1 Introduction
Skew brace structure plays a key role in the combinatorial theory of the Yang–Baxter equation. Skew left braces, introduced in [Reference Guarnieri and Vendramin8], 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 [Reference Ballester-Bolinches and Esteban-Romero1, Reference Caranti and Stefanello3, Reference Childs4])
Skew left braces classify solutions of the Yang–Baxter equation (see [Reference Guarnieri and Vendramin8]). 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 [Reference Guarnieri and Vendramin8, 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 [Reference Tsang9], 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 [Reference Tsang9, 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 [Reference Tsang9, 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 [Reference Tsang9, Reference Tsang10].
In [Reference Tsang11], 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 [Reference Tsang9] 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:
-
(a)
$XY=X{\operatorname {\mathrm {Inn}}(K)}=Y{\operatorname {\mathrm {Inn}}(K)}$
, -
(b) there are two subgroups N and M of
$\operatorname {\mathrm {Inn}}(K)$
such that
$N \unlhd X$
and
$M \unlhd Y$
, -
(c) there exists an isomorphism
$\gamma \colon Y/M\longrightarrow X/N$
such that
$$\begin{align*}\operatorname{\mathrm{Inn}}(K) = \{xy^{-1}\mid x\in X, \, y\in Y,\, \gamma(yM) = xN\},\end{align*}$$
-
(d)
$\lvert K\rvert =\lvert X\rvert \lvert M\rvert =\lvert Y\rvert \lvert N\rvert $
.
In this case,
-
(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 [Reference Tsang9, Reference Tsang10] 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 [Reference Tsang9, 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 [Reference Doerk and Hawkes6, Chapter A, Definition 19.2]). By [Reference Doerk and Hawkes6, 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 [Reference Ballester-Bolinches, Esteban-Romero and Pérez-Calabuig2, Proposition 2.5] and so it is isomorphic to the multiplicative group of a brace with additive group K (see [Reference Guarnieri and Vendramin8, 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
$$ \begin{align} \mathsf{D}\mathsf{U}_2^{(5)}\mathsf{D} ^5\mathsf{U}_2=\mathsf{R}^{15}\mathsf{T},\end{align} $$
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 [Reference Conway, Curtis, Norton, Parker and Wilson5], 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.
Acknowledgements
We would like to thank the referee for his thoughtful comments toward improving our article.
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