# Constructing skew left braces whose additive group has trivial centre

A. Ballester-Bolinches Departament de Matemàtiques, Universitat de València,
Dr. Moliner, 50, 46100 Burjassot, València, Spain<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>ORCID 0000-0002-2051-9075,
0000-0002-2321-8139, 0000-0003-4101-8656 R. Esteban-Romero00footnotemark: 0 P.
Jiménez-Seral Departamento de Matemáticas, Universidad de Zaragoza, Pedro
Cerbuna, 12, 50009 Zaragoza, Spain<EMAIL_ADDRESS>ORCID 0000-0003-4809-1784
V. Pérez-Calabuig-1-1footnotemark: -1

###### Abstract

A complete description of all possible multiplicative groups of finite skew
left braces whose additive group has trivial centre is shown. 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.

_Keywords: skew left brace, trifactorised group, trivial centre_

_Mathematics Subject Classification (2020): 16T25, 81R50, 20C35, 20C99,
20D40._

##  1 Introduction

An important algebraic structure that plays a key role in the combinatorial
theory of Yang-Baxter equation is the skew brace structure. Skew left braces,
introduced in [10], can be regarded as extensions of Jacobson radical rings
and show connections with several areas of mathematics such as triply
factorised groups and Hopf-Galois structures (see [1, 4, 5])

Skew left braces classify solutions of the Yang-Baxter equation (see [8, 10]).
This justifies the search for constructions of skew braces and classification
results.

Recall that a _skew left brace_ is a set endowed with two group structures
$(B,+)$ 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_ :

$\lambda\colon
a\in(B,\cdot)\longmapsto\lambda_{a}\in\operatorname{Aut}(B,+)),\quad\lambda_{a}(b)=-a+ab,\
\text{for all $a,b\in B$.}$

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 $\operatorname{Hol}(G)=[G]\operatorname{Aut}(G)$. 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 $\operatorname{Hol}(K)$ isomorphic to $(B,{\cdot})$
(see [10, Theorem 4.2]). If we consider the subgroup
$S=KH\leq\operatorname{Hol}(K)$, then

$S=KH=KE=HE,$

where $E=\\{(0,\lambda_{b})\mid b\in B\\}$ and $\operatorname{C}_{E}(K)=K\cap
E=H\cap E=1$. We call $\mathsf{S}(B)=(S,K,H,E)$ the _small trifactorised
group_ associated with $B$.

In [12], 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 [12, 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
$\operatorname{Aut}(K)$ that are quotients of $G$ satisfying

$XY=X{\operatorname{Inn}(K)}=Y{\operatorname{Inn}(K)}.$

She looks 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 [12, Proposition 2.7]).

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

The above two theorems are the key to prove the main results of [12, 13].

In [14], Tsang posed the following question:

######  Question 3.

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

The aim of this paper is to give a complete characterisation 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 [12] 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)

$XY=X{\operatorname{Inn}(K)}=Y{\operatorname{Inn}(K)}$,

  2. (b)

$\operatorname{Inn}(K)$ has two subgroups $N$ and $M$ such that
$N\trianglelefteq X$ and $M\trianglelefteq Y$ and there exists an isomorphism
$\gamma\colon Y/M\longrightarrow X/N$ such that whenever $\gamma(yM)=xN$, it
follows that $xy^{-1}\in\operatorname{Inn}(K)$. Furthermore, if
$z\in\operatorname{Inn}(K)$, there exist $x\in X$, $y\in Y$ such that
$xN=\gamma(yM)$ and $xy^{-1}=z$,

  3. (c)

$\lvert K\rvert=\lvert X\rvert\lvert M\rvert=\lvert Y\rvert\lvert N\rvert$.

In this case,

  1. (d)

$C$ has two normal subgroups $T$ and $V$ with $T\cap V=1$, $X\cong C/T$ and
$Y\cong C/V$.

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

######  Corollary 4.

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

###### Proof.

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

$Y/M\cong
Y{\operatorname{Inn}(K)}/{\operatorname{Inn}(K)}=X{\operatorname{Inn}(K)}/{\operatorname{Inn}(K)}\cong
X/N,$

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
$ab^{-1}\in\operatorname{Inn}(K)$. Observe that $a$ and $b$ are unique as
$X\cap Y=1$. Then, the groups $X$ and $Y$ satisfy Statements (a)–(c) of
Theorem A, and therefore, there exists a brace whose additive group is $K$ and
a multiplicative group isomorphic to a subdirect product of $X$ and $Y$. ∎

Corollary 4 also allows to give a considerably shorter proof of the main
results of [12, 13] 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 $\operatorname{Aut}(K)$ such
that $X\cap Y=1$ and $XY=X{\operatorname{Inn}(K)}=Y{\operatorname{Inn}(K)}$.
Therefore, Codes 2, 3, and 4 in the proof of [12, Theorem 1.3] can be avoid,
as well as checking in every case that the subgroup $X$ splits over
$X\cap{\operatorname{Inn}(K)}$.

In Section 3 we present a worked example of a construction of a brace with
additive group $K=\operatorname{PSL}_{2}(25)$ by means of subgroups $X$ and
$Y$ of $\operatorname{Aut}(K)$ 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 $H=\\{(b,\lambda_{b})\\!\mid b\in B\\}$ the regular subgroup of
$\operatorname{Hol}(K)$ appearing in the small trifactorised group
$\mathsf{S}(B)=(S,K,H,E)$ associated with $B$. Recall that
$E=\\{(0,\lambda_{b})\\!\mid b\in B\\}\leq\operatorname{Hol}(K)$, and
$S=KH=KE=HE$ with $K\cap E=H\cap E=1$.

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

---
$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(X)}$$\textstyle{(H\cap
K){\operatorname{C}_{H}(K)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(Y)}$$\textstyle{{\operatorname{C}_{H}(K)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H\cap
K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1}$ 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 $\operatorname{Ker}{\pi}|_{H}=K\cap
H\trianglelefteq H$ and
$\operatorname{Ker}{\alpha|_{H}}=\operatorname{C}_{H}(K)\trianglelefteq H$.
Moreover, it holds that

$\displaystyle\operatorname{Ker}{\pi}|_{H}\cap\operatorname{Ker}{\alpha|_{H}}=K\cap
H\cap\operatorname{C}_{H}(K)=K\cap H\cap\operatorname{C}_{S}(K)=$
$\displaystyle=H\cap\operatorname{C}_{K}(K)=H\cap\operatorname{Z}(K)=1\quad\text{(see
Figure\leavevmode\nobreak\ \ref{fig-subdirect}).}$

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

$\displaystyle\alpha(G)$ $\displaystyle=\alpha(HE)=\alpha(KH)=\alpha(KE)$
$\displaystyle=\alpha(H)\alpha(E)=\alpha(K)\alpha(H)=\alpha(K)\alpha(E)$
$\displaystyle=XY=(\operatorname{Inn}(K))X=(\operatorname{Inn}(K))Y.$

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

$(b,\lambda_{b})(k,1)(b,\lambda_{b})^{-1}=(b+\lambda_{b}(k)-b,1)=(k,1),$

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

$\displaystyle Y/M$
$\displaystyle\cong(H/\operatorname{Ker}{\pi}|_{H})/(R/\operatorname{Ker}{\pi}|_{H})\cong
H/R$
$\displaystyle\cong(H/\operatorname{Ker}\alpha|_{H})/(R/\operatorname{Ker}\alpha|_{H})\cong
X/N;$

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 $a\in\gamma(\lambda_{b}M)$, we have
that $a\lambda_{b}^{-1}\in\alpha_{b}N\subseteq\operatorname{Inn}(K)$.
Furthermore, given $x\in\operatorname{Inn}(K)$, 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 $\operatorname{Ker}{\pi|}_{H}\cap{\operatorname{Ker}\alpha|_{H}}=(H\cap
K)\cap\operatorname{C}_{H}(K)=1$, we have that $\lvert R\rvert={\lvert H\cap
K\rvert}\lvert{\operatorname{C}_{H}(K)}\rvert$ and $\lvert N\rvert=\lvert
R/(H\cap K)\rvert=\lvert\operatorname{C}_{H}(K)\rvert$, $\lvert M\rvert=\lvert
R/{\operatorname{C}_{H}(K)}\rvert=\lvert H\cap K\rvert$. As $\lvert
X\rvert=\lvert K\rvert/\lvert H\cap K\rvert$ and $\lvert Y\rvert=\lvert
K\rvert/\lvert{\operatorname{C}_{H}(K)}\rvert$, the claim about the order
follows.

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

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

be a subdirect product of $X$ and $Y$ with amalgamated factor group $Y/M\cong
X/N$ (see [7, Chapter A, Definition 19.2]). By [7, Chapter A, Proposition
19.1], and the hypothesis, we have that $\lvert W\rvert=\lvert K\rvert$. Since
$\operatorname{Z}(K)$ is trivial, the map $\zeta\colon
K\longrightarrow\operatorname{Inn}(K)$, where $\zeta(k)$ is the inner
automorphism of $K$ induced by $k$, is an isomorphism. By hypothesis, the map
$W\longrightarrow\operatorname{Inn}(K)$ given by $(x,y)\longmapsto xy^{-1}$ is
surjective. Since $\lvert W\rvert=\lvert\operatorname{Inn}(K)\rvert=\lvert
K\rvert$, it is a bijection. We can consider $H=\\{(b,y)\mid(x,y)\in
W,\,\zeta(b)=xy^{-1}\\}\subseteq\operatorname{Hol}(K)$. 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 $\zeta(b_{1})=x_{1}y_{1}^{-1}$ with $(x,y)$,
$(x_{1},y_{1})\in B$. Then

$\zeta(b+y(b_{1}))=\zeta(b)\zeta(y(b_{1}))=\zeta(b)y\zeta(b_{1})y^{-1}=xy^{-1}yx_{1}y_{1}^{-1}y^{-1}=(xx_{1})(yy_{1})^{-1}$

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

$\zeta(y^{-1}(-b))=y^{-1}\zeta(-b)y=y^{-1}\zeta(b)^{-1}y=y^{-1}yx^{-1}y=x^{-1}{(y^{-1})}^{-1}$

with $(x^{-1},y^{-1})=(x,y)^{-1}\in W$. We conclude that $H$ is a subgroup of
$\operatorname{Hol}(K)$. As the projection onto its first component is
surjective, it turns out that it $H$ is a regular subgroup of
$\operatorname{Hol}(G)$ by [2, Proposition 2.5] and so it is isomorphic to the
multiplicative group of a brace with additive group $K$ (see [10, Theorem
4.2]).

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

$\displaystyle\gamma(b,y)\gamma(b_{1},y_{1})$
$\displaystyle=(\zeta(b)y,y)(\zeta(b_{1})y_{1},y_{1})=(x,y)(x_{1},y_{1})=(xx_{1},yy_{1}),$
$\displaystyle\gamma((b,y)(b_{1},y_{1}))$
$\displaystyle=\gamma(b+y(b_{1}),yy_{1})=(\zeta(b+y(b_{1}))yy_{1},yy_{1})$
$\displaystyle=(\zeta(b)y\zeta(b_{1})y^{-1}yy_{1},yy_{1})=(xy^{-1}yx_{1}y_{1}^{-1}y_{1},yy_{1})$
$\displaystyle=(xx_{1},yy_{1}).$

We conclude that $\gamma$ is a group homomorphism. Assume that
$\gamma(b,y)=(\zeta(b)y,y)=(1,1)$, with $\zeta(b)=xy^{-1}$ and $(x,y)\in H$,
then $y=1$ and so $\zeta(b)=x=1$, what implies that $b=0$. Consequently,
$\gamma$ is injective. As $W$ and $H$ are finite and have the same order, we
obtain that $\gamma$ is an isomorphism. ∎

##  3 A worked example

In general, we do not have that $X\cap Y=1$. Let us consider
$K=\operatorname{PSL}_{2}(25)$. Its automorphism group
$A=\operatorname{Aut}(K)$ is generated by $\operatorname{Inn}(K)$, the
diagonal automorphism $d$ induced by the conjugation by the matrix

$\mathsf{D}=\begin{bmatrix}\zeta&0\\\
0&1\end{bmatrix}\in\operatorname{GL}_{2}(25),$

where $\zeta$ is a primitive $24$th-root of unity of $\operatorname{GF}(25)$,
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

$\mathsf{C}_{1}=\begin{bmatrix}\zeta^{11}&0\\\
\zeta^{14}&\zeta^{13}\end{bmatrix},\qquad\mathsf{C}_{2}=\begin{bmatrix}1&0\\\
\zeta&1\end{bmatrix},\qquad\mathsf{C}_{3}=\begin{bmatrix}1&0\\\
1&1\end{bmatrix}$

respectively, and, if $c_{0}$ is the inner automorphism of $K$ induced by

$\mathsf{C}_{0}=\begin{bmatrix}\zeta^{7}&0\\\
\zeta^{17}&\zeta^{17}\end{bmatrix},$

the element $c_{0}fd$. We have that $c_{1}$ has order $12$, $\langle
c_{2},c_{3}\rangle$ is an elementary abelian group of order $25$, and
$(c_{0}fd)^{2}$ is induced by the matrix $\mathsf{R}^{2}\mathsf{C}_{1}^{3}$ of
order $8$, with

$\mathsf{R}=\begin{bmatrix}\zeta&0\\\
0&\zeta\end{bmatrix}\in\operatorname{Z}(\operatorname{GL}_{2}(25)).$

The group $\langle c_{1},c_{2},c_{3},c_{0}fd\rangle$ has order $600$.

It also possesses a subgroup $Y$ generated by the inner automorphisms $u_{1}$
and $u_{2}$ induced by the conjugation by

$\mathsf{U_{1}}=\begin{bmatrix}3&\zeta^{21}\\\
\zeta^{14}&\zeta^{22}\end{bmatrix},\qquad\mathsf{U_{2}}=\begin{bmatrix}3&2\\\
0&2\end{bmatrix},$

respectively, and $u_{0}fd$, where $u_{0}$ is the inner automorphism induced
by the conjugation by

$\mathsf{U_{0}}=\begin{bmatrix}\zeta&4\\\ \zeta^{7}&\zeta^{8}\end{bmatrix}.$

Then $(u_{0}fd)^{2}$ is induced by $\mathsf{R}^{12}\mathsf{U}_{2}$, that is,
$(u_{0}fd)^{2}=u_{2}$, $(u_{0}fd)^{2}$ has order $2$, $u_{1}$ has order $13$,
and $\langle u_{1},(u_{0}fd)^{2}\rangle$ has order $52$. By [6], $X$ and $Y$
are maximal subgroups of the almost simple group $\operatorname{Inn}(K)\langle
fd\rangle$.

We note that
$\operatorname{Inn}(K)X=\operatorname{Inn}(K)Y=\operatorname{Inn}(K)\langle
fd\rangle$. Moreover, $\lvert{X\cap Y}\rvert$ divides $\gcd(\lvert
X\rvert,\lvert Y\rvert)=4$ and, since $X$ and $Y$ have cyclic Sylow
$2$-subgroups, $X\cap Y$ must be cyclic and so $\lvert X\cap Y\rvert\leq 4$.
But if $\lvert X\cap Y\rvert=4$, then $X\cap Y$ is contained in
$X\cap\operatorname{Inn}(K)$, but it is not contained in
$Y\cap\operatorname{Inn}(K)$. This shows that $\lvert X\cap Y\rvert\leq 2$. As
$XY\subseteq\operatorname{Inn}(K)\langle fd\rangle$,

$15\,600=\lvert\operatorname{Inn}(K)\langle fd\rangle\rvert\geq\lvert
XY\rvert=\frac{\lvert X\rvert\lvert Y\rvert}{\lvert X\cap
Y\rvert}=15\,600\cdot\frac{2}{\lvert X\cap Y\rvert}\geq 15\,600,$

and so $XY=\operatorname{Inn}(K)\langle fd\rangle$ and $\lvert X\cap
Y\rvert=2$. In fact, the inner automorphism $t$ induced by the matrix

$\mathsf{T}=\begin{bmatrix}2&\zeta^{8}\\\ 0&3\end{bmatrix}$

belongs to $X\cap Y$, as
$\mathsf{T}=\mathsf{C}_{1}^{6}\mathsf{C}_{2}\mathsf{C}_{3}$ induces by
conjugation a non-trivial automorphism of $X$ and
$\mathsf{T}=\mathsf{U}_{1}^{3}\mathsf{U}_{2}^{3}$ induces a non-trivial
automorphism by conjugation on $\operatorname{SL}_{2}(25)$ that belongs to
$Y$.

Let $N=\langle c_{2},c_{3},c_{1}^{2}\rangle\trianglelefteq X$, $M=\langle
u_{1}\rangle\trianglelefteq Y$. Then $\lvert N\rvert=150$, $\lvert
M\rvert=13$, $N\leq X\cap\operatorname{Inn}(K)$, $M\leq
Y\cap\operatorname{Inn}(K)$, $Y/M\cong X/N\cong C_{4}$, 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
$\gamma((u_{0}fd)^{r}M)=(c_{0}fd)^{r}N$ for $0\leq r\leq 4$, and it is clear
that

$\displaystyle(c_{0}fd)(u_{0}fd)^{-1}=c_{0}u_{0}^{-1}\in\operatorname{Inn}(K),$
$\displaystyle(c_{0}fd)^{2}(u_{0}fd)^{-2}\in\operatorname{Inn}(K),$
$\displaystyle(c_{0}fd)^{3}(u_{0}fd)^{-3}=(c_{0}fd)^{2}(c_{0}u_{0}^{-1})(u_{0}fd)^{-2}\in\operatorname{Inn}(K).$

Let $z\in XY\cap\operatorname{Inn}(K)$. Then there exist $x\in X$, $y\in Y$
with $z=xy^{-1}=(xt)(yt)^{-1}$, where $\langle t\rangle=X\cap Y$. We observe
that $t\in N$, as $\mathsf{T}=\mathsf{C}_{1}^{6}\mathsf{C}_{2}\mathsf{C}_{3}$,
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 $xN=(c_{0}fd)^{r}N$ and
$yM=(u_{0}fd)^{s}M$. We also observe that $x\in\operatorname{Inn}(K)$ if, and
only if, $y\in\operatorname{Inn}(K)$. 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 $x\in\operatorname{Inn}(K)\setminus N$, then $r=2$; and if
$x\notin\operatorname{Inn}(K)$, then $r\in\\{1,3\\}$. Analogously, if $y\in
M$, then $s=0$; if $y\in\operatorname{Inn}(K)\setminus M$, then $s=2$; and if
$y\notin\operatorname{Inn}(K)$, then $s\in\\{1,3\\}$. We also have that
$tM=(u_{0}fd)^{2}M$ and that $tN=N$, as $t\in\operatorname{Inn}(K)$, $t\in N$,
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
$y\in\operatorname{Inn}(K)$ and so, $xN=N$ and $yM=(u_{0}fd)^{2}M$.
Consequently, $xtN=N$, $ytN=N$, and $\gamma(ytN)=xtN$. Suppose that $x\notin
N$ and $y\in M$. We have that $x\in\operatorname{Inn}(K)$ and so,
$xN=(c_{0}fd)^{2}N$ and $yM=M$. It follows that $xtN=(c_{0}fd)^{2}N$ and
$ytM=(u_{0}fd)^{2}M$, that is, $\gamma(ytM)=xtN$. Suppose that $x$,
$y\in\operatorname{Inn}(K)$, $x\notin N$, and $y\notin M$. Then
$xN=(c_{0}fd)^{2}N$, $yM=(u_{0}fd)^{2}M$, and $\gamma(yM)=xN$. Finally,
suppose that $x$ and $y\notin M$. Then $xN=(c_{0}fd)^{r}N$ and
$yM=(u_{0}fd)^{s}M$, with $r$, $s\in\\{1,3\\}$. If $r=s$, then
$\gamma(yM)=xN$. If $r\neq s$, then $xtN=(c_{0}fd)^{r}N$ and
$ytM=(u_{0}fd)^{s+2}M$, 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 [9] all this information about these subgroups.

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