Constructing skew left braces whose additive group has trivial centre

Abstract

A complete description of all possible multiplicative groups of finite skew left braces whose additive group has trivial centre is given. As a consequence, some earlier results of Tsang can be improved and an answer to an open question set by Tsang at Ischia Group Theory 2024 Conference is provided.

1 Introduction

Skew brace structure plays a key role in the combinatorial theory of the Yang–Baxter equation. Skew left braces, introduced in [8], can be regarded as extensions of Jacobson radical rings and show connections with several areas of mathematics such as triply factorized groups and Hopf–Galois structures (see [1, 3, 4])

Skew left braces classify solutions of the Yang–Baxter equation (see [8]). This connection to the Yang–Baxter equation motivates the search for constructions of skew braces and classification results.

Recall that a skew left brace is a set endowed with two group structures $(B,+)$ , not necessarily abelian, and $(B,\cdot )$ which are linked by the distributive-like law ${a(b+c)}=ab-a+ac$ for a, b, $c\in B$ .

In the sequel, the word brace refers to a skew left brace.

Given a brace B, there is an action of the multiplicative group on the additive group by means of the so-called lambda map:

$$\begin{align*}\lambda \colon a \in (B,\cdot) \longmapsto \lambda_a \in \operatorname{\mathrm{Aut}}(B,+), \quad \lambda_a(b)=-a+ab, \ \text{for all }a,b\in B.\end{align*}$$

Braces can be described in terms of regular subgroups of the holomorph of the additive group. Recall that the holomorph of a group G is the semidirect product $\operatorname {\mathrm {Hol}}(G) = [G]\operatorname {\mathrm {Aut}}(G)$ . Let B be a brace and set $K = (B,+)$ . Then $H=\{ (a,\lambda _a) \mid a\in B\}$ is a regular subgroup of the holomorph $\operatorname {\mathrm {Hol}}(K)$ isomorphic to $(B, {\cdot })$ (see [8, Theorem 4.2]). If we consider the subgroup $S = KH \leq \operatorname {\mathrm {Hol}}(K)$ , then

$$\begin{align*}S=KH=KE=HE,\end{align*}$$

where $E = \{(0,\lambda _b)\mid b \in B\}$ and $\operatorname {\mathrm {C}}_{E}(K)= K\cap E =H\cap E =1$ . We call $\mathsf {S}(B)=(S, K, H, E)$ the small trifactorized group associated with B.

In [9], Tsang showed it is possible to construct finite braces by just looking at the automorphism group of the additive group instead of looking at the whole holomorph. This is a significant improvement both from an algebraic and computational approach.

Theorem 1 (see [9, Corollary 2.2])

If the finite group G is the multiplicative group of a brace with additive group K, then there exist two subgroups X and Y of $\operatorname {\mathrm {Aut}}(K)$ that are quotients of G satisfying

$$\begin{align*}XY=X{\operatorname{\mathrm{Inn}}(K)}=Y{\operatorname{\mathrm{Inn}}(K)}.\end{align*}$$

She looked for a sort of converse of the above theorem in the case of finite braces with an additive group of trivial centre, and proved the following.

Theorem 2 (see [9, Proposition 2.7])

Suppose that the centre of a finite group $(K, {+})$ is trivial and let P be a subgroup of $\operatorname {\mathrm {Aut}}(K)$ containing $\operatorname {\mathrm {Inn}}(K)$ . If $P=XY$ is a factorization by two subgroups X and Y such that $X\cap Y=1$ , $X{\operatorname {\mathrm {Inn}}(K)}=Y{\operatorname {\mathrm {Inn}}(K)}=P$ and X splits over $X\cap \operatorname {\mathrm {Inn}}(K)$ , then there exists a brace B whose additive group is isomorphic to $(K, {+})$ and whose multiplicative group is isomorphic to a semidirect product $[X\cap \operatorname {\mathrm {Inn}}(K)]Y$ for a suitable choice of the action $\alpha \colon Y\longrightarrow \operatorname {\mathrm {Aut}}(X\cap \operatorname {\mathrm {Inn}}(K))$ .

The above two theorems are the key to prove the main results of [9, 10].

In [11], Tsang posed the following question.

Question 3 Is it possible to extend Theorem 2 by dropping the assumption that X splits over $X\cap \operatorname {\mathrm {Inn}}(K)$ ?

The aim of this article is to give a complete characterization of the multiplicative groups of a brace with additive group of trivial centre. As a consequence, we present an improved version of Theorem 2 (on which the main result of [9] heavily depends), and we give an affirmative answer to Question 3.

Theorem A Let K be a finite group with trivial centre. For every brace B with additive group $K = (B,+)$ and multiplicative group $C = (B,\cdot )$ , there exist subgroups X and Y of $Aut(K)$ satisfying the following properties:

1. (a) $XY=X{\operatorname {\mathrm {Inn}}(K)}=Y{\operatorname {\mathrm {Inn}}(K)}$ ,

2. (b) there are two subgroups N and M of $\operatorname {\mathrm {Inn}}(K)$ such that $N \unlhd X$ and $M \unlhd Y$ ,

3. (c) there exists an isomorphism $\gamma \colon Y/M\longrightarrow X/N$ such that

   $$\begin{align*}\operatorname{\mathrm{Inn}}(K) = \{xy^{-1}\mid x\in X, \, y\in Y,\, \gamma(yM) = xN\},\end{align*}$$

4. (d) $\lvert K\rvert =\lvert X\rvert \lvert M\rvert =\lvert Y\rvert \lvert N\rvert $ .

In this case,

1. (e) C has two normal subgroups T and V with $T\cap V=1$ , $X\cong C/T$ and $Y\cong C/V$ , that is, C is a subdirect product of X and Y.

Conversely, for every pair X, Y of subgroups of $\operatorname {\mathrm {Aut}}(K)$ satisfying conditions (a)–(d), there exists a brace B with $K = (B,+)$ and $C= (B,\cdot )$ satisfying (e).

Corollary 4 Let K be a finite group with trivial centre. Suppose that there exist subgroups X, Y of $\operatorname {\mathrm {Aut}}(K)$ such that $X \cap Y = 1$ and $XY=X{\operatorname {\mathrm {Inn}}(K)}=Y{\operatorname {\mathrm {Inn}}(K)}$ . Then there exists a brace with additive group K and a multiplicative group that is isomorphic to a subdirect product of X and Y.

Proof Assume that $X\cap Y=1$ . Consider $N=X\cap {\operatorname {\mathrm {Inn}}(K)}$ , $M=Y\cap {\operatorname {\mathrm {Inn}}(K)}$ . Then $\lvert X\rvert \lvert M\rvert =\lvert K\rvert $ as $\lvert X\rvert \lvert Y\rvert =\lvert {\operatorname {\mathrm {Inn}}(K)}\rvert \lvert Y\rvert /\lvert Y\cap {\operatorname {\mathrm {Inn}}(K)}\rvert $ . Analogously, $\lvert Y\rvert \lvert N\rvert =\lvert K\rvert $ . Moreover, since

$$\begin{align*}Y/M\cong Y{\operatorname{\mathrm{Inn}}(K)}/{\operatorname{\mathrm{Inn}}(K)}=X{\operatorname{\mathrm{Inn}}(K)}/{\operatorname{\mathrm{Inn}}(K)}\cong X/N,\end{align*}$$

we have an isomorphism $\gamma \colon Y/M\longrightarrow X/N$ given by $\gamma (bM)=aN$ , where $b \in Y$ , $a \in X$ such that $ab^{-1} \in \operatorname {\mathrm {Inn}}(K)$ . Since $X \cap Y = 1$ , for each $k \in K$ , conjugation by k can be expressed as $ab^{-1}$ , for a unique $a\in X$ and $b\in Y$ . Then, the groups X and Y satisfy Statements (a)–(d) of Theorem A, and therefore, there exists a brace whose additive group is K and whose multiplicative group is isomorphic to a subdirect product of X and Y.

Corollary 4 also allows to give a considerably shorter proof of the main results of [9, 10] about the almost simple groups K that can appear as additive groups of braces with soluble multiplicative group. By Corollary 4, it is enough to find two subgroups X and Y of $\operatorname {\mathrm {Aut}}(K)$ such that $X \cap Y = 1$ and $XY=X{\operatorname {\mathrm {Inn}}(K)}=Y{\operatorname {\mathrm {Inn}}(K)}$ . Therefore, Codes 2, 3, and 4 in the proof of [9, Theorem 1.3] can be avoided, as well as checking in every case that the subgroup X splits over $X\cap {\operatorname {\mathrm {Inn}}(K)}$ .

In Section 3, we present a worked example of a construction of a brace with additive group $K = \operatorname {\mathrm {PSL}}_2(25)$ by means of subgroups X and Y of $\operatorname {\mathrm {Aut}}(K)$ satisfying all conditions of Theorem A but $X \cap Y \neq 1$ .

2 Proof of Theorem A

Proof of Theorem A

Suppose that B is a brace with additive group K and lambda map $\lambda $ . Let $H=\{(b, \lambda _b)\!\mid b\in B\}$ be the regular subgroup of $\operatorname {\mathrm {Hol}}(K)$ appearing in the small trifactorized group $\mathsf {S}(B)=(S, K, H, E)$ associated with B. Recall that H is isomorphic to the multiplicative group $(C,\cdot )$ of B, $E= \{(0, \lambda _b)\!\mid b \in B\} \leq \operatorname {\mathrm {Hol}}(K)$ , and $S = KH = KE = HE$ with $K\cap E=H\cap E= 1$ .

Observe that S acts on K by means of the homomorphism $\pi \colon (b, \omega )\in S \mapsto \omega \in \operatorname {\mathrm {Aut}}(K)$ . On the other hand, S also acts on K by conjugation. In fact, this action naturally induces a homomorphism $\alpha \colon S \rightarrow \operatorname {\mathrm {Aut}}(K)$ . In particular, for every $b\in B$ and every $k \in K$ , $(0,\lambda _b)(k,1)(0,\lambda _b)^{-1}=(\lambda _b(k), 1)$ , that is, $\alpha (0,\lambda _b) = \lambda _b = \pi (0,\lambda _b)$ . Thus, $\alpha (E) = \pi (E) = \pi (H)$ .

Figure 1 Structure of the multiplicative group in Theorem A.

The restrictions of $\pi $ and $\alpha $ to H induce two actions of H on K, with respective kernels $\operatorname {\mathrm {Ker}}{\pi }|_H=K\cap H\trianglelefteq H$ and $\operatorname {\mathrm {Ker}}{\alpha |_H}=\operatorname {\mathrm {C}}_H(K)\trianglelefteq H$ . Moreover, it holds that

$$ \begin{align*} \operatorname{\mathrm{Ker}}{\pi}|_H\cap \operatorname{\mathrm{Ker}}{\alpha|_H} & = K\cap H\cap \operatorname{\mathrm{C}}_H(K)= K\cap H\cap \operatorname{\mathrm{C}}_S(K) \\ & = H\cap \operatorname{\mathrm{C}}_{K}(K)=H\cap \operatorname{\mathrm{Z}}(K)=1 \quad \text{(see Figure~1).} \end{align*} $$

Let $X:=\alpha (H)$ and $Y := \pi (H) = \alpha (E) = \{\lambda _b\mid b\in B\}$ such that $X \cong H/{\operatorname {\mathrm {C}}_H(K)}$ and $Y \cong H/(K\cap H)$ . Since $\alpha (K)=\operatorname {\mathrm {Inn}}(K)$ , we have that

$$ \begin{align*} \alpha(S)&=\alpha(HE)=\alpha(KH)=\alpha(KE)\\ &=\alpha(H)\alpha(E)=\alpha(K)\alpha(H)=\alpha(K)\alpha(E)\\ &=XY=(\operatorname{\mathrm{Inn}}(K))X=(\operatorname{\mathrm{Inn}}(K))Y. \end{align*} $$

Take $R:=(H\cap K){\operatorname {\mathrm {C}}_H(K)} \unlhd H$ . Then, $N:=\alpha (R)\trianglelefteq \alpha (H)=X$ and $ M:= \pi (R)\trianglelefteq \pi (H) = Y$ . It follows that $N =\alpha (H\cap K)\le \alpha (K)=\operatorname {\mathrm {Inn}}(K)$ . On the other hand, $M = \pi (\operatorname {\mathrm {C}}_H(K))$ and if $(b,\lambda _b)\in {\operatorname {\mathrm {C}}_H(K)}$ , then for every $k\in K$ ,

$$\begin{align*}(b, \lambda_b)(k,1)(b,\lambda_b)^{-1}=(b+\lambda_b(k)-b,1)=(k,1), \end{align*}$$

that is, $\lambda _b$ coincides with the inner automorphism of K induced by $-b$ . Thus, $M \le \operatorname {\mathrm {Inn}}(K)$ . Moreover, we see that

$$ \begin{align*} Y/M &\cong (H/\operatorname{\mathrm{Ker}}{\pi}|_H)/(R/\operatorname{\mathrm{Ker}}{\pi}|_H)\cong H/R\\ &\cong(H/\operatorname{\mathrm{Ker}}\alpha|_H)/(R/\operatorname{\mathrm{Ker}}\alpha|_H)\cong X/N; \end{align*} $$

here the isomorphism $\gamma \colon Y/M\longrightarrow X/N$ is given by $\gamma (\lambda _b M)=\alpha _b\lambda _b N$ , where $\alpha _b$ is the inner automorphism of K induced by b. Given $a\in \gamma (\lambda _bM)$ , we have that $a\lambda _b^{-1}\in \alpha _bN\subseteq \operatorname {\mathrm {Inn}}(K)$ . Furthermore, given $x\in \operatorname {\mathrm {Inn}}(K)$ , we have that $x=\alpha _b$ for some $b\in B$ and so $\gamma (\lambda _b M)=\alpha _b\lambda _b N=x\lambda _b N$ with $(\alpha _b\lambda _b)\lambda _b^{-1}=x$ .

Since $\operatorname {\mathrm {Ker}}{\pi |}_H\cap {\operatorname {\mathrm {Ker}}\alpha |_H}=(H\cap K)\cap \operatorname {\mathrm {C}}_H(K)=1$ , we have that $\lvert R\rvert ={\lvert H\cap K\rvert }\lvert {\operatorname {\mathrm {C}}_H(K)}\rvert $ and $\lvert M\rvert =\lvert R/(H\cap K)\rvert =\lvert \operatorname {\mathrm {C}}_H(K)\rvert $ , $\lvert N\rvert =\lvert R/{\operatorname {\mathrm {C}}_H(K)}\rvert =\lvert H\cap K\rvert $ . As $\lvert X\rvert =\lvert K\rvert /\lvert {\operatorname {\mathrm {C}}_H(K)}\rvert $ and $\lvert Y\rvert =\lvert K\rvert /\lvert H\cap K\rvert $ , the claim about the order follows.

Item (e) follows by the fact that H is isomorphic to the multiplicative group $(C,\cdot )$ of B, so that T and V are respectively isomorphic to $\operatorname {\mathrm {Ker}} \alpha |_H$ and $\operatorname {\mathrm {Ker}} \pi |_H$ .

Now, suppose that $\operatorname {\mathrm {Aut}}(K)$ possesses subgroups X and Y satisfying conditions (a)–(d). Let

$$\begin{align*}W=\{(x,y)\mid x\in X,\, y\in Y,\, \gamma(yM)=xN\}\end{align*}$$

be a subdirect product of X and Y with amalgamated factor group $Y/M\cong X/N$ (see [6, Chapter A, Definition 19.2]). By [6, Chapter A, Proposition 19.1], and the hypothesis, we have that $\lvert W\rvert =\lvert K\rvert $ . Since $\operatorname {\mathrm {Z}}(K)$ is trivial, the map $\zeta \colon K\longrightarrow \operatorname {\mathrm {Inn}}(K)$ , where $\zeta (k)$ is the inner automorphism of K induced by k, is an isomorphism. By hypothesis, the map $W\longrightarrow \operatorname {\mathrm {Inn}}(K)$ given by $(x,y)\longmapsto xy^{-1}$ is surjective. Since $\lvert W\rvert =\lvert \operatorname {\mathrm {Inn}}(K)\rvert =\lvert K\rvert $ , it is a bijection. We can consider $H=\{(b,y)\mid (x,y)\in W,\, \zeta (b)=xy^{-1}\}\subseteq \operatorname {\mathrm {Hol}}(K)$ . Given $(b, y)$ , $(b_1, y_1)\in H$ , we have that $(b, y)(b_1,y_1)=(b+y(b_1),yy_1)$ , $\zeta (b)=xy^{-1}$ , and $\zeta (b_1)=x_1y_1^{-1}$ with $(x,y)$ , $(x_1, y_1)\in B$ . Then

$$\begin{align*}\zeta(b+y(b_1))=\zeta(b)\zeta(y(b_1))=\zeta(b)y\zeta(b_1)y^{-1}=xy^{-1}yx_1y_1^{-1}y^{-1}=(xx_1)(yy_1)^{-1}\end{align*}$$

with $(xx_1, yy_1)=(x,y)(x_1,y_1)\in W$ . Furthermore, if $(b, y)\in H$ , with $\zeta (b)=xy^{-1}$ , we have that $(b, y)^{-1}=(y^{-1}(-b), y^{-1})$ and

$$\begin{align*}\zeta(y^{-1}(-b))=y^{-1}\zeta(-b)y=y^{-1}\zeta(b)^{-1}y=y^{-1}yx^{-1}y=x^{-1}{(y^{-1})}^{-1}\end{align*}$$

with $(x^{-1}, y^{-1})=(x,y)^{-1}\in W$ . We conclude that H is a subgroup of $\operatorname {\mathrm {Hol}}(K)$ . As the projection onto its first component is surjective, it turns out that it H is a regular subgroup of $\operatorname {\mathrm {Hol}}(K)$ by [2, Proposition 2.5] and so it is isomorphic to the multiplicative group of a brace with additive group K (see [8, Theorem 4.2]).

We finish the proof by showing that the map $\phi \colon H\to W$ given by $(b, y)\longmapsto (\zeta (b)y, y)$ , where $\zeta (b)=xy^{-1}$ and $(x, y)\in W$ , is an isomorphism. Indeed, if $\zeta (b)=xy^{-1}$ , $\zeta (b_1)=x_1y_1^{-1}$ , where $(x,y)$ , $(x_1,y_1)\in W$ , we have that

$$ \begin{align*} \phi(b,y)\phi(b_1,y_1)&=(\zeta(b)y,y)(\zeta(b_1)y_1,y_1)=(x,y)(x_1,y_1)=(xx_1,yy_1),\\ \phi((b,y)(b_1,y_1))&=\phi(b+y(b_1),yy_1)=(\zeta(b+y(b_1))yy_1,yy_1)\\ &=(\zeta(b)y\zeta(b_1)y^{-1}yy_1,yy_1)=(xy^{-1}yx_1y_1^{-1}y_1,yy_1)\\ &=(xx_1,yy_1). \end{align*} $$

We conclude that $\phi $ is a group homomorphism. Assume that $\phi (b, y)=(\zeta (b)y,y)=(1,1)$ , with $\zeta (b)=xy^{-1}$ and $(x,y)\in W$ , then $y=1$ and so $\zeta (b)=x=1$ , which implies that $b=0$ . Consequently, $\phi $ is injective. As W and H are finite and have the same order, we obtain that $\phi $ is an isomorphism. Since C is isomorphic to H we have just proved that (e) holds for C.

3 A worked example

In general, we do not have that $X\cap Y=1$ . Let us consider $K=\operatorname {\mathrm {PSL}}_2(25)$ . Its automorphism group $A=\operatorname {\mathrm {Aut}}(K)$ is generated by $\operatorname {\mathrm {Inn}}(K)$ , the diagonal automorphism d induced by the conjugation by the matrix

$$\begin{align*}\mathsf{D}= \begin{bmatrix} \zeta&0\\ 0&1 \end{bmatrix}\in\operatorname{\mathrm{GL}}_2(25),\end{align*}$$

where $\zeta $ is a primitive $24$ th-root of unity of $\operatorname {\mathrm {GF}}(25)$ , and the field automorphism f. The group A possesses a subgroup X generated by the inner automorphisms $c_1$ , $c_2$ , and $c_3$ induced by the matrices

$$\begin{align*}\mathsf{C}_1= \begin{bmatrix} \zeta^{4}&0\\ 0&\zeta^{20} \end{bmatrix},\qquad\mathsf{C}_2= \begin{bmatrix} 1&0\\ \zeta&1 \end{bmatrix},\qquad \mathsf{C}_3= \begin{bmatrix} 1&0\\ 1&1 \end{bmatrix}, \end{align*}$$

respectively, and $df$ . We have that $c_1$ has order $3$ , $\langle c_2, c_3\rangle $ is an elementary abelian group of order $25$ , $c_1$ normalises $\langle c_2, c_3\rangle $ , $(df)c_1(df)^{-1}=c_1^{-1}$ , $df$ has order $8$ , and $df$ normalizes $\langle c_2, c_3\rangle $ . Then the group $\langle df, c_1, c_2, c_3\rangle $ has order $600$ .

Let $u_1$ and $u_2$ be the inner automorphisms induced by the conjugation by

$$\begin{align*}\mathsf{U_1}=\begin{bmatrix} \zeta^3&\zeta^{16}\\ \zeta^{13}&\zeta^{11} \end{bmatrix},\qquad \mathsf{U_2}= \begin{bmatrix} \zeta^{5}&\zeta^5\\ \zeta^9&\zeta^{22} \end{bmatrix}. \end{align*}$$

Let $Y=\langle u_1, dfu_2\rangle $ . We have that $u_1$ has order $13$ . Let

$$\begin{align*}\mathsf{R}= \begin{bmatrix} \zeta&0\\ 0&\zeta \end{bmatrix}\in\operatorname{\mathrm{Z}}(\operatorname{\mathrm{GL}}_2(25)),\qquad \mathsf{T}= \begin{bmatrix} 3&0\\ 4&2 \end{bmatrix} \end{align*}$$

and let t be the automorphism induced by conjugation by $\mathsf {T}$ . Then $(dfu_2)^2=dfu_2dfu_2=d^{f}{{u_2}}{}d^5u_2$ is the automorphism induced by conjugation by

(1) $$ \begin{align} \mathsf{D}\mathsf{U}_2^{(5)}\mathsf{D} ^5\mathsf{U}_2=\mathsf{R}^{15}\mathsf{T},\end{align} $$

where $\mathsf {U}_2^{(5)}$ denotes the matrix whose entries are obtained from the entries of $\mathsf {U}_2$ by applying the Frobenius field automorphism, that is, $(dfu_2)^2=t$ . As $(\mathsf {R}^{15}\mathsf {T})^2=\mathsf {R}^3$ , we conclude that $dfu_2$ has order $4$ . We can also check that $(dfu_2)u_1(dfu_2)^{-1}=u_1^8$ . It follows that Y has order $52$ .

By [5], X and Y are maximal subgroups of the almost simple group $\operatorname {\mathrm {Inn}}(K)\langle df\rangle $ . Observe that $(df)^4c_3^2=(dfdf)^2c_3^2=(dd^5)^2c_3^2=d^{12}c_3^2$ is induced by $\mathsf {D}^{12}\mathsf {C}_3^2=\mathsf {R}^{18}\mathsf {T}$ , consequently, $(df)^4c_3^2=t$ . This, together with Equation (1), shows that $t\in X\cap Y$ . We note that $\operatorname {\mathrm {Inn}}(K)X=\operatorname {\mathrm {Inn}}(K)Y=\operatorname {\mathrm {Inn}}(K)\langle df\rangle $ . Moreover, $\lvert {X\cap Y}\rvert $ divides $\gcd (\lvert X\rvert , \lvert Y\rvert )=4$ . If $\lvert X\cap Y\rvert = 4$ , then $X\cap Y$ is contained in $X\cap \operatorname {\mathrm {Inn}}(K)$ , but it is not contained in $Y\cap \operatorname {\mathrm {Inn}}(K)$ . This shows that $\lvert X\cap Y\rvert \le 2$ . Hence $\lvert X\cap Y\rvert = 2$ . As $XY\subseteq \operatorname {\mathrm {Inn}}(K)\langle df\rangle $ ,

$$\begin{align*}15\,600=\lvert \operatorname{\mathrm{Inn}}(K)\langle df\rangle\rvert\ge \lvert XY\rvert=\frac{\lvert X\rvert \lvert Y\rvert}{\lvert X\cap Y\rvert}=15\,600\cdot \frac{2}{\lvert X\cap Y\rvert}=15\,600,\end{align*}$$

and so $XY=\operatorname {\mathrm {Inn}}(K)\langle df\rangle $ .

Let $N=\langle c_1, c_2, c_3, (df)^2\rangle \trianglelefteq X$ , $M=\langle u_1\rangle \trianglelefteq Y$ . Then $\lvert N\rvert =150$ , $\lvert M\rvert =13$ , $N\le X\cap \operatorname {\mathrm {Inn}}(K)$ , $M\le Y\cap \operatorname {\mathrm {Inn}}(K)$ , $Y/M\cong X/N\cong C_4$ , and $\lvert K\rvert =\lvert X\rvert \lvert M\rvert = \lvert Y\rvert \lvert N\rvert $ . The isomorphism between $Y/M$ and $X/N$ is given by $\gamma ((dfu_2)^rM)=(df)^rN$ for $0\le r < 4$ , and, since $d^6\in \operatorname {\mathrm {Inn}}(K)$ , it is clear that

$$ \begin{align*} (df)(dfu_2)^{-1}&=c_0u_0^{-1}\in \operatorname{\mathrm{Inn}}(K),\\ (df)^2(dfu_2)^{-2}&=d^6t^{-1}\in \operatorname{\mathrm{Inn}}(K),\\ (df)^3(dfu_2)^{-3}&=(df)(d^6t^{-1}u_2^{-1})(df)^{-1}\in\operatorname{\mathrm{Inn}}(K). \end{align*} $$

Let $z\in XY\cap \operatorname {\mathrm {Inn}}(K)$ . Recall that $X\cap Y=\langle t\rangle $ . Then there exist $x\in X$ , $y\in Y$ with $z=xy^{-1}=(xt)(yt)^{-1}$ . We observe that $t=(df)^4c_3^2\in N$ , but $t\notin M$ by order considerations. Given $x\in X$ , $y\in Y$ , there exist r, $s\in \{0,1,2,3\}$ such that $xN=(df)^rN$ and $yM=(dfc_3)^sM$ . We also observe that $x\in \operatorname {\mathrm {Inn}}(K)$ if, and only if, $y\in \operatorname {\mathrm {Inn}}(K)$ . To prove that we can choose $x\in X$ , $y\in Y$ such that $z=xy^{-1}$ and $\gamma (yM)=xN$ , it is enough to prove that for such a choice we have that $z=xy^{-1}$ and $r=s$ . Note that if $x\in N$ , then $r=0$ ; if $x\in \operatorname {\mathrm {Inn}}(K)\setminus N$ , then $r=2$ ; and if $x\notin \operatorname {\mathrm {Inn}}(K)$ , then $r\in \{1,3\}$ . Analogously, if $y\in M$ , then $s=0$ ; if $y\in \operatorname {\mathrm {Inn}}(K)\setminus M$ , then $s=2$ ; and if $y\notin \operatorname {\mathrm {Inn}}(K)$ , then $s\in \{1,3\}$ . We also have that $tM=(dfu_2)^2M$ and that $tN=N$ , as $t\in \operatorname {\mathrm {Inn}}(K)$ , $t\in N$ , but $t\notin M$ . If $x\in N$ and $y\in M$ , we can choose $r=s=0$ and $\gamma (yM)=xN$ . Suppose that $x\in N$ and $y\notin M$ . Then $y\in \operatorname {\mathrm {Inn}}(K)$ and so, $xN=N$ and $yM=(dfu_2)^2M$ . Consequently, $xtN=N$ , $ytN=N$ , and $\gamma (ytN)=xtN$ . Suppose that $x\notin N$ and $y\in M$ . We have that $x\in \operatorname {\mathrm {Inn}}(K)$ and so, $xN=(df)^2N$ and $yM=M$ . It follows that $xtN=(df)^2N$ and $ytM=(dfu_2)^2M$ , that is, $\gamma (ytM)=xtN$ . Suppose that x, $y\in \operatorname {\mathrm {Inn}}(K)$ , $x\notin N$ , and $y\notin M$ . Then $xN=(df)^2N$ , $yM=(dfu_2)^2M$ , and $\gamma (yM)=xN$ . Finally, suppose that x and $y\notin M$ . Then $xN=(df)^rN$ and $yM=(dfu_2)^sM$ , with r, $s\in \{1,3\}$ . If $r=s$ , then $\gamma (yM)=xN$ . If $r\ne s$ , then $xtN=(df)^rN$ and $ytM=(dfu_2)^{s+2}M$ , with $r\equiv s+2\pmod {4}$ . Thus $\gamma (ytM)=xtN$ .

It follows that X, Y satisfy all conditions of Theorem A. We can also check with GAP [7] all this information about these subgroups.