma 2 (applied several times) every element from the set E(2x, 2y, z) can be written as a linear combination of elements from E(x, y, z), with coefficients which do

4 not depend on x, y and z. We can then organize these coefficients in a matrix, which we will call M = {mi,j}, so that we obtain

MVx,y,z = V2x,2y,z.

Notice that elements from E(x, y, z) (and from E(2x, 2y, z)) which are shorter appear before elements which are longer in our vectors Vx,y,z and V2x,2y,z. Therefore by Lemma 2 it follows that M is an upper triangular matrix. Observe that the first four elements in the vector V2x,2y,z are (2x) ∗ ((2y) ∗ z), ((2x) ∗ (2y)) ∗ z, (2y) ∗ ((2x) ∗ z) and ((2y) ∗ (2x)) ∗ z (arranged in some order). We can assume that (2x) ∗ ((2y) ∗ z) is the first entry in the vector V2x,2y,z (so x ∗ (y ∗ z) is the first entry in the vector Vx,y,z). By Lemma 2 applied several times, (2x) ∗ ((2y) ∗ z) can be written as 4(x ∗ (y ∗ z)) plus elements from E(x, y, z) of degree larger than 3 (so these elements are products of more than three elements from the set {x, y, z}). It follows that the first diagonal entry in M equals 4, so m1,1 = 4. Observe that the following diagonal entries will be equal to 4 or more, because, for example, (2x) ∗ ((2x) ∗ ((2x) ∗ y)) can be written using Lemma 2 as 8(x ∗ (x ∗ (x ∗ y))) plus elements of degree larger than 4. Therefore M is an upper triangular matrix with all diagonal entries of the form 2i, where 1

1 1 n V n n =( M) V . 2n 2 x,2 y,z 2 x,y,z

p−1 p−1 p−1 Observe that 2 x = x and 2 y = y because 2 = 1 in Fp. Therefore

1 1 p−1 V = V p−1 p−1 =( M) V . x,y,z 2p−1 2 x,2 y,z 2 x,y,z

Notice that there is a vector V with entries in Fp such that

T T n C(n)= V V2nx,2ny,z = V M Vx,y,z for each n, where V T is the transposition of V . Now we calculate

p−2 p−2 1 1 C(n)= V T ( M nV ). X 2n X 2n x,y,z n=0 n=0 p−2 1 n 1 −1 1 p−1 Notice that Pn=0 2n M =(I − 2 M) (I − ( 2 M) ), therefore

p−2 1 1 1 C(n)= V T (I − M)−1((I − ( M)p−1)V )=0. X 2n 2 2 x,y,z n=0 This concludes the proof.

We will now prove the main result of this section.

5 Theorem 4. Let A be an Fp-brace which is strongly nilpotent of degree k. Assume that 2k

p−2 1 a · b = ((2ia) ∗ b), X 2i i=0 for a, b ∈ A, where 2ia denotes the sum of 2i copies of element a, and 2−i denotes the i inverse of 2 in Fp. Then

(a · b) · c − a · (b · c)=(b · a) · c − (b · a) · c

for every a, b, c ∈ A.

Proof. By Lemma 2 applied several times we get

(x + y) ∗ z = x ∗ z + y ∗ z + x ∗ (y ∗ z) − (x ∗ y) ∗ z + d(x, y, z),

(y + x) ∗ z = x ∗ z + y ∗ z + y ∗ (x ∗ z) − (y ∗ x) ∗ z + d(y, x, z), T where d(x, y, z) = E Vx,y,z for some vector E with entries in Fp which does not depend of x, y, z, and where Vx,y,z is as in Notation 2 (moreover d(x, y, z) is a combination of elements with at least 3 occurences of elements from the set {x, y}). It follows that

x ∗ (y ∗ z) − (x ∗ y) ∗ z − y ∗ (x ∗ z)+(y ∗ x) ∗ z = d(y, x, z) − d(x, y, z).

Let a, b, c ∈ A and let m, n be natural numbers. Applying it to x = 2ia, y = 2jb, z = c we get (2ia) ∗ ((2jb) ∗ c) − ((2ia) ∗ (2jb)) ∗ c + d(2ia, 2jb, c)= = (2jb) ∗ ((2ia) ∗ c) − ((2jb) ∗ (2ia)) ∗ c + d(2jb, 2ia, c). Notice that

p−2 p−2 p−2 1 1 1 a · (b · c)= a · ((2jb) ∗ c)= ((2ia) ∗ ((2jb) ∗ c)). X 2j X 2i X 2j j=0 i=0 j=0

Consequently, p−2 1 a · (b · c)= ((2ia) ∗ ((2jb) ∗ c)). X 2i+j i,j=0 On the other hand

p−2 p−2 1 1 (a · b) · c =( ((2ia) ∗ b)) · c = (((2ia) ∗ b) · c), X 2i X 2i i=0 i=0 where the last equation follows from Proposition 3. Consequently,

p−2 p−2 1 1 (a · b) · c = ((2j((2ia) ∗ b)) ∗ c)= ((2ia) ∗ (2jb)) ∗ c, X 2i+j X 2i+j i,j=0 i,j=0

6 1 Recall a previous equation, multiplied by 2i+j on both sides: 1 1 1 (2ia) ∗ ((2jb) ∗ c) − ((2ia) ∗ (2jb)) ∗ c + d(2ia, 2jb, c)= 2i+j 2i+j 2i+j 1 1 1 = (2jb) ∗ ((2ia) ∗ c) − ((2jb) ∗ (2ia)) ∗ c + d(2jb, 2ia, c). 2i+j 2i+j 2i+j By summing the above equation for all 0 ≤ i, j ≤ p − 2 and subtracting the previous equations we obtain that

p−2 1 (a · b) · c − a · (b · c)+ d(2ia, 2jb, c)= X 2i+j i,j=0

p−2 1 (b · a) · c − b · (a · c)+ d(2jb, 2ia, c). X 2i+j i,j=0

p−2 1 i j So it remains to show that Pi,j=0 2i+j d(2 a, 2 b, c) = 0, for all a, b, c ∈ A (and hence p−2 1 j i Pi,j=0 2i+j d(2 b, 2 a, c) = 0). p−2 1 i j Proof that Pi,j=0 2i+j d(2 a, 2 b, c) = 0. The proof uses a similar idea as the proof of Proposition 3, but we include all the details here for convenience. Notice that for some m, d(a, b, c) = w(a, b, c)+ v(a, b, c) where w(a, b, c) contains all the products of elements a, b, c which appear as summands in d(a, b, c) and in which a appears at least twice, and v(a, b, c) is a sum of products which are summands in d(a, b, c) and in which a appears only once (and hence b appears at least twice). It suffices to show that p−2 1 w(2ia, 2jb, c) = 0 and p−2 1 v(2ia, 2jb, c) = 0. Observe that it suffices Pi,j=0 2i+j Pi,j=0 2i+j p−2 1 i ′ p−2 1 ′ j ′ ′ to show that Pi=0 2i w(2 a, b ,c) = 0 and Pj=0 2j v(a , 2 b, c) = 0 for any a, a , b, b ,c ∈ A. We will first show that p−2 1 w(2ia, b′,c) = 0. Pi=0 2i F ′ T ′ Observe that there is a vector W with entries in p such that w(a, b ,c) = W Va,b′,c ′ where Va,b′,c is a vector constructed as in Notation 1 but only including as entries these products from E(a, b′,c) in which a appears at least twice. By using Lemma 2 several times (similarly as in the proof of Proposition 3) there exists a matrix M such that ′ ′ i V2a,b′,c = MVa,b′,c and M is upper triangular with diagonal entries 2 for i ≥ 2 (because a ′ appears at least twice in each product which is an entry in Va,b′,c) and for i

p−2 p−2 p−2 1 i ′ 1 T ′ T 1 i ′ w(2 a, b ,c)= W V i ′ = W M V ′ . X 2i X 2i 2 a,b ,c X 2i a,b ,c i=0 i=0 i=0

p−1 Notice that 2 = 1 in Fp, therefore

′ ′ p−1 ′ Va,b′,c = V2p−1a,b′,c = M Va,b′,c.

1 F Notice that 2p−1 = 1 in p. It follows that

p−2 1 i ′ 1 −1 1 p−1 p−1 ′ M V ′ =(I − M) (( ) M − I)V ′ = X 2i a,b ,c 2 2 a,b ,c i=1

7 1 −1 p−1 =(I − M) (M − I)V ′ =0. 2 a,b ,c It follows that p−2 1 w(2ia, b′,c)=0. X 2i i=0 p−2 1 ′ j ′ The proof that Pj=0 2j v(a , 2 b, c) = 0 for all a ,b,c ∈ A is similar. Observe that there ′ F ′ ′T ′′ ′′ is a vector W with entries in p such that v(a ,b,c)= W Va′,b,c, where Va′,b,c is a vector constructed as in Notation 1 but only including as entries those products in which b appears at least twice. By applying Lemma 2 several times, there exists a matrix M¯ ′′ ¯ ′′ ¯ i such that Va′,2b,c = MVa′,b,c and M is upper triangular with diagonal entries 2 for i ≥ 2 ′′ (because b appears at least twice in each product which is an entry in Va′,b,c) and for i ≤ k. Similarly as before,

′′ ′′ p−1 ′′ 1 p−1 ′′ V ′ = V ′ p−1 = M¯ V ′ = M¯ V ′ . a ,b,c a ,2 b,c a ,b,c 2p−1 a ,b,c It follows that

p−2 p−2 p−2 1 ′ j 1 ′T ′′ ′T 1 j ′′ v(a , 2 b, c)= W V ′ j = W M¯ V ′ X 2j X 2j a ,2 b,c X 2j a ,b,c i=0 i=0 i=0 hence p−2 1 ′ j 1 −1 1 p−1 ′′ v(a , 2 b, c)=(I − M¯ ) ( M¯ − I)V ′ =0. X 2j 2 2p−1 a ,b,c j=0

We obtain the following corollary:

Corollary 5. Let A be an Fp brace of degree k which is strongly nilpotent . Assume that 2k

p−2 1 a · b = ((2ia) ∗ b), X 2i i=0 for a, b ∈ A where 2ia denotes the sum of 2i copies of element a, and 2−1 denotes the i inverse of 2 in Fp. Define a ⊙ b = −(a · b), then A with operations + and ⊙ is a pre- Lie algebra over the field Fp.

Proof. By Proposition 3, (a ⊙ b) ⊙ c =(−(a · b) ⊙ c)= −(−(a · b) · c)=(a · b) · c, similarly, a ⊙ (b ⊙ c) = −(a · (−(b · c)) = a · (b · c). by Theorem 4, A with operations +, ⊙ is a pre-Lie algebra. The following question remains unresolved:

Question 1. Let A be a finite Fp-brace which is not right nilpotent. Does it follow that, for a sufficiently large p, A with operations + and ⊙ as above is a pre-Lie algebra?

8 There are limitations to our approach, as we were not able to remove the assumption that braces and pre-Lie algebras are right nilpotent. We do not know for sure that there are no underlying necessary assumptions in Rump’s sentence quoted at the beginning of this paper as his statement was not stated as a theorem but only as a comment. Notice that in [23] a pre-Lie algebra was attached to every strongly nilpotent brace as an associated graded structure of this brace, but it was not possible to reverse this approach since one pre-Lie algebra could be associated to many non-isomorphic braces in this way.

3 From pre-Lie algebras to braces

Observe that if the pre-Lie algebra is nilpotent we can use the group of flows of a pre- Lie algebra to obtain the passage from finite nilpotent pre-Lie algebras of cardinality pn and right nilpotent Fp-braces in a manner similar to [36]. Upon closer inspection this gives the same brace when we use Lazard’s correspondence and later change the obtained group with 1-cocycle into brace suggested by Rump in [31] on pages 135, 141. The correspondence betwen groups with 1-cocycle and braces is mentioned in Theorem 2.1 in [12]. As mentioned by Rump in a private correspondence, the addition in the pre-Lie algebra and in the corresponding brace is always the same, so we only need to define the multiplication ◦ in the brace. Let A with operations + and · be a nilpotent pre-Lie algebra (over field Fp) of nilpo- tency index k. Recall that a pre-Lie algebra A is nilpotent of nilpotency index n if the product of any n elements is zero in this pre-Lie algebra, and n is minimal possible. Let p be a prime number larger than k and let Fp be the field of p elements. Define the Fp-brace (A, +, ◦) with the same addition as in the pre-Lie algebra A and with the multiplication ◦ defined as in the group of flows as follows.

1. Let a ∈ A, and let La : A → A denote the left multiplication by a, so La(b)= a · b. Define Lc · Lb(a)= Lc(Lb(a)) = c · (b · a). Define 1 1 eLa (b)= b + a · b + a · (a · b)+ a · (a · (a · b)) + ··· 2! 3! where the sum ‘stops’ at place k, since the nilpotency index of A is k. This is well defined since p>k. 2. We can formally consider element 1 such that 1 · a = a · 1= a in our pre-Lie algebra (as in [30]) and define 1 1 W (a)= eLa (1) − 1= a + a · a + a · (a · a)+ ··· 2! 3! where the sum ‘stops’ at place k. Notice that W : A → A is a bijective function, provided that A is a nilpotent pre-Lie algebra. 3. Let Ω : A → A be the inverse function to the function W , so Ω(W (a)) = W (Ω(a)) = a. Following [30] the first terms of Ω are 1 1 1 Ω(a)= a − a · a + (a · a) · a + a · (a · a)+ ... 2 4 12 where the sum stops at place k. In [30] the formula for Ω is given using Bernoulli numbers. This assures that p does not appear in a denominator.

9 4. Define a ◦ b = a + eLΩ(a) (b). Here the addition is the same as in the pre-Lie algebra A. It was shown in [1] that (A, ◦) is a group. The same argument will work in our case, as (W (a) ◦ W (b)) ◦ W (c) = W (a) ◦ (W (b) ◦ W (c)) for a, b, c ∈ A by BCH formula (at this stage the result is related to Lazard’s correspondence). We can immediately see that (A, +, ◦) is a left brace because

a ◦ (b + c)+ a = a + eLΩ(a) (b + c)+ a =(a + eLΩ(a) (b))+(a + eLΩ(a) (c)) = a ◦ b + a ◦ c.

4 The correspondence is one-to-one

In this chapter we show that the correspondence between strongly nilpotent Fp-braces of nilpotency index k and nilpotent pre-Lie algebras over Fp of nilpotency index k is one- to-one for p> 2k, where p is a prime number. We begin with the following proposition.

Proposition 6. Let (A, +, ·) be a nilpotent pre-Lie algebra over a field Fp of nilpotency index k, where 2k < p. Let (A, +, ◦) be the brace obtained as in Section 3, so (A, ◦) is the formal group of flows of the pre-Lie algebra A. Then A has the nilpotency index k. Moreover, the following holds,

p−2 1 a · b = − ((2ia) ∗ b), X 2i i=0 i i −i 1 for a, b ∈ A, where 2 a denotes the sum of 2 copies of element a, and 2 = 2i denotes i the inverse of 2 in Fp. Proof. Notice that, by using formulas from Section 3, we see that since (A, ◦) is the group of flows of the pre-Lie algebra A, then

a ∗ b = a · b + X αww w∈Pa,b

where αw ∈ Fp and Pa,b is the set of all products of elements a and b from (A, ·) with b appearing only at the end, and a appearing at least two times in each product. Moreover, αw does not depend on a and b, but only on their arrangement in word w as an element of set Pa,b. This follows from the construction of Ω(a), which is a sum of a and a linear combination of all possible products of more than one element a with any distribution of brackets, which can be proved by induction. Notice that each word w will be a product of at most k elements because pre-Lie algebra A has nilpotency index k. Let w ∈ Pa,b, then w is a product of some elements a and element b. We define the word w2i to be the word obtained if at each place where a appears in w we write 2ia instead of a. It follows that: i i (2 a) ∗ b = (2 a) · b + X αww2i . w∈Pa,b Consequently, p−2 p−2 1 i 1 i ((2 a) ∗ b)= [(2 a) · b + α w i ]. X 2i X 2i X w 2 i=0 i=0 w∈Pa,b

10 Notice that p−2 1 (2ia) · b =(p − 1)a · b = −a · b. X 2i i=0

Therefore, it suffices to show that for every w ∈ Pa,b we have

p−2 1 w i =0. X 2i 2 i=0

i j We know that every pre-Lie algebra is distributive, hence w2i = (2 ) w where j is the number of occurences of a in the product which gives w. It suffices to show that p−2 1 ij F F p−2 1 ji Pi=0 2i (2 ) = 0 in p. Because 2 ≤ j

Theorem 7. Let p be a prime number and Fp be the field consisting of p elements. Let (A, +, ◦) be a strongly nilpotent brace of nilpotency index k where 2k

p−2 1 a · b = − (2ia) ∗ b. X 2i i=0

Then (A, ◦) is the group of flows of the pre-Lie algebra A and (A, +, ◦) can be obtained as in Section 3 from pre-Lie algebra (A, +, ·).

Proof. Let Ea,b ⊆ A denote the set consisting of any product of elements a and one element b at the end of each product under the operation ∗, in any order, with any distribution of brackets, each product consisting of at least 2 elements a. Observe that by Lemma 2 applied several times

a · b = a ∗ b + X αww w∈Ea,b where αw ∈ Fp do not depend on a, b, but only on their arrangement in word w as an element of Ea,b. Moreover, each w is a product of at least 3 elements from the set {a, b}. Observe that coefficients αw do not depend on the brace A as they were constructed using the formula from Lemma 2 which holds in every strongly nilpotent brace. Therefore, for any given n, the same formula will hold for all braces of nilpotency index not exceeding n. Therefore a ∗ b = a · b − αww, and now we can use this formula several times Pw∈Ea,b to write every element from Ea,b as a product of elements a and b under the operation ·. In this way we can recover the brace (A, +, ◦) from the pre-Lie algebra (A, +, ·).

11 Therefore two distinct strongly nilpotent braces cannot give the same pre-Lie algebra p−2 1 i using the formula a · b = − Pi=0 2i (2 a) ∗ b. Notice that, because we know that pre-Lie algebra (A, +, ·) can be obtained as in Theorem 4 from the brace which is it’s group of flows (by Theorem 6), it follows that (A, ◦) is the group of flows of pre-Lie algebra A. Therefore we obtain the following corollary.

Corollary 8. [15] Let k be a natural number, and p be a prime number such that p> 2k. Then there is one-to-one correspondence between the set of strongly nilpotent Fp-braces of nilpotency index k and the set of nilpotent pre-Lie algebras over Fp of nilpotency index k.

Proof. For every pre-Lie algebra of nilpotency index k we can attach the brace which is its group of flows and form a pair. Since the group of flows is uniquely defined, every pre-Lie algebra will be in exactly one pair. Moreover, every brace will be in some pair, by Theorem 7. Observe that every brace will be in exactly one pair, as otherwise there would be two distinct pre-Lie algebras which give the same group of flows. However, by Proposition 6 we can apply the formula

p−2 1 a · b = − ((2ia) ∗ b), X 2i i=0 to recover these pre-Lie algebras from this brace. Because the formula defines uniquely the underlying pre-Lie algebra every brace is in at most one pair.

Remark regarding connections with the BCH formula and with Lazard’s correspon- dence: Let (A, +, ·) be a finite nilpotent pre-Lie algebra, and (A, ◦) be its group of flows. Notice that the formula for the multiplication in the group of flows can also be written using the Baker-Campbell-Hausdorff formula and Lazard’s correspondence:

W (a) ◦ W (b)= W (C(a, b)), where C(a, b) is obtained using the Baker-Campbell-Hausdorff series in the Lie algebra L(A) [1], [30]. Therefore there is a group isomorphism between the group G(L(A)) obtained from L(A) by using the Baker-Campbell-Hausdorff (BCH) formula and the group of flows (A, ◦) of a pre-Lie algebra A given by the map p : a → W (a). It follows from the formula W (a)◦W (b)= W (C(a, b)). Recall that the Lie algebra L(A) is obtained from a pre-Lie algebra A by taking [a, b]= a · b − b · a, and has the same addition as A.

• Applying the inverse of the BCH formula to the group G(L(a)) gives the map G(L(A)) → L(A). Notice that when we apply the inverse of the BCH formula to the group of flows (A, ◦) we are applying the inverse of the BCH formula to the group isomorphic to G(L(A)). This gives the Lie algebra L(A, ◦) which is isomorphic to L(A), because the addition and multiplication constructed using the inverse of the BCH formula only depends on the group multiplication.

12 • Therefore the formula p−2 1 a · b = − (2ia) ∗ b X 2i i=0 from Theorem 7 applied to the multiplicative group of some strongly nilpotent brace A would give a pre-Lie algebra A whose Lie algebra is isomorphic to the Lie algebra obtained from the inverse of the BCH formula applied to the multiplicative group (A, ◦) of brace A (because the formula from Theorem 7 reverses the formula which gives the group of flows).

• Therefore, for multiplicative groups of strongly nilpotent braces, this may be useful for calculations related to the inverse of the BCH formula.

5 Applications: braces of cardinality p4

Let A be a brace with the usual operations +, ◦, ∗ where a ◦ b = a ∗ b + a + b. Recall that [1] [i+1] i [j] [i+1−j] [i+1] [i] [2] A = A and A = Pj=1 A ∗ A . Notice that A ⊆ A since A ⊆ A. We begin with the following lemma. Lemma 9. Let A be a strongly nilpotent brace of cardinality p4 for some prime number p. Let k be a natural number and suppose that A[k] =0, then A[6] =0. Proof. Consider sets A, A[2], A[3], A[4]. Then we have the following two cases: • A =6 A[2] =6 A[3] =6 A[4]. Then A[i]/A[i+1] have cardinality p for i = 1, 2, 3, 4, since the cardinality A is p4 so it could not be bigger (it could not be smaller, since every subgroup of the additive group is a p-group). In this case we see that the cardinality of A2 is p3 and the cardinality of A[3] is p2 and the cardinality of A[4] is p. Suppose that A[6] =6 0 then A[4] = A[6] and A[5] ⊆ A[6]. It follows that A[6] = A[5] ∗ A + A[4] ∗ A[3] + A[3] ∗ A[4] + A ∗ A[5] ⊆ A[7]. Continuing in this way we get that A[4] ⊆ A[6] = 0, a contradiction.

• A[i] = A[i+1] for some i ∈ {1, 2, 3}. Notice that A ⊆ A2 implies A[j] ⊆ A[j+1] for every j, consequently A = A[k] = 0, a contradiction. Similarly A[2] ⊆ A[3] implies A[j] ⊆ A[j+1] for every j > 1, consequently A[2] = A[k] = 0, hence A[6] = 0 as required. Consequently we only need to consider the case when A[3] = A[4]. Let x,y,z,t ∈ A. Observe that since A is a strongly nilpotent brace we can apply Lemma 2 for a = x∗y, b = z,c = t and then for a = z, b = x∗y,c = t and subtracting we get that ((x∗y)∗z)∗t−(x∗y)∗(z ∗t)−(z ∗(x∗y))∗t+z ∗((x∗y)∗t) ∈ A[5] (this also follows from the fact that associated graded structures of braces are pre-Lie algebras [23]). Therefore (x ∗ y) ∗ (z ∗ t) ⊆ A[3] ∗ A + A ∗ A[3] ⊆ A[5]. Therefore we get that A[3] ⊆ A[4] implies A[4] ⊆ A[5]. Now we can use this to show by induction that A[j] ⊆ A[j+1] for all j > 3, consequently A[3] ⊆ A[4] ⊆ ... ⊆ A[k] = 0, therefore A[5] = 0, as required.

i Let A be a pre-Lie algebra, and by A we denote the linear space over Fp spanned by all 1 i+1 i j i+1−j products of i or more elements from A. Observe that A = A and A = Pj=1 A ·A . Our next result is as follows.

13 6 Lemma 10. Let p be a prime number larger than 2 and let Fp be the field consisting 4 of p elements. Let (A, +, ◦) be a right nilpotent Fp-brace of cardinality p . Then (A, ◦) is the group of flows of some nilpotent pre-Lie algebra (A, +, ·). Moreover the product of any 6 elements in this pre-Lie algebra A is zero. Proof. By a result by Rump, every brace of order pn is left nilpotent [32]. By a result from [35] a brace which is right nilpotent and left nilpotent is strongly nilpotent, therefore our brace A is strongly nilpotent. By Lemma 9 A has nilpotency index 6 or less. The result now follows from Theorem 7. Lemma 11. Let assumptions and notation be as in Lemma 10. Suppose that the A is generated as a brace by one element x. Then (A, ◦) is the group of flows of a nilpotent pre-Lie algebra generated by one element x. Proof. Let A be the pre-Lie algebra obtained as in Proposition 7 from brace A. Then by Theorem 7 (A, ◦) is the group of flows of the pre-Lie algebra (A, +, ·). By the definition of the group of flows elements x, x∗x, x∗(x∗x), (x∗x)∗x, (x∗x)∗(x∗x), x∗(x∗(x∗x)), x∗((x∗ x)∗x), ((x∗x)∗x)∗x, (x∗(x∗x))∗x can be presented as linear combination of products x, x·x, x·(x·x), (x·x)·x, (x·x)·(x·x), x·(x·(x·x)), x·((x·x)·x), ((x·x)·x)·x, (x·(x·x))·x. Since A[6] = 0 by Lemma 10 these elements span brace (A, +, ◦) and hence span pre- Lie algebra A as a linear space over Fp. So the pre-Lie algebra (A, +, ·) is generated as pre-Lie algebra by element x and A6 = 0.

Lemma 12. Let (A, +, ·) be a pre-Lie algebra over Fp generated as a pre-Lie algebra by one element x. Suppose that A has cardinality p4 and A6 = 0 and A4 =6 0. Then A3 · A2 =0, A4 · A =0, A · A4 =0. Moreover, x2 · (x · x2)=0. Proof. Reasoning as in Lemma 9 we get that if A5 =6 0 then A4 = A5, and A6 = 0. Observe that we have the following pre-Lie algebra relations: Let a ∈ A3, then

(x · a) · x − x · (a · x)=(a · x) · x − a · x2.

Notice that A4 = A5 implies A·A4 ⊆ A·A5 ⊆ A6 = 0, similarly A4 ·A ⊆ A5 ·A ⊆ A6 = 0. This along with the above relation imply A3 ·A2 = 0. We also have the following relation:

(x · x2) · x2 − x · (x2 · x2)=(x2 · x) · x2 − x2 · (x · x2),

which implies x2 · (x · x2) ∈ A3 · A2 + A · A4 ⊆ A6 =0.

Let (A, +, ·) be a pre-Lie algebra over Fp. For α ∈ Fp and a ∈ A, by αa we will denote the sum of α elements a, and we denote a2 = a · a.

Proposition 13. Let A be a pre-Lie algebra over Fp generated as a pre-Lie algebra by one element x. Suppose that A has cardinality p4 and A5 =6 0 and A6 = 0. Then the following holds: • Elements x, x2, x2 · x, x2 · (x2 · x) form a base of the pre-Lie A as a vector space over Fp.

14 • All products of 5 or more elements x are zero, except of the element x2 · (x2 · x) =06 .

• The following relations hold,

x · (x · x2)=0, (x · x2) · x =0,

x2 · x2 = x · (x2 · x) − (x2 · x) · x,

• For some α,β,γ ∈ Fp the following relations hold:

(x2 · x) · x = β(x · (x · x2)), x · (x2 · x)= γ(x · (x · x2)),

x · x2 = α(x2 · (x2 · x)).

Therefore, every element from A4 equals element x2 · (x2 · x) multiplied by some element from Fp. Notice that the above relations give a well defined pre-Lie algebra.

Proof. By Lemma 12 the only possible non-zero product of five or more elements x is x2 ·(x2 ·x). Reasoning as in Lemma 9 we obtain that A3/A4 has dimension 1 and A4 = A5 4 5 2 2 also has dimension 1 as a vector space over field Fp. Therefore, A = A = Fp(x ·(x ·x)). 2 2 2 2 Therefore, (x · x) · x = β(x · (x · x )), x · (x · x)= γ(x · (x · x )), for some α, β ∈ Fp. Notice that A2 is not a subset of A3 = A2 · A + A · A2, since we could then substitute A3 instead of A2 on the right hand side several times and obtain that A2 ⊆ A3 ⊆ A4 ⊆ A5 ⊆ A6 = 0. Similarly A is not a subset of A2. 3 4 2 2 4 Because A /A has dimension 1, then either x · x − αx · x ∈ A for some α ∈ Fp or x · x2 ∈ A4. If x2 · x − αx · x2 ∈ A4 then x2 · (x2 · x) − αx2 · (x · x2) ∈ A6, and since x2 · (x · x2) = 0 by Lemma 12 then we would get A2 · A3 = 0. This and Lemma 12 would imply A5 = 0. So, since A5 =6 0 then x · x2 ∈ A4 = A5. Notice that x · (x · x2) ∈ A6 =0 and (x · x2) · x ∈ A6 = 0 since x · x2 ∈ A5. To obtain relation (x2 · x) · x + x2 · x2 = x · (x2 · x) we can use the pre Lie algebra relation (x2 · x) · x − (x · x2) · x = x2 · x2 − x · (x2 · x) and notice that x · x2 ∈ A4 = A5 and so (x · x2) · x ∈ A6 =0. This implies the relations from our proposition. To see that the relations assumed in our theorem give a well defined pre-Lie algebra observe that every element from Ai can be writen as sums of elements from the base x, x2, x2 · x, x2 · (x2 · x) which are also from Ai so the degree will stay the same or increase (by the degree we mean the number of occurence of x in any product). We will check that the algebra is well defined by using the multiplication table, by considering all products (a · b) · c and a · (b · c) of elements from our base, and use the multiplication table to substitute sums of elements from the base for each product a · b and b · c, and then use this to calculate (a · b) · c and a · (b · c). Because any product of 6 or more elements x will be zero, and by substituting elements from our base we cannot decrease the degree (the number of appearance of x in each product), then we need to only consider products a · (b · c) and (a · b) · c where a, b, c are elements from our base, and x appears at most 5 times in each product a · (b · c). Therefore it is easy to check with the multiplication table that all of the pre-Lie algebra relations (a · b) · c − a · (b · c)=(b · a) · c − b · (a · c),

15 are satisfied as we only need to consider the case when a = x, b = x · x and c ∈{x, x · x} and the case when a = x2 · x, b = x, c = x. Therefore, every structure of this type is a well defined pre-Lie algebra.

Proposition 14. Let A be a pre-Lie algebra over Fp generated as a pre-Lie algebra by one element x. Suppose that A has cardinality p4 and A4 =6 0 and A5 = 0. Then the following holds:

2 • Elements x, x , a, b form a base of the pre-Lie A as a vector space over Fp for some a ∈ {x · x2, x2 · x} and b ∈ {x · a, a · x} where b ∈ A4,b∈ / A3. Moreover, every element from Ai will be a sum of elements from this basis which belong to Ai for each i. • All products of 5 or more elements from A are zero. • All products in A or four or more elements from A span a one dimensional vector space over Fp spanned by element b. • The following relation holds:

(x2 · x) · x − (x · x2) · x = x2 · x2 − x · (x2 · x).

• For some α, β ∈ Fp, not both zero, we have

α(x · x2)+ β(x2 · x) ∈ A4,

and consequently the following relations hold in A4:

αx · (x · x2)+ βx · (x2 · x)=0

α(x · x2) · x + β(x2 · x) · x =0

Notice that the above relations give a well defined pre-Lie algebra. Proof. We can use a similar proof as in Proposition 13. We can take any non-zero product of some copies of element x from A4 to be element b. Notice that either b = a·x or b = x·a for some a ∈ A3, and we can add this a to the base. Notice that A3/A4 has dimension 1 as a vector space over Fp, which gives α, β. The fact that this pre-Lie algebra is well defined follows from the fact that every element from Ai will be a sum of elements from thie basis which belong to Ai for each i. Therefore, by substituting elements from basis for a · b in a product (a · b) · c we cannot decrease the degree of this product. Therefore, reasoning similarly as in Proposition 13 we only need to check that the pre-Lie algebra relations

(a · b) · c − a · (b · c)=(b · a) · c − b · (a · c),

are satisfied for a = x, b = x2,c = x. Notice that this relation holds by assumptions.

4 Proposition 15. Let A be a pre-Lie algebra over field Fp such that A = 0 and A is generated by element x ∈ A as a pre-Lie algebra. Suppose that A has cardinality p4. 2 2 2 Then elements x, x , x · x, x · x form a base of A as a vector space over Fp, moreover all products of more than 3 elements from A are zero and there are no other relations in this pre-Lie algebra A. Notice that this gives a well defined pre-Lie algebra.

16 Proof. Notice that elements x and x2 appear in every product of monomials so x and x2 cannot be sums of products of more than 2 elements x, as by substituting such relations in every product of the right hand side several times we would get x, x2 ∈ A[6] = 0. Therefore, if elements x · x2 and x2x are linearly independent then x, x2, x · x2, x2 · x will be a basis of A as a linear space over Fp. Notice that we cannot have any relation involving x · x2 and x2 · x, because then our pre-Lie algebra would have dimension less than 3 over Fp. By constructing a multiplication table, as in the proof of Proposition 13, we see that our pre-Lie algebra is well defined. The results in this section yield the following corollary:

6 Corollary 16. Let p be a prime number larger than 2 and let Fp be the field consisting 4 of p elements. Let (A, +, ◦) be a right nilpotent Fp-brace of cardinality p . Then A is the group of flows of one of pre-Lie algebras from Propositions 13, 14 and 15.

6 Left nilpotent radical

In associative algebra sums of nilpotent ideals are nilpotent ideals. It was shown in [13] that a sum of a finite number of left nilpotent ideals in a pre-Lie algebra is a left nilpotent ideal. In this section we obtain an analogon of this result for braces. Recall that an ideal I in a brace (A, +, ◦) is left nilpotent if In = 0 for some n where I1 = I and Ii+1 = I ∗ Ii. We will first prove two supporting lemmas: Lemma 17. Let (A, +, ◦) be a left brace and let I,J be left nilpotent ideals in A. Denote I + J = {a + b : a ∈ A, b ∈ B}. Then for every c ∈ A we have

(I + J) ∗ c ⊆ I ∗ c + J ∗ c + I ∗ (J ∗ c).

Proof. Let λx(y)= x ∗ y + y, then it is known that λx◦z(y)= λx(λz(y)) for all x, y, z ∈ A. Let a−1 be the inverse of a in the group (A, ◦). Observe that for any c ∈ A we have

(I + J) ∗ c ⊆ I ∗ c + J ∗ c + I ∗ (J ∗ c).

′ ′ Indeed, let a ∈ I, b ∈ J then, (a + b) ∗ c = (a + λa(λa−1 (b)) ∗ c = (a + b + a ∗ b ) ∗ c = ′ ′ ′ a ∗ c + b ∗ c + a ∗ (b ∗ c) where b = λa−1 (b) ∈ J since b ∈ J and J is an ideal in A. Lemma 18. Let (A, +, ◦) be a left brace and let I,J be left nilpotent ideals in A. Denote, I + J = {a + b : a ∈ A, b ∈ B}. Then for every c ∈ A we have I ∗ (J ∗ (c)) ⊆ I ∗ c + I ∗ (−c)+I ∗(I ∗c)+J ∗(I ∗c)+I ∗(J ∗(I ∗c))+I ∗(I ∗(−c))+J ∗(I ∗(−c))+I ∗(J ∗(I ∗(−c))). Proof. It suffices to show that

I ∗ (J ∗ (c)) ⊆ J ∗ (I ∗ c)+ I ∗ c + I ∗ (−c)+(I + J) ∗ (I ∗ c)+(I + J) ∗ (I ∗ (−c))

and then apply Lemma 17 to get:

(I + J) ∗ (I ∗ c) ⊆ I ∗ (I ∗ c)+ J ∗ (I ∗ c)+ I ∗ (J ∗ (I ∗ c)).

It remains to show that

I ∗ (J ∗ (c)) ⊆ J ∗ (I ∗ c)+(I + J) ∗ (I ∗ c)+ I ∗ c +(I + J) ∗ (I ∗ (−c)) + I ∗ (−c).

17 Let a ∈ I, b ∈ J, observe that (a + b + a ∗ b) ∗ c = a ∗ c + b ∗ c + a ∗ (b ∗ c)

′ ′ ′ (a + b + a ∗ b) ∗ c = ((a + b)+ λa+b(a )) ∗ c =(a + b) ∗ c + a ∗ c +(a + b) ∗ (a ∗ c), ′ where a = λ(a+b)−1 (a ∗ b) ∈ I ∩ J, since a ∗ b ∈ I ∩ J. Therefore, (a + b) ∗ c − a ∗ c − b ∗ c = a ∗ (b ∗ c) − a′ ∗ c − (a + b) ∗ (a′ ∗ c). By applying it to a := b and b := a we get (b + a) ∗ c − b ∗ c − a ∗ c = b ∗ (a ∗ c) − a′′ ∗ c − (b + a) ∗ (a′′ ∗ c), for some a′′ ∈ J ∩ I. By comparing the above equations we obtain a ∗ (b ∗ c)= b ∗ (a ∗ c)+ a′ ∗ c +(a + b) ∗ (a′ ∗ c) − a′′ ∗ c − (b + a) ∗ (a′′ ∗ c). Since a ∗ (−b)= −(a ∗ b) in every brace and a, b, c were arbitrary this implies I ∗ (J ∗ (c)) ⊆ J ∗ (I ∗ c)+ I ∗ c + I ∗ (−c)+(I + J) ∗ (I ∗ c)+(I + J) ∗ (I ∗ (−c)).

Theorem 19. Let (A, +, ◦) be a left brace and let I,J be left nilpotent ideals in A. Then I + J = {a + b : a ∈ A, b ∈ B} is a left nilpotent ideal in A.

Proof. Notice first that I + J is an ideal in A by Lemma 3.7 [28]. Denote En(I,J) to be the sum of elements which belong to some sets Pn ∗ (Pn−1 ∗ (···∗ (P1 ∗ c))) and some sets Pn ∗ (Pn−1 ∗ (···∗ (P1 ∗ (−c)))) where each Pi ∈ {I,J}. By Lemma 17 we get that En(I + J, I + J) ⊆ Pn≤k≤2n Ek(I,J) for each n (it can be proved by induction on n). To show that I + J is nilpotent it suffices to show that for a sufficiently large k, all α α Ek(I,J)=0. Let α be such that I , J = 0. Let Tk,i,n consist of elements from sets Pn ∗ (Pn−1 ∗ (···∗ (P1 ∗ c))) and sets Pn ∗ (Pn−1 ∗ (···∗ (P1 ∗ (−c)))) for which P1 = ... = Pk = I and Pk+1 = ... = Pk+i = J and Pk+i+1 = I. Notice that for k = 0 we have P1 = J. Observe that Tk,α+i,n = 0 for every i, k, n ≥ 1 because Pk+α+1 ∗(Pk+α ∗(...∗(P1 ∗c))) = α Pk+α+1 ∗(Pk+α ∗(...∗(Pk+2 ∗D))) = 0 for D = Pk+1 ∗(Pk ∗(...∗(P1 ∗c))) ∈ J since J =0 (and Pk+α+1 = Pk+α = ... = Pk+1 = J). Similarly Tα+k,i,n = 0 for every i, n, k ≥ 1. We will use a similar argument as in [13]. Suppose that i> 1. Notice that Lemma 18 applied to ideals Pk+i+1 = I and Pk+i = J and to C = Pk+i−1 ∗ (Pk+i−2 ∗ (···∗ (P1 ∗ c))) yields Pk+i+1 ∗ (Pk+i ∗ C)= I ∗ (J ∗ C) ⊆ I ∗ C + I ∗ (−C)+ I ∗ (I ∗ C)+ J ∗ (I ∗ C)+ I ∗ (J ∗ (I ∗ C)) + I ∗ (I ∗ (−C)) + J ∗ (I ∗ (−C)) + I ∗ (J ∗ (I ∗ (−C))) ⊆ Tk,i−1, provided that i> 1. Therefore Tk,i,n ⊆ Tk,i−1,n−1 + Tk,i−1,n + Tk,i−1,n+1 for i> 1. If i = 1 then we obtain Tk,1,n ⊆ Pj≤α,m∈{n−1,n,n+1} Tk+1,j,m + Tk+2,j,m + ··· . Applying it several times (at 2 most (α + 1) times) we can obtain k>α, so we eventually obtain zero, since Tk,i,n =0 for k>α, n> 2(α + 1)2. Notice that every left nilpotent ideal is a solvable ideal, and that a sum of a finite number of left nilpotent ideals in a brace is a left nilpotent ideal. Therefore every finite brace contains the largest nilpotent ideal, and this ideal is also contained in the largest solvable ideal of this brace (it is known that a sum of two solvable ideals in a brace is a solvable ideal [28]). This mirrors the situation for pre-Lie algebras from [13]. We obtain the following.

18 Corollary 20. If A is a finite brace, then A contains the largest left nilpotent ideal, which is the sum of all left nilpotent ideals in A. We will call this ideal the left nilpotent radical of A.

In [28], the Wedderburn radical of a brace was defined as a sum of all ideals in A which are both left nilpotent and right nilpotent. In Lemma 6.4 [28] it was shown that the Wedderburn radical in any brace A is solvable. We get the following result.

Corollary 21. If A is a finite brace, then the Wedderburn radical of A is left nilpotent.

This suggests the following (open) questions.

Question 2. Let A be a finite brace. Is the Wedderburn radical of A strongly nilpotent?

Question 3. Let A be a left brace, and I,J be two strongly nilpotent ideals in A. Is I +J a strongly nilpotent ideal in R?

Question 4. Let A be a left brace, and I,J be two right nilpotent ideals in A. Is I + J a right nilpotent ideal in R?

The above questions have some similarity to the Koethe conjecture in ring theory, which states that a sum of two nil right ideals in a ring is nil.

Acknowledgments. The author is grateful to Michael West and Alicja Smoktunow- icz for their help in preparing the Introduction. This research was supported by EPSRC grant EP/R034826/1.

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