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In mathematics, a skew left brace ${\displaystyle B}$ is a set with two operations, ${\displaystyle \cdot }$  and ${\displaystyle \circ }$  so that ${\displaystyle B}$ with the operation ${\displaystyle \cdot }$  is a group, often called the additive group, ${\displaystyle B}$ with the operation ${\displaystyle \circ }$  is a group, often called the circle group, and a certain compatibility condition, analogous to distributivity, connects the two group operations.    First defined in 2017, skew left braces yield set-theoretic solutions of the Yang-Baxter equation (See Yang-Baxter equation) and also have a close connection with Hopf-Galois structures on classical Galois extensions of fields.

A skew left brace is a set ${\displaystyle 'B}$  with two operations, ${\displaystyle \cdot }$  and ${\displaystyle \circ }$  , somewhat analogous to a ring (see Ring (mathematics)).    The set ${\displaystyle B}$  with the operation ${\displaystyle \cdot }$ , denoted    ${\displaystyle (B,\cdot )}$ ,    is a group with identity element 1, often called the \emph{additive group}, ${\displaystyle B}$  with the operation ${\displaystyle \circ }$ , denoted ${\displaystyle (B,\circ )}$ ,    is a group, often called the \emph{circle group}, and the compatibility condition $${\displaystyle a\circ (b\cdot c)=(a\circ b)\cdot a^{-1}\cdot (a\circ c)}$$  holds for all ${\displaystyle a}$ , ${\displaystyle b}$ , ${\displaystyle c}$ ' in ${\displaystyle B}$ .    Here ${\displaystyle a^{-1}}$  denotes the inverse of ${\displaystyle a}$  in the group ${\displaystyle (B,\cdot )}$  , and ${\displaystyle {\overline {a}}}$  will denote the inverse of ${\displaystyle a}$  in the group ${\displaystyle (B,\circ )}$ .   

The identity element ${\displaystyle I}$  of the group ${\displaystyle (B,\circ )}$  is equal to the identity element 1 of ${\displaystyle (B,\cdot )}$ :    for setting ${\displaystyle a=I,c=1}$  in the compatibility condition shows that ${\displaystyle b=b\cdot I^{-1}}$  for all ${\displaystyle b}$  in ${\displaystyle B}$ , hence by uniqueness of identity in ${\displaystyle (B,\cdot )}$  ,    ${\displaystyle I^{-1}=1}$ , hence ${\displaystyle I=1^{-1}=1}$ .    A skew left brace ${\displaystyle B}$  whose additive group ${\displaystyle (B,\cdot )}$  is abelian: ${\displaystyle a\cdot b=b\cdot a}$  for all ${\displaystyle a,b}$  in ${\displaystyle B}$ , is called a left brace. 

Examples. The most trivial example of a skew brace is a group ${\displaystyle G=(G,\cdot )}$  made into a skew brace ${\displaystyle (G,\cdot ,\circ )}$  where ${\displaystyle \cdot =\cdot =\circ }$  (see Group (mathematics)).    Less trivial: if ${\displaystyle N}$  is a non-abelian group, then defining ${\displaystyle \circ }$  on ${\displaystyle N}$  by ${\displaystyle a\circ b=b\cdot a}$  makes ${\displaystyle N}$  into a skew brace.

Radical ring.    A ring ${\displaystyle A}$  without unity is a set with two operations, ${\displaystyle +}$  (addition) and ${\displaystyle \cdot }$  (multiplication), where ${\displaystyle a\cdot b}$  is typically written ${\displaystyle ab}$ , and ${\displaystyle a\cdot a\cdot \ldots \cdot a}$  (${\displaystyle n}$  factors) is denoted ${\displaystyle a^{n}}$ . With those operations ${\displaystyle A}$  satisfies all of the properties of a ring (associativity of multiplication, left and right distributivity of multiplication over addition, etc.) except that ${\displaystyle A}$  has no multiplicative identity element.

• Given any ring ${\displaystyle A}$ , the circle operation ${\displaystyle \circ }$  on ${\displaystyle A}$  is defined by ${\displaystyle a\circ b=a+b+a\cdot b}$ .    It is easy to check that the operation ${\displaystyle \circ }$  is associative, and ${\displaystyle a\circ 0=0\circ a=a}$ , so ${\displaystyle (A,\circ )}$ , the set ${\displaystyle A}$  with the circle operation ${\displaystyle \circ }$ , is a monoid with identity element equal to the additive identity element 0 of the ring ${\displaystyle A}$ : (see monoid).   

The ring ${\displaystyle A}$  is a radical ring if and only if ${\displaystyle (A,\circ )}$  is a group:    that is, given any ${\displaystyle a}$  in ${\displaystyle A}$ , there is some ${\displaystyle bin<math>A}$  so that ${\displaystyle a\circ b=b\circ a=0}$ : see Jacobson radical, where such an element is called right quasi-regular and left quasi-regular. Thus ${\displaystyle A}$  is a radical ring if and only if ${\displaystyle A}$  is equal to its Jacobson radical ${\displaystyle J(A)}$ .    The group ${\displaystyle (A,\circ )}$  is sometimes called the \emph{adjoint group} of ${\displaystyle A}$ . See Radical ring.

A radical ring is a skew left brace. If ${\displaystyle A}$  is a radical ring, then with the operations ${\displaystyle +}$  and ${\displaystyle \circ }$ , ${\displaystyle A}$  is a left brace.    The defining property:    for all ${\displaystyle a,b,c}$  in ${\displaystyle A}$ , ${\displaystyle a\circ (b+c)}$  = ${\displaystyle (a\circ b)-a+(a\circ c)}$  becomes, after replacing ${\displaystyle x\circ y}$  by ${\displaystyle x+y+xy}$  for all ${\displaystyle x,y}$  in ${\displaystyle A}$ , $${\displaystyle a+b+c+a(b+c)}$$  = $${\displaystyle a+b+ab-a+a+c+ac,}$$  and the two sides are equal if and only if the left distributive law ${\displaystyle a(b+c)=ab+ac}$  holds.

${\displaystyle A}$  is also a right brace, because the corresponding defining property for a right brace is equivalent to right distributivity.      Conversely, suppose ${\displaystyle B}$  is a set with two operations ${\displaystyle +}$  and ${\displaystyle \circ }$      which make ${\displaystyle B}$  into both a left brace and a right brace. Define ${\displaystyle \cdot }$  on ${\displaystyle B}$  by    ${\displaystyle x\cdot y=x\circ y-x-y}$ . Then the set ${\displaystyle B}$  with operations ${\displaystyle +}$  and ${\displaystyle \cdot }$  satisfies both the left and right distributive laws, and so is a ring ${\displaystyle (B,+,\cdot )}$ ; and the circle operation on ${\displaystyle B}$  defined from the multiplication ${\displaystyle \cdot }$  is the original circle operation, hence ${\displaystyle (B,\circ )}$  is a group, so ${\displaystyle B}$  is a radical ring.    Therefore the set of radical rings is a subset of the set of left braces.    For some counts of radical rings with n elements for various numbers n, see Radical rings.

Some history A set-theoretic solution of the Yang-Baxter equation (see Yang-Baxter equation) is a pair (${\displaystyle X}$ , ${\displaystyle r}$ ) where ${\displaystyle X}$  is a set and ${\displaystyle r:X\times X\to X\times X}$  is a bijective map such that $${\displaystyle (r\times id)(id\times r)(r\times id)=(id\times r)(r\times id)(id\times r):X\times X\times X\to X\times X\times X}$$ . The question of finding set-theoretic solutions of the Yang-Baxter equation was first raised by V. G. Drinfel'd in 1990 [Dr92].    W. Rump [Ru07] defined a left brace as a generalization of a radical ring with the property that a left brace yields a set-theoretic solution of the Y-B equation. Skew left braces (where the “additive” group need not be abelian) were first defined in 2017 by L. Guarneri and L. Vendramin [GV17], who showed that every skew left brace yields a solution of the Y-B equation, and conversely, that every non-degenerate solution of the Yang-Baxter equation corresponds to a unique skew left brace.

Skew left braces and the Yang-Baxter equation

Given a skew brace ${\displaystyle A}$ , define ${\displaystyle \sigma _{a}}$ :   ${\displaystyle A\to A}$  by ${\displaystyle \sigma _{a}(b)=a^{-1}(a\circ b)}$ . Then ${\displaystyle a\circ b=a\sigma _{a}(b)}$  . Define  ${\displaystyle R}$  : ${\displaystyle A}$  × ${\displaystyle A}$  → ${\displaystyle A}$  × ${\displaystyle A}$  by ${\displaystyle R(a,b)}$  =  ${\displaystyle (\sigma _{a}(b),\tau _{b}(a))}$  =  ${\displaystyle (\sigma _{a}(b),{\overline {\sigma _{a}(b)}}\circ a\circ b)}$  where ${\displaystyle \tau _{b}(a)}$  =  ${\displaystyle {\overline {\sigma _{a}(b)}}\circ a\circ b}$ .    Then for all ${\displaystyle a}$ , ${\displaystyle b}$  in ${\displaystyle A}$ , ${\displaystyle \sigma _{a}}$      and ${\displaystyle \tau _{b}}$  are one-to-one maps from ${\displaystyle A}$  to ${\displaystyle A}$ , and we have:

([GV], Theorem 3.1)    If ${\displaystyle A}$  is a skew left brace, then ${\displaystyle A}$  yields a solution    ${\displaystyle R(a,b)}$  of the Yang-Baxter equation:    for all ${\displaystyle a,b,c}$  in ${\displaystyle A}$ , $${\displaystyle (R\times id)(id\times R)(R\times id)(a,b,c)=(id\times R)(R\times id)(id\times R)(a,b,c)}$$ .

Since ${\displaystyle \sigma _{a}}$  and ${\displaystyle \tau _{b}}$  are one-to-one maps from ${\displaystyle A}$  to ${\displaystyle A}$  for all ${\displaystyle a}$ , ${\displaystyle b}$  in ${\displaystyle A}$ , the solution ${\displaystyle R}$  of the Yang-Baxter equation is called nondegenerate.

Here is a proof of this result, adapted from [LYZ00] to the skew brace setting.

There are three equalities to show:

(L:) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \sigma_{\sigma_x(y)}(\sigma_{\tau_y(x)}(z))= \sigma_x(\sigma_y(z)), }

(C:) Failed to parse (syntax error): {\displaystyle \tau_{\sigma_{\tau_y(x)}(z)}(\sigma_x(y)) =    \sigma_{\tau_{\sigma_y(z)}(x)}(\tau_z(y)) }

and

(R:) Failed to parse (syntax error): {\displaystyle \tau_z(\tau_y(x))) = \tau_{\tau_z(y)}(\tau_{\sigma_y(z)}(x)) )} .

Here is how it goes:

Given a skew brace ${\displaystyle B(\circ ,\cdot )}$  (${\displaystyle \cdot }$  is usually omitted), the maps ${\displaystyle \sigma _{x}(y)=x^{-1}(x\circ y)}$  and ${\displaystyle \tau _{y}(x)={\overline {\sigma _{x}(y)}}\circ x\circ y}$  satisfy the following three properties (c.f. [GV19]):

i) ${\displaystyle \sigma }$  is a homomorphism from ${\displaystyle (B,\circ )}$  to ${\displaystyle Perm(A)}$ : for ${\displaystyle x}$ , ${\displaystyle y}$ , ${\displaystyle z}$  in ${\displaystyle A}$ , $${\displaystyle \sigma _{x\circ y}(z)=\sigma _{x}(\sigma _{y}(z)):(B,\circ )\to Perm(A)}$$ ; (this is equivalent to the defining equation for a left brace).

ii) ${\displaystyle \tau }$  is an anti-homomorphism from ${\displaystyle (B,\circ )}$  to ${\displaystyle Perm(A)}$ : $${\displaystyle \tau _{x\circ y}=\tau _{y}\tau _{x}:(B,\circ )\to Aut(A)}$$ .

iii) ${\displaystyle \sigma _{u}(v)\circ \tau _{v}(u)=u\circ v}$ . Thus, since $${\displaystyle R(u,v)=(\sigma _{u}(v),\tau _{v}(u))}$$  then if ${\displaystyle R(u,v)=(y,z)}$ , then ${\displaystyle u\circ v=y\circ z}$ . These properties suffice to show that ${\displaystyle R}$  satisfies $${\displaystyle (R\times id)(id\times R)(R\times id)(a,b,c)=(id\times R)(R\times id)(id\times R)(a,b,c)}$$ .

The left side of (*) is: $${\displaystyle (R\times id)(id\times R)(R\times id)(a,b,c)=(id\times R)(R\times id)(id\times R)(a,b,c)}$$ .

Now

$${\displaystyle (R\times id)(id\times R)(R\times id))(a,b,c)=(R\times id)(id\times R)(d,e,c)=(R\times id)(d,f,g)=(h,k,g)}$$ 

where ${\displaystyle k=\tau _{f}(d),h=\sigma _{d}(f),g=\tau _{c}(e),f=\sigma _{e}(c),e=\tau _{b}(a),d=\sigma _{a}(b)}$  and $${\displaystyle a\circ b=d\circ e,e\circ c=f\circ g,d\circ f=h\circ k}$$  by property iii).

The right side of (*) is:

Failed to parse (syntax error): {\displaystyle (id × R)(R × id)(id × R)(a, b, c) = (id × R)(R × id)(a, q, r) = (id × R)(s, t, r)= (s, v, w)} ,

where ${\displaystyle w=\tau _{r}(t),v=\sigma _{t}(r),t=\tau _{q}(a),s=\sigma _{a}(q),r=\tau _{c}(b),q=\sigma _{b}(c)}$ , and $${\displaystyle b\circ c=q\circ r,a\circ q=s\circ t,t\circ r=v\circ w}$$  by property iii).

To show that ${\displaystyle h=s}$ : $${\displaystyle s=\sigma _{a}(q)=\sigma _{a}(\sigma _{b}(c))=\sigma _{a\circ b}(c)}$$ ; $${\displaystyle h=\sigma _{d}(f)=\sigma _{d}(\sigma _{e}(c))=\sigma _{d\circ e}(c)}$$ , and ${\displaystyle d\circ e=\sigma _{a}(b)\circ \tau _{b}(a)=a\circ b}$ . So ${\displaystyle h=\sigma _{d\circ e}(c)=\sigma _{a\circ b}(c)=s}$ .

To show that ${\displaystyle w=g}$ : $${\displaystyle g=\tau _{c}(e)=\tau _{c}(\tau _{b}(a))=\tau _{b\circ c}(a)}$$ ; $${\displaystyle w=\tau _{r}(t)=\tau _{r}(\tau _{q}(a))=\tau _{q\circ r}(a)}$$ ; and $${\displaystyle q\circ r=\sigma _{b}(c)\circ \tau _{c}(b)=b\circ c}$$ . So ${\displaystyle w=\tau _{q\circ r}(a)=\tau _{b\circ c}(a)=g}$ .

To show that ${\displaystyle k=v}$ : For any ${\displaystyle u,v}$ , if ${\displaystyle R(u,v)=(x,y)}$ , then ${\displaystyle x\circ y=u\circ v}$ . So the left side of equation (*) is ${\displaystyle (h,k,g)}$ ; the right side is ${\displaystyle (s,v,w)}$ , and $${\displaystyle s\circ v\circ w=s\circ t\circ r=a\circ q\circ r=a\circ b\circ c=d\circ e\circ c=d\circ f\circ g=h\circ k\circ g}$$ . Since ${\displaystyle w=g}$ , and ${\displaystyle h=s}$  in the group ${\displaystyle (B,o)}$ , it follows that ${\displaystyle k=v}$ . That completes the proof. For an illustration of what these equalities look like for A a radical algebra, see Radical Rings.

Skew Braces and Hopf-Galois theory.    Generalizing work of [FCC12] on Hopf-Galois structures on Galois extensions of fields whose Galois group is a finite abelian p-group for some prime p, D. Bachiller [Bac16], [Bac18] observed a connection between left braces ${\displaystyle B=(B,\circ ,+)}$  and Hopf-Galois structures of type ${\displaystyle (B,+)}$  on Galois extensions of fields with Galois group isomorphic to ${\displaystyle (B,\circ )}$ . N. Byott and L. Ventramin [SV18] then showed that every left skew brace ${\displaystyle (B,\circ ,\cdot )}$  yields at least one Hopf-Galois structure of type ${\displaystyle (B,\cdot )}$  on a classical Galois extension of fields with Galois group ${\displaystyle N}$  isomorphic to (B, circ)</math>, and conversely, given a ${\displaystyle N}$ -Galois extension of fields which also has a Hopf Galois structure of type ${\displaystyle N}$ , then there exists a skew left brace ${\displaystyle (B,\circ ,\cdot )}$  with additive group ${\displaystyle (B,\cdot )}$  isomorphic to ${\displaystyle N}$  and circle group ${\displaystyle (B,\circ )}$  isomorphic to ${\displaystyle N}$ . A given skew brace with circle group ${\displaystyle N}$  may yield more than one Hopf-Galois structure on a Galois extension with Galois group ${\displaystyle G}$ :    the number of Hopf-Galois structures corresponding to a given skew brace relates to the sizes of the automorphism groups of automorphisms of ${\displaystyle G}$  and of ${\displaystyle N}$  and is described in the appendix to [SV18].)

Classification results. A natural question arising in skew brace theory, and independently, in Hopf-Galois theory, is to ask, given a pair ${\displaystyle (G,N)}$  of finite groups, is there a skew brace ${\displaystyle (B,\circ ,\cdot )}$  so that ${\displaystyle (B,\circ )\cong G}$  and ${\displaystyle (B,\cdot )\cong N}$ ?   In skew brace theory the question was typically posed:    given a skew brace ${\displaystyle B}$  with additive group ${\displaystyle (B,\cdot )}$  isomorphic to ${\displaystyle N}$ , what are the possible isomorphism types of groups ${\displaystyle G}$  so that ${\displaystyle (B,\circ )}$  is isomorphic to ${\displaystyle G}$ ? In Hopf-Galois theory the question was typically posed: given a Galois group ${\displaystyle G}$ , what are the possible types ${\displaystyle N}$  of Hopf-Galois structures on a ${\displaystyle G}$ -Galois extension?    Here are some results on this question.    (All groups are finite.)

   • If ${\displaystyle G}$  is a cyclic group of order p^n</math>  where p</math> is an odd prime, then ${\displaystyle N}$  must be isomorphic to ${\displaystyle G}$ . [Kohl98]
   • If ${\displaystyle N}$  is a cyclic group of order  p^n</math> where p</math> is an odd prime, then ${\displaystyle G}$  must be isomorphic to ${\displaystyle N}$  [Ru07a].  Neither of these results hold if ${\displaystyle p=2}$ : see [By07] and [Ru07a], [Ru19].
   • If ${\displaystyle p}$  is odd, ${\displaystyle N}$  is an abelian p</math>-group of order p^n</math> and of p</math>-rank m</math> where m+1 < p</math>, and ${\displaystyle N}$  is an abelian p-group, then ${\displaystyle G}$  is isomorphic to ${\displaystyle N}$ . [Fe03, FCC12] This was generalized in [Bac16] to yield that if B</math> = (B, \circ, \cdot)</math> is a brace with additive group (B, \cdot)</math> = ${\displaystyle N}$  an abelian group of p</math>-rank m</math> with m+1 < p</math>, then for any b</math> in B</math>, the order of b</math> in (B, \cdot)</math> is equal to the order of b</math> in (B, \circ) = ${\displaystyle N}$ .  In particular, if ${\displaystyle N}$  is an elementary abelian p</math>-group of order p^m</math> for p</math> an odd prime and m+1< p</math>, then ${\displaystyle G}$  must have exponent p.</math> 
   •  If ${\displaystyle N}$  is abelian, then ${\displaystyle G}$  is solvable. [ESS99], [Byo13].
   •  If ${\displaystyle G}$  is a simple group, then ${\displaystyle N}$  must be isomorphic to ${\displaystyle G}$  [By04].
   • If ${\displaystyle G}$  = S_n</math>, the symmetric group and n = 5</math> or > 6</math>, then ${\displaystyle N}$  must be isomorphic to ${\displaystyle G}$ </math> or to  A_n \times C_2</math> (the direct product of the alternating group A_n</math> and the cyclic group of order 2 [Ts19].
   • If ${\displaystyle G}$  is abelian, then ${\displaystyle N}$  must be a metabelian group [By15], [Nas19], [TQ20]
   • If ${\displaystyle G}$ </math> is a nilpotent group, then ${\displaystyle N}$  is a solvable group [TQ20].
   • If ${\displaystyle G}$  is solvable, then ${\displaystyle N}$  need not be solvable [By15]: there exists a skew brace with circle group isomorphic to ${\displaystyle G=A_{4}\times C_{5}}$  and additive group ${\displaystyle N=A_{5}}$ , where ${\displaystyle A_{n}}$ is the alternating group (the group of even permutations on ${\displaystyle n}$  symbols) and ${\displaystyle C_{n}}$  is the cyclic group of order ${\displaystyle n}$ .
   • If ${\displaystyle G}$  is a nilpotent group of class 2, then there exists a brace with circle group ${\displaystyle G}$ : in fact, a radical algebra with circle group ${\displaystyle G}$  [AW73].
   • If for some m dividing the order of ${\displaystyle N}$ , if ${\displaystyle N}$  has more characteristic subgroups of order ${\displaystyle m}$  than ${\displaystyle G}$  has subgroups of order ${\displaystyle m}$ , then there is no skew brace with additive group isomorphic to ${\displaystyle N}$  and circle group isomorphic to ${\displaystyle G}$  [Koh19].
   • If ${\displaystyle G}$  is a non-cyclic abelian p-group of order  ${\displaystyle p^{n}}$   with ${\displaystyle n>2}$  and ${\displaystyle p}$  odd, then there is a skew brace with circle group ${\displaystyle G}$  and non-abelian additive group. [BC12]

An open conjecture of N. Byott states that if ${\displaystyle G}$  is insolvable, then ${\displaystyle N}$  cannot be solvable: see [TQ20]. Given groups ${\displaystyle G}$  and N of order ${\displaystyle n}$  for which there is a skew brace with additive group ${\displaystyle N}$  and circle group ${\displaystyle G}$ , there are also many results on the number of isomorphism types of skew braces ${\displaystyle B}$ with additive and circle groups isomorphic to ${\displaystyle N}$  and ${\displaystyle G}$ , resp. In particular, for ${\displaystyle n}$  squarefree, see [AB21].

Bi-skew braces

A set ${\displaystyle B}$  with two group structures ${\displaystyle (B,\circ )}$  and ${\displaystyle (B,\cdot )}$  so that ${\displaystyle B}$  is a skew brace with either group acting as the circle group is called a bi-skew brace.    One large set of examples are ${\displaystyle (B,\circ ,\cdot )}$  where ${\displaystyle B}$ is a nilpotent algebra ${\displaystyle (B,\cdot ,+)}$  of index 3 (that is, for all ${\displaystyle a,b,c}$  in ${\displaystyle B}$ , ${\displaystyle a\cdot b\cdot c=0}$ ).    Another set of examples are ${\displaystyle (B,\circ ,\cdot )}$  where ${\displaystyle (B,\cdot )}$  is a semidirect product ${\displaystyle H\rtimes K}$  for ${\displaystyle H}$ , ${\displaystyle K}$  finite subgroups of ${\displaystyle (B,\cdot )}$ , and ${\displaystyle (B,\circ )}$  is the direct product ${\displaystyle H\times K}$  . The concept of bi-skew brace, from [Ch19], has been extended to the concept of brace block, a collection of different group operations on a set ${\displaystyle G}$  so that every ordered pair of operations on ${\displaystyle G}$  makes ${\displaystyle G}$  into a skew brace. See [Koc21] and [CS21] for examples and theory.

See also

   • Jacobson radical 

   • Yang-Baxter equation.

Notes For expositions of Hopf-Galois structures and their applications to Galois module theory, see [Ch00] and [CGKKKTU21].

References

[AB21] Alabdali, A., Byott, N. Skew braces of squarefree order, J. Algebra and Its Applications 20, No. 7 (2021).

[AW73] Ault, J., Watters, J.. Circle groups of nilpotent rings, American Math. Monthly 80 (1973), 48-52.

[Bac16]    Bachiller, D., Counterexample to a conjecture about braces,    Journal of Algebra, vol    453 (2016), 160-176.

[Bac18]    Bachiller, D.,    Solutions of the Yang–Baxter equation associated to skew left braces, with applications to racks, Journal of Knot Theory and Its Ramifications, Vol. 27 (2018),

[BC12] Byott, N. P., Childs, L. N., Fixed point free pairs of homomorphisms and Hopf-Galois structures, New York J. Math 18 (2012), 707--731.

[By04] Byott, N. P., Hopf-Galois structures on field extensions with simple Galois groups, Bull. London Math. Soc. 36 (2004), 23--29

[By13] Byott, N. P., Nilpotent and abelian Hopf-Galois structures on field extensions, Journal of Algebra 318 (2013), 131-139.

[By15] Byott, N. P.,Solubility criteria for Hopf-Galois structures, New York J. Math 21 (2015), 883-903.

[CS21] Caranti, A., Stefanello, L., From endomorphisms to bi-skew braces, regular subgroups, the Yang--Baxter equation, and Hopf-Galois structures, J. Algebra 587 (2021), 462--487.

[Ch00] Childs, L. N. Taming Wild Extensions: Hopf Algebras and Local Galois Module Theory, American Math. Soc. Math. Surveys and Monographs, vol. 80 (2000).

[Ch19] Childs, L. N., Bi-skew braces and Hopf-Galois structures, New York J. Math. 25 (2019), 574--588.

[CGKKKTU21] Childs, L. N., Greither, C., Keating, K. P., Koch, A., Kohl, T., Truman, P. J., Underwood, R. G., Hopf Algebras and Galois Module Theory, American Math. Soc. Math. Surveys and Monographs, vol. 260 (2021).

[Dr92]    Drinfel'd, V., On some unsolved problems in quantum group theory, Lecture Notes in Mathematics 1510 (1992), 1--8.

[ESS99] Etinghof, P., Schedler, T., Soloviev, A., Set-theoretical solutions to the quantum Yang-Baxter equation, Duke Math. J. 100 (1999), 169-209.

[Fe03] Featherstonhaugh, S. C., Hopf algebra structures on abelian Galois extensions of fields, Ph. D. thesis, Univ. at Albany, 2003.

[FCC12]    Featherstonhaugh, S. C., Caranti, A., Childs, L. N., Abelian Hopf Galois structures on prime-power Galois field extensions, Trans. Amer. Math. Soc. 364 (2012), 3675--3684.

[GV17]    Guarnieri, L., Ventramin, L., Skew braces and the Yang-Baxter equation, Math. Comp. 86 (2017), 2519--2534.

[Koc21] Koch, A., Abelian maps, brace blocks, and solutions to the Yang-Baxter equation, arXiv:2102.06104

[Koh98] Kohl, T., Classification of the Hopf Galois structures on prime power radical extensions, J. Algebra 207 (1998), 525-546.

[Koh19] Kohl, T. Characteristic subgroups lattices and Hopf-Galois structures, Intern. J. Algebra Computation 29 (2019), 391--405.

[LYZ00] Lu, J-H., Yang, M., Zhu, Y.-C., On the set-theoretical Yang-Baxter equation, Duke Math. J. 104 (2000), 1--18.

[Nas19] Nasybullow, L., Connections between properties of the additive and multiplicative groups of a two-sided skew brace, J. Algebra 540 (2019), 156-167.

[Ru07]    Rump, W., Braces, radical rings, and the quantum Yang-Baxter equation, J. Algebra 307 (2007), 153--170.

[Ru07a] Rump, W., Classification of cyclic braces, J. Pure and Applied Algebra 209 (2007), 671--685.

[Ru19] Rump, W., Classification of cyclic braces II, Trans. Amer. Math. Soc. 372 (2019), 305-328.

[SV18]    Smoktunowicz, A., Vendramin, L., On skew braces (with an appendix by N. Byott and L. Vendramin), J. Combinatorial Algebra 2 (2018), 47--86.

[Tsa19] Tsang, C., Hopf-Galois structures on a Galois S_n-extension, J. Algebra (2019), 349--361.

[TQ20] Tsang, C., Qin, C., On the solvability of regular subgroups in the holomorph of a finite solvable group, Internat. J. Algebra Comput. 30 (2020), 253--265.