Proof. Let \(M\leq S\times T\) and \(N\trianglelefteq M\) be such that \(M/N\cong L.\) Set \[K_S=M\cap (S\times 1), \qquad K_T=M\cap (1\times T).\] Since \(K_S\trianglelefteq M\) and \(K_T\trianglelefteq M\) we have that \(K_SN/N\) and \(K_TN/N\) are normal subgroups of \(M/N\). Since \(M/N\cong L\) and \(L\) is simple, each of these two normal subgroups is either trivial or equal to \(M/N\). Moreover, the subgroups \(K_S\) and \(K_T\) commute elementwise, because \(K_S\leq S\times 1\) and \(K_T\leq 1\times T\). Hence their images \(K_SN/N\) and \(K_TN/N\) commute elementwise in \(M/N\). Since \(L\) is non-abelian, these two images cannot both be equal to \(M/N\). Assume first that \(K_SN/N=M/N.\) Then \(M=K_SN\). By the second isomorphism theorem, we obtain \[M/N=K_SN/N\cong K_S/(K_S\cap N).\] Since \(K_S\leq S\times 1\cong S\), the quotient \(K_S/(K_S\cap N)\) is a section of \(S\). Thus \(L\) is isomorphic to a section of \(S\). Similarly, if \(K_TN/N=M/N,\) then \(M=K_TN\), and again by the second isomorphism theorem, \[M/N=K_TN/N\cong K_T/(K_T\cap N).\] Since \(K_T\leq 1\times T\cong T\), this shows that \(L\) is isomorphic to a section of \(T\). It remains to consider the case \(K_SN/N=1\) and \(K_TN/N=1.\) Equivalently, \(K_S\leq N\) and \(K_T\leq N.\) Let \(\pi_S:S\times T\to S\) and \(\pi_T:S\times T\to T\) be the two canonical projections. We restrict them to \(M\). Then \[\ker(\pi_S|_M)=M\cap (1\times T)=K_T\leq N,\] and \[\ker(\pi_T|_M)=M\cap (S\times 1)=K_S\leq N.\]

We claim that \(\pi_S|_M\) induces an isomorphism \[M/N\cong \pi_S(M)/\pi_S(N).\] Indeed, since \(\pi_S(N)\trianglelefteq \pi_S(M)\), the map \[\overline{\pi}_S:M/N\longrightarrow \pi_S(M)/\pi_S(N), \qquad mN\longmapsto \pi_S(m)\pi_S(N),\] is well defined and surjective. Its kernel is \[\ker \overline{\pi}_S = \{\,mN\in M/N : \pi_S(m)\in \pi_S(N)\,\}.\] If \(mN\in \ker\overline{\pi}_S\), then there exists \(n\in N\) such that \(\pi_S(m)=\pi_S(n).\) Hence \(mn^{-1}\in \ker(\pi_S|_M)=K_T\leq N.\) Since \(n\in N\), it follows that \(m\in N\). Therefore \(\ker\overline{\pi}_S=1.\) Thus \[M/N\cong \pi_S(M)/\pi_S(N),\] which is a section of \(S\). Analogously, using the projection \(\pi_T\), one obtains \[M/N\cong \pi_T(M)/\pi_T(N),\] which is a section of \(T\). Therefore, also in this last case, \(L\) is isomorphic to a section of \(S\) and to a section of \(T\). Hence, in all cases, \(L\) is isomorphic to a section of \(S\) or to a section of \(T\), as required. ◻

Proof. This is the consequence of Byott’s result used in [7]. ◻

Proof. Write \(G=G_1\times \cdots \times G_n\) where \(G_i\cong \operatorname{PSL}_{2}(7)\) for each \(i\). Let \(H\leq G\) be a subgroup such that \([G:H]=3^m\). If \(G_i\leq H\) for every \(i\), then \[G=\langle G_1,\dots,G_n\rangle\leq H,\] and hence \(H=G\), a contradiction. Therefore \(G_j\nleq H\) for some \(j\). It follows that \(G_j\cap H < G_j,\) so \([G_j:G_j\cap H]>1\). On the other hand, \([G_j:G_j\cap H]=[G_jH:H]\) so this index divides \([G:H]=3^m\). Hence \([G_j:G_j\cap H]\) is a non-trivial power of \(3\). This is impossible, since \(\operatorname{PSL}_{2}(7)\) has no subgroups of index equal to a non-trivial power of \(3\). Therefore no such subgroup \(H\) exists. ◻

Proof. Since \((B,+)\) is solvable, there is \(n\in \mathbb{N}_{>0}\) such that \((B,+)^{(n)}\leq H\). By [8] there is \(m\in \mathbb{N}_{>0}\) such that \((B,\cdot)^{(n+m)}\leq H\). ◻

Proof of Theorem 1. Suppose, for contradiction, that the statement is false, and let \(B\) be a counterexample. Write \(|B|=2^\alpha m,\) with \(\alpha\in \mathbb{N}_{>0}\) and \(m\in \mathbb{N}_{>0}\) odd. Since \((B,+)\) is supersolvable, [15] ensures the existence of a characteristic subgroup \(H\leq (B,+)\) of order \(m\). Hence \((H,\cdot)\) is a \(2\)-complement of \((B,\cdot)\). Write \((B,\cdot)=G_1\times G_2,\) where \(G_{1} \simeq S\) and \(G_{2} \simeq T\). Let \(i \in \{1,2\}\). Since each \(G_i\) is normal in \((B,\cdot)\), we have \(G_iH\leq (B,\cdot)\), and \([G_i:G_i\cap H]=[G_iH:H]\) is a power of \(2\). If \(G_i\leq H\) for some \(i\), then \(G_i\) has odd order. By the Feit–Thompson theorem (see the main Theorem of [16]), \(G_i\) is solvable, a contradiction. Therefore \(G_j\nleq H\) for every \(j\), and so \(G_j\) has a subgroup of index a power of \(2\). Since \(G_j\cap H\leq H\), it follows that \(G_j\cap H\) is a \(2\)-complement of \(G_j\). Thus \(S\) and \(T\) are non-abelian simple groups admitting a \(2\)-complement. By [5], this implies that \[S\cong \mathop{\mathrm{PSL}}_n(q), \qquad\text{where } \frac{q^n-1}{q-1} = 2^{t}\] for some \(t \in \mathbb{N}\) and \[T\cong \mathop{\mathrm{PSL}}_m(q'), \qquad\text{where } \frac{(q')^m-1}{q'-1} = 2^{t'}\] for some \(t'\in \mathbb{N}\). By [7], \((B,\cdot)\) has a section isomorphic to \(\operatorname{PSL}_{2}(7)\). By Lemma 1 we can assume without loss of generality that \(S\) also has a section isomorphic to \(\operatorname{PSL}_{2}(7)\). Write \(q=p^f\) for a prime \(p\) and a positive integer \(f\). Reducing modulo \(2\) we see that \(p\) has to be odd. Dividing both sides by \(p^f-1\), we obtain \[\frac{p^{fn}-1}{p^f-1}=2^t.\] The left-hand side is a geometric series of \(n\) terms with ratio \(p^f\): \[(p^f)^{n-1}+(p^f)^{n-2}+\cdots+p^f+1=2^t.\] Since \(p^f\) is odd, each term of the sum is odd. Hence, in order for the sum to be a power of \(2\), the number of terms \(n\) must be even. Let \(n=2k\). Then \[p^{2fk}-1=(p^{fk}-1)(p^{fk}+1),\] so that the equation becomes \[\frac{(p^{fk}-1)(p^{fk}+1)}{p^f-1}=2^t \qquad\Longrightarrow\qquad \left(\frac{p^{fk}-1}{p^f-1}\right)(p^{fk}+1)=2^t.\] For the product of two integers to be a power of \(2\), both factors must themselves be powers of \(2\). In particular, we must have \[2^s-p^{fk}=1\] for some integer \(s\leq t\). By the Catalan–Mihăilescu theorem (see the main theorem of [17]) we must have \(fk=1\). Hence \(f=1\), \(k=1\), and \(n=2\). Hence \[S\cong \mathop{\mathrm{PSL}}_2(p).\] Similarly, writing \(q'=r^{f'}\) we get: \[T\cong \mathop{\mathrm{PSL}}_2(r),\] where \(r+1\) is a power of \(2\). Now let \(K\leq S\) and \(N\trianglelefteq K\) be such that \[K/N\cong \operatorname{PSL}_{2}(7).\] Since \(\operatorname{PSL}_{2}(7)\) is non-solvable, the group \(K\) is also non-solvable. By the classification of the subgroups of \(\mathop{\mathrm{PSL}}_2(q)\) (see [15]), it follows that \(K\) must itself be simple, and hence \(N=1\). Therefore, \[\operatorname{PSL}_{2}(7)\leq \mathop{\mathrm{PSL}}_2(p).\] Applying [15] once again, we conclude that \(p=7\). Thus \(S \simeq \operatorname{PSL}_{2}(7)\). We distinguish two cases separately:

The case r=7. In this case \[(B,\cdot)\cong \operatorname{PSL}_2(7)\times \operatorname{PSL}_2(7),\] Since \(|\operatorname{PSL}_{2}(7)|=2^3\cdot 3\cdot 7\), we have \(v_3(|B|)=2\). Assume that \[3\mid [(B,+):(B,+)'].\] Then by [5] there exists a characteristic subgroup \(K\) of \((B,+)\) such that \[(B,+)/K\cong C_3^m, \qquad m\in \{1,2\}.\] Hence \((B,\cdot)\) admits a subgroup of index \(3^m\), which is impossible by Lemma 3. Therefore \[3\nmid [(B,+):(B,+)'].\] Thus every Sylow \(3\)-subgroup of \((B,+)\) is contained in \((B,+)'\). By [15], the derived subgroup \((B,+)'\) is nilpotent. Hence \((B,+)\) has a normal Sylow \(3\)-subgroup. Let \(P_3\) denote the Sylow \(3\)-subgroup of \((B,+)\). Since \(P_3\) is the unique Sylow \(3\)-subgroup of \((B,+)\), it is characteristic in \((B,+)\). Therefore its centralizer \(C_{(B,+)}(P_3)\) is also characteristic in \((B,+)\). Thus, by [18], \[(B,+)/C_{(B,+)}(P_3)\leq \mathop{\mathrm{Aut}}(P_3).\] We now show that \(\mathop{\mathrm{Aut}}((B,+)/C_{(B,+)}(P_3))\) is solvable. If \(P_3\) is cyclic, then \[\mathop{\mathrm{Aut}}(P_3)\cong \mathop{\mathrm{Aut}}(C_9)\cong C_6,\] and therefore \((B,+)/C_{(B,+)}(P_3)\) is cyclic and \(\mathop{\mathrm{Aut}}((B,+)/C_{(B,+)}(P_3))\) is solvable. Suppose instead that \[P_3\cong C_3\times C_3.\] Then \[\mathop{\mathrm{Aut}}(P_3)\cong \mathop{\mathrm{GL}}_2(3).\] Thus \((B,+)/C_{(B,+)}(P_3)\) is a subgroup of \(\mathop{\mathrm{GL}}_2(3)\). Using GAP [19], one checks that \(\mathop{\mathrm{Aut}}(H)\) is solvable for every subgroup \(H\leq \mathop{\mathrm{GL}}_2(3)\). Thus \(\mathop{\mathrm{Aut}}((B,+)/C_{(B,+)}(P_3))\) is solvable. In any case, by Lemma 4 there exists \(k\in \mathbb{N}\) such that \[(B,\cdot)^{(k)}\leq C_{(B,+)}(P_3).\] Since \((B,\cdot)\) is perfect, we conclude that \[C_{(B,+)}(P_3)=B,\] which implies that \(P_3\leq Z(B,+)\). Thus \(P_3\) is abelian. By [15] we have \[P_3\cap (B,+)'\cap Z(B,+)=1.\] Since \(P_3\leq (B,+)'\), it follows that \[P_3\cap Z(B,+)=1,\] which contradicts the fact that \(P_3\) is central.

The case r > 7. Let \(R\) be a Sylow \(r\)-subgroup of \((B,+)\). Since \(r>7\) and \(r+1\) is a power of \(2\), the prime \(r\) is the largest prime divisor of \(|B|\), and it divides \(|B|\) exactly once. Hence \(|R|=r.\) Since \((B,+)\) is supersolvable, its Sylow subgroup corresponding to the largest prime divisor of \(|B|\) is normal. Thus \(R\) is a characteristic subgroup of \((B,+)\). We claim that \[R\leq (B,+)'.\] Suppose not. Then \(r \mid |(B,+)/(B,+)'|\). By [5] there is a characteristic \(K\) of \((B,+)\) such that \[(B,+)/K\cong C_r.\] Since \(K\) is characteristic in \((B,+)\), it is invariant under all \(\lambda_b\), \(b\in B\). Hence \(K\) is a left ideal of \(B\). Therefore \((K,\cdot)\) is a subgroup of \((B,\cdot)\) of index \(r\). But \(G:=\operatorname{PSL}_2(7)\times \operatorname{PSL}_2(r)\) has no subgroup of index \(r\). Suppose otherwise, and let \(M\leq G\) with \([G:M]=r\). Suppose, by contradiction, that \(M\leq G\) and \([G:M]=r\). Write \[S=\operatorname{PSL}_2(7),\qquad T=\operatorname{PSL}_2(r).\] Identifying \(S\) and \(T\) with \(S\times 1\) and \(1\times T\), respectively, we have \[[S:S\cap M]=[SM:M]\] and \[[T:T\cap M]=[TM:M].\] Both indices divide \([G:M]=r\). Hence each of them is either \(1\) or \(r\). They cannot both be \(1\), because otherwise \(S\leq M\) and \(T\leq M\), and therefore \(G=ST\leq M\), contradicting \(M<G\). Thus either \(S\) has a subgroup of index \(r\), or \(T\) has a subgroup of index \(r\). This is impossible. Indeed, \(S=\operatorname{PSL}_2(7)\) has order \(168\), so it has no subgroup of index \(r>7\). On the other hand, by the classification of the subgroups of \(\operatorname{PSL}_2(r)\) (see [15]), the smallest index of a proper subgroup of \(\operatorname{PSL}_2(r)\) is \(r+1\). Hence \(T=\operatorname{PSL}_2(r)\) has no subgroup of index \(r\). This contradiction proves that \(G\) has no subgroup of index \(r\). Therefore no such \(K\) exists, and we have proved that \(R\leq (B,+)'.\) Now \(R\) is characteristic in \((B,+)\). Hence the lambda action induces a homomorphism \[\lambda_R: (B,\cdot) \longrightarrow \operatorname{Aut}(R), \qquad b \mapsto \lambda_{b}|_{R}.\] Since \(R\cong C_r\), we have \(\operatorname{Aut}(R)\cong C_{r-1},\) which is abelian. On the other hand, \((B,\cdot)\) is perfect. Hence every homomorphism from \((B,\cdot)\) to an abelian group is trivial. Therefore \(\lambda_b(a)=a\) for every \(b\in B\) and every \(a\in R\). Thus \[b\cdot a=b+a\] for every \(b\in B\) and every \(a\in R\). Applying the same argument to the opposite skew brace, we also obtain \[b\cdot a=a+b\] for every \(b\in B\) and every \(a\in R\). Hence \[b+a=a+b\] for all \(b\in B\) and all \(a\in R\). Therefore \(R\leq Z(B,+).\) By [15] we have \[R \cap (B,+)'\cap Z(B,+)=1.\] Since \(R\leq (B,+)'\), it follows that \[R\cap Z(B,+)=1,\] which contradicts the fact that \(R\) is central. Therefore \((B,+)\) cannot be supersolvable. ◻