2 Preliminaries

In this section, we briefly recall some facts on cycle sets needed in what follows. A set X with a binary operation $\cdot $ is said to be a cycle set [Reference Rump21] if the left multiplication $\sigma (x)$ : $y\mapsto x\cdot y$ is invertible, and the equation (1.1) holds for all $x,y,z\in X$ . So $\sigma $ defines a map

(2.1) $$ \begin{align} \sigma: X\rightarrow\mathfrak{S}(X) \end{align} $$

into the permutation group $\mathfrak {S}(X)$ . The subgroup $G(X)$ of $\mathfrak {S}(X)$ generated by the image of $\sigma $ is said to be the permutation group of X. The image $\sigma (X)$ admits a well-defined operation

$$ \begin{align*}\sigma(x)\cdot\sigma(y):=\sigma(x\cdot y),\end{align*} $$

so that equation (1.1) is retained for $\sigma (X)$ , the retraction [Reference Etingof, Schedler and Soloviev10, Reference Rump21] of X. In general, $\sigma (X)$ is not a cycle set, unless X is nondegenerate, that is, the square map $x\mapsto x\cdot x$ is bijective. The retraction $\sigma (X)$ is then again nondegenerate, so that the retraction process can be iterated. Finite cycle sets are always nondegenerate [Reference Rump21, Theorem 2].

A brace A is a cycle set with an abelian group structure satisfying

(2.2) $$ \begin{align} a\cdot (b+c) &= (a\cdot b) + (a\cdot c), \end{align} $$

(2.3) $$ \begin{align} (a+b)\cdot c &= (a\cdot b)\:\cdot\: (a\cdot c), \end{align} $$

for all $a,b,c\in A$ . Note that equation (1.1) follows by equation (2.3) and the commutativity of $(A;+)$ . Equation (2.2) says that $\sigma (a)$ is a group automorphism for all $a\in A$ . As a cycle set, any brace A is nondegenerate. Moreover, it is a group with respect to the operation

(2.4) $$ \begin{align} a\circ b:=a^b+b, \end{align} $$

where $a^b:=\sigma (b)^{-1}(a)$ . The group $A^\circ :=(A;\circ )$ is called the adjoint group of A. Typical examples of braces are radical rings R (=Jacobson radicals of rings, viewed as pseudorings) with the operation $a\cdot b:=b(1+a)^{-1}$ . Then, the adjoint group is given by Jacobson’s circle operation [Reference Jacobson17]

$$ \begin{align*}a\circ b=a+ab+b.\end{align*} $$

In accordance with Jacobson’s notation [Reference Jacobson17], we write $a'$ for the inverse in the adjoint group of a brace. Note that the unit element of $A^\circ $ is 0.

For any nondegenerate cycle set X, the permutation group $G(X)$ is the adjoint group of a unique brace $A(X)$ such that $\sigma \colon X\rightarrow A(X)$ is a cycle set morphism. A subset of a brace A is said to be a cycle base [Reference Rump22, Definition 4] if it is invariant under the adjoint group $A^\circ $ and generates the additive group of A. By [Reference Rump28, Proposition 10], a brace A is generated by an element $e\in A$ if and only if $X:=\{e^a\:|\:a\in A\}$ is a cycle base of A. Thus, each transitive cycle base (i.e., such that $A^\circ $ acts transitively on it) is given by a generator $e\in A$ .

A right ideal [Reference Rump22] of a brace A is an additive subgroup which is invariant under the action of $A^\circ $ . By equation (2.4), any right ideal is a subbrace. For radical rings, this concept coincides with the usual one. We say that a brace A is a torsion brace if each $a\in A$ is of finite order in $(A;+)$ . For a torsion brace A, equation (2.2) implies that the primary decomposition of the additive group is a decomposition into right ideals.

The fundamental groupoid [Reference Rump26] $\pi _1(X)$ of a cycle set X has X as set of objects, and pairs $(a,x)\in G(X)\times X$ as morphisms from x to $a(x)$ . Composition is the obvious one:

$$ \begin{align*}x\longrightarrow b(x)\longrightarrow ab(x),\end{align*} $$

given by $(a,b(x))(b,x):=(ab,x)$ . Inverses are $(a,x)^{-1}:=(a^{-1},a(x))$ . The isomorphism classes of objects in $\pi _1(X)$ form a set $C(X)$ , consisting of the $G(X)$ -orbits of X. Any morphism $f\colon X\rightarrow Y$ of cycle sets induces a map $C(f)\colon C(X)\rightarrow C(Y)$ . If f is surjective, it gives rise to a commutative diagram

with a surjective group homomorphism $G(f)$ . We call $f\colon X\twoheadrightarrow Y$ a covering [Reference Rump26] if $C(f)$ and $G(f)$ are bijective. By [Reference Rump26, Theorem 6], any cycle set X admits a covering $\widetilde {X}\twoheadrightarrow X$ which is universal in the sense that it factors uniquely though each covering $Y\twoheadrightarrow X$ , so that $\widetilde {X}\rightarrow Y$ is again a covering. Cycle sets X with no proper coverings $Y\twoheadrightarrow X$ are characterized by their fundamental groupoid which is skeletal.

A cycle set X is said to be indecomposable if $G(X)$ acts transitively on X, that is, $|C(X)|=1$ . Then, $\widetilde {X}$ is indecomposable, too. Thus, $G(X)=G(\widetilde {X})$ acts freely and transitively on $\widetilde {X}$ . Being quite analogous to simply connected topological spaces, such cycle sets are called uniconnected [Reference Rump27].

3 Bracial self-maps of groups

Let $\varepsilon \colon G\rightarrow G$ be a self-map of a group G. We define the kernel of $\varepsilon $ to be the subgroup

$$ \begin{align*}\mbox{Ker}\,\varepsilon:=\{h\in G\:|\;\forall\,g\in G\colon\varepsilon(hg)= \varepsilon(g)\}.\end{align*} $$

For a group homomorphism $\varepsilon $ , this concept coincides with the usual one. Note, however, that $\mbox {Ker}\,\varepsilon $ need not be normal for an arbitrary self-map $\varepsilon $ .

Thus, B is uniquely determined by its group structure and the self-map

$$ \begin{align*}\varepsilon(a):=(a\odot 1)^{-1}.\end{align*} $$

We call $\varepsilon $ the structure map and $\varepsilon (B)$ the image of B. The kernel $\mbox {Ker}\,B:=\mbox {Ker}\,\varepsilon $ of $\varepsilon $ will be called the kernel of B. In terms of $\varepsilon $ , we have

Proposition 1 A self-map $\varepsilon \colon B\rightarrow B$ of a group B defines a bracial cycle set if and only if it satisfies the equation

(3.1) $$ \begin{align} \varepsilon\bigl(a\varepsilon(b)^{-1}\bigr)\varepsilon(b)= \varepsilon\bigl(b\varepsilon(a)^{-1}\bigr)\varepsilon(a). \end{align} $$

Proof. For $a,b,c\in B$ , we have

$$\begin{align*} (a\odot b)\odot(a\odot c)&=b\varepsilon(a)^{-1}\odot c\varepsilon(a)^{-1}=c\varepsilon(a)^{-1}\varepsilon(b\varepsilon(a)^{-1})^{-1}\\&= c(\varepsilon\bigl(b\varepsilon(a)^{-1}\bigr)\varepsilon(a))^{-1}.\end{align*}$$

Equation (3.1) says that this expression is symmetric in a and b.▪

Consider the subgroup $\langle \varepsilon (B)\rangle $ of B generated by the image of B. For $a,b\in B$ , we have $a\odot b=b(a\odot 1)=b\varepsilon (a)^{-1}$ . Therefore, the right regular representation of B induces a group isomorphism

(3.2) $$ \begin{align} \pi\colon\langle\varepsilon(B)\rangle\mathop{\longrightarrow}\limits^{\sim}\, G(B) \end{align} $$

onto the permutation group of B, which maps $a\in \langle \varepsilon (B)\rangle $ to the right multiplication $b\mapsto ba^{-1}$ . On the other hand, we have a map

$$ \begin{align*}\sigma\colon B\rightarrow G(B)\end{align*} $$

with $\sigma (a)(b):=a\odot b$ . So the isomorphism (3.2) yields a factorization

(3.3)

As a consequence, we get a classification of uniconnected cycle sets.

Proof. The first statement follows by Theorem 3.1. Conversely, let $(B;\odot )$ be any uniconnected cycle set. Choose an element $1\in B$ . Then, there is a bijection $\gamma \colon G(B)\mathop{\longrightarrow}\limits^{\sim}\, B$ with $\gamma (\alpha ):=\alpha (1)$ . So the map $\sigma \colon B\rightarrow G(B)$ with $\sigma (a)(b)=a\odot b$ satisfies $\sigma (a)=\gamma ^{-1}(a\odot 1)$ . Define a multiplication in B as follows:

(3.4) $$ \begin{align} ab:=\gamma^{-1}(b)(a). \end{align} $$

For given $a,b\in B$ , consider the automorphisms $\alpha :=\gamma ^{-1}(a)$ and $\beta :=\gamma ^{-1}(b)$ in $G(B)$ . Because $\gamma (\alpha \beta )=\alpha \gamma (\beta )$ , we have $\gamma \bigl (\gamma ^{-1}(b)\gamma ^{-1}(a)\bigr )= \gamma ^{-1}(b)(a)=ab$ , which yields $\gamma ^{-1}(ab)=\gamma ^{-1}(b) \gamma ^{-1}(a)$ . Thus, equation (3.4) defines a group structure on B with an isomorphism

$$ \begin{align*}\gamma\colon G(B)^{\text{op}}\mathop{\longrightarrow}\limits^{\sim}\, B.\end{align*} $$

In particular, we obtain $\langle \varepsilon (B)\rangle =B$ . For $a,b\in B$ , equation (3.4) gives $a\odot b=\sigma (a)(b)=\gamma ^{-1}(a\odot 1) (b)=b(a\odot 1)$ . Whence B is a bracial cycle set. ▪

In [Reference Rump27, Theorem 2], nondegenerate uniconnected cycle sets are classified in terms of braces with a transitive cycle base. The following example shows that uniconnected cycle sets need not be nondegenerate.

Example 1 Let G be a right $\ell $ -group [Reference Rump23], that is, a group with a lattice order, satisfying

$$ \begin{align*}a\leqslant b\;\Longrightarrow\; ac\leqslant bc\end{align*} $$

for all $a,b,c\in G$ . Consider the self-map $\varepsilon \colon G\rightarrow G$ given by $\varepsilon (a):=a\vee 1$ . Then, $\varepsilon (a\varepsilon (b)^{-1})\varepsilon (b)=\bigl (a(b\vee 1)^{-1}\vee 1\bigr )(b\vee 1)=a\vee b\vee 1$ , which is symmetric in a and b. Hence, $\varepsilon $ makes G into a bracial cycle set. The kernel of $\varepsilon $ is $\{1\}$ , while the image $\varepsilon (G)$ is the positive cone of G. Therefore, the corresponding cycle set $(G;\odot )$ with $a\odot b:=b (a\vee 1)^{-1}$ is uniconnected. However, $(G;\odot )$ is degenerate: For $a\geqslant 1$ , we have $a\odot a=a(a\vee 1)^{-1}=1$ . So the square map $a\mapsto a\odot a$ is not injective if $|G|\ne 1$ .

The diagram (3.3) shows that the structure map $\varepsilon $ of a bracial cycle set B gives the retraction $\varepsilon (B)$ of B. By [Reference Rump21, Section 6], this implies that the cycle set structure of B induces a well-defined binary operation on $\varepsilon (B)$ :

(3.5) $$ \begin{align} \varepsilon(a)\cdot\varepsilon(b):=\varepsilon(b\varepsilon(a)^{-1}). \end{align} $$

Indeed, $\varepsilon (b)=\varepsilon (c)$ implies that $\varepsilon (b\varepsilon (a)^{-1}) \varepsilon (a)=\varepsilon (a\varepsilon (b)^{-1})\varepsilon (b)=\varepsilon (a\varepsilon (c)^{-1}) \varepsilon (c)=\varepsilon (c\varepsilon (a)^{-1})\varepsilon (a)$ , which yields $\varepsilon (b\varepsilon (a)^{-1})=\varepsilon (c\varepsilon (a)^{-1})$ . Thus, $\varepsilon (a) \cdot \varepsilon (b)$ does not depend on the choice of b. So the operation (3.5) satisfies equation (1.1), which makes $\varepsilon (B)$ into a cycle set if B is nondegenerate. In general, however, $(\varepsilon (B);\cdot )$ is not a subcycle set of $(B;\odot )$ .

Now, we turn our attention to the nondegenerate case. Note that a bracial cycle set B is nondegenerate if and only if the map $a\mapsto a \varepsilon (a)^{-1}$ is bijective.

Definition 3.2 We call a self-map $\varepsilon \colon G\rightarrow G$ of a group G right invariant if the implication

(3.6) $$ \begin{align} \varepsilon(a)=\varepsilon(b)\;\Longrightarrow\;\varepsilon(ac)=\varepsilon(bc) \end{align} $$

holds for $a,b,c\in G$ .

Let $\varepsilon \colon G\rightarrow G$ be a self-map with kernel H. Because $ac=(ab^{-1}) bc$ , condition (3.6) says that $\varepsilon (a)=\varepsilon (b)$ if and only if $ab^{-1}\in H$ . So the fibers of $\varepsilon $ are the right cosets of H. Therefore, $\varepsilon $ induces a bijection $\overline {\varepsilon }\colon G\backslash H\mathop{\longrightarrow}\limits^{\sim}\, \varepsilon (G)$ from the set $G\backslash H$ of right cosets onto the image of $\varepsilon $ . In Example 2, but not in Example 1, the map $\varepsilon $ is right invariant.

Proposition 2 Let G be a group with a right invariant self-map $\varepsilon \colon G\rightarrow G$ . Then,

(3.7) $$ \begin{align} g\cdot\varepsilon(a):=\varepsilon(ag^{-1}) \end{align} $$

(with $g,a\in G$ ) defines a transitive action of G on the image of $\varepsilon $ .

Proof. The diagram (3.3) shows that $\varepsilon (a)=\varepsilon (b)$ is equivalent to $\sigma (a)=\sigma (b)$ . Assuming this, we have to show that $\sigma (ac)= \sigma (bc)$ holds for all $c\in B$ . Because B is uniconnected, it is enough to verify the equivalence

$$ \begin{align*}\sigma(a)=\sigma(b)\;\Longleftrightarrow\;\sigma(c\odot a)=\sigma(c\odot b).\end{align*} $$

Indeed, $\sigma (a)=\sigma (b)$ implies $(c\odot a) \odot (c\odot d)=(a\odot c)\odot (a\odot d)= (b\odot c)\odot (b\odot d)=(c\odot b)\odot (c\odot d)$ for all $d\in B$ . Thus, $\sigma (c\odot a)=\sigma (c\odot b)$ . Conversely, assume that $\sigma (c\odot a)=\sigma (c\odot b)$ . Then, $(a\odot c)\odot (a\odot c)= (c\odot a)\odot (c\odot c)=(c\odot b)\odot (c\odot c)=(b\odot c)\odot (b\odot c)$ . Because B is nondegenerate, we obtain $a\odot c=b\odot c$ . As B is uniconnected, this implies that $\sigma (a)=\sigma (b)$ .▪

For a nondegenerate bracial cycle set B, the group $G(B)$ is the adjoint group of a brace. So the group isomorphism (3.2) also makes $\langle \varepsilon (B)\rangle $ into a brace $A(B)$ . Taking $\pi $ as an identification, we get a factorization

(3.8)

which identifies $\varepsilon (B)$ with the image of $\sigma $ .

Proof. By [Reference Rump28, Proposition 10], $X:=\{e^a\:|\:a\in B\}$ is a transitive cycle base of B. So [Reference Rump27, Theorem 2] shows that the operation

$$ \begin{align*}a\odot b:=b\circ(e^a)'\end{align*} $$

makes B into a nondegenerate uniconnected cycle set. In particular, $(B;\odot )$ is a bracial cycle set with $\varepsilon (a)=(a\odot 0)'=e^a$ . Thus, $\varepsilon (B)=X$ and $\varepsilon (0)=e$ . By Propositions 2 and 3, the structure map $\varepsilon $ is unique: $\varepsilon (a)=\varepsilon (0\circ a)= \varepsilon (0)^a=e^a$ .

Conversely, let B be a nondegenerate uniconnected cycle set. Because $\varepsilon (B)$ is the retraction of B, the map $\varepsilon \colon B\twoheadrightarrow \varepsilon (B)$ is a cycle set morphism. So the cycle set morphism (3.8) shows that $\iota $ is a morphism of cycle sets. Thus, $\varepsilon (B)$ is a cycle base of $A(X)$ . Equation (3.5) shows that it is transitive.

By equation (3.5), the action (3.7) coincides with the adjoint action in $A(B)$ . Hence, $\varepsilon (a)=a^{-1}\cdot \varepsilon (1)=\varepsilon (1)^a$ for all $a \in A(B)$ . Thus, $a\odot b=b\varepsilon (a)^{-1}=b(\varepsilon (1)^a)^{-1}$ . By [Reference Rump27, Theorem 2], the proof is complete. ▪