Proof. Let $(G,H_1,H_2)$ be a counterexample such that first $|G|$ and then $|G:H_1|+|G:H_2|$ is minimal. The minimality of G implies that $G={\langle }H_1,H_2{\rangle }$ . As $(G,H_1,H_2)$ is a counterexample, we have moreover $H_1\not \leq H_2$ and $H_2\not \leq H_1$ . Similarly, one observes that

(2.1) $$ \begin{align} G\neq S\text{ and }p\text{ divides the order of }G. \end{align} $$

Assume $H_1\,{\trianglelefteq }\, G$ . Then $G={\langle }H_1,H_2{\rangle }=H_1H_2$ , $H_1\cap S\,{\trianglelefteq }\, S$ and ${\langle }H_1\cap S,H_2\cap S{\rangle }=(H_1\cap S)(H_2\cap S)$ . As $S\cap H_1\cap H_2\in \operatorname {Syl}_p(H_1\cap H_2)$ , the order formula for products of subgroups (cf. [Reference Kurzweil and Stellmacher14, 1.1.6]) yields that $(H_1\cap S)(H_2\cap S)$ is a Sylow p-subgroup of $G=H_1H_2$ and thus equal to S. This contradicts the assumption that $(G,H_1,H_2)$ is a counterexample and shows thus that

(2.2) $$ \begin{align} H_1\text{ is not normal in }G. \end{align} $$

As $H_i\neq G$ is subnormal in G for each $i=1,2$ , there exists $M_i\,{\trianglelefteq }\, G$ such that $H_i\leq M_i$ and $G/M_i$ is simple. Set

$$\begin{align*}H:={\langle}H_1,H_2\cap M_1{\rangle}\end{align*}$$

Since $H_1$ and $H_2\cap M_1$ are subnormal in G, it follows from Wielandt’s Join Theorem that H is subnormal in G and thus $ H\cap S\in \operatorname {Syl}_p(H)$ . As $H\leq M_1<G$ , the minimality of $|G|$ implies that $(H,H_1,H_2\cap M_1)$ is not a counterexample. Hence,

(2.3) $$ \begin{align} H\cap S={\langle}H_1\cap S,H_2\cap M_1\cap S{\rangle}. \end{align} $$

Assume now that $H_2\cap M_1\not \leq H_1$ and so $|G:H|<|G:H_1|$ . As $G={\langle }H,H_2{\rangle }$ , the minimality of $|G:H_1|+|G:H_2|$ implies then that $S={\langle }H\cap S,H_2\cap S{\rangle }$ . So $S={\langle }H_1\cap S,H_2\cap S{\rangle }$ by (2.3), contradicting the assumption that $(G,H_1,H_2)$ is a counterexample. This proves that $H_2\cap M_1\leq H_1$ , which yields $H_1\cap H_2=H_2\cap M_1\,{\trianglelefteq }\, H_2$ . A symmetric argument shows that $H_1\cap H_2=H_1\cap M_2\,{\trianglelefteq }\, H_1$ and thus

$$\begin{align*}H_1\cap H_2=H_1\cap M_2=H_2\cap M_1\,{\trianglelefteq}\, {\langle}H_1,H_2{\rangle}=G.\end{align*}$$

Consider now $\overline {G}:=G/H_1\cap H_2$ . If $H_1\cap H_2\neq 1$ , then the minimality of $|G|$ implies that $(\overline {G},\overline {H_1},\overline {H_2})$ is not a counterexample and hence

$$\begin{align*}\overline{S}={\langle}\overline{H_1}\cap \overline{S},\overline{H_2}\cap\overline{S}{\rangle}={\langle}\overline{H_1\cap S},\overline{H_2\cap S}{\rangle}.\end{align*}$$

This implies however that $S={\langle }H_1\cap S,H_2\cap S{\rangle }$ , again a contradiction. Thus, we have

$$\begin{align*}1=H_1\cap H_2=H_1\cap M_2=H_2\cap M_1.\end{align*}$$

Let $i\in \{1,2\}$ and set $j:=3-i$ . Recall that $G/M_j$ is simple and has thus no non-trivial subnormal subgroup. Since $H_i\not \leq M_j$ and $H_i\cap M_j=1$ , it follows that

$$\begin{align*}G/M_j=H_iM_j/M_j\cong H_i/H_i\cap M_j\cong H_i.\end{align*}$$

In particular, $H_i$ is simple. As $H_i$ is subnormal, it follows that $H_i$ is a component of G if $H_i$ is non-abelian. Otherwise, $H_i$ is cyclic of prime order q, and then contained in $O_q(G)$ (cf. [Reference Kurzweil and Stellmacher14, 6.3.1]). Since any two distinct components centralize each other, and every component centralizes

it follows that $H_1$ and $H_2$ either centralize each other, or ${\langle }H_1,H_2{\rangle }\leq O_q(G)$ for some prime q. As $G={\langle }H_1,H_2{\rangle }$ , we get a contradiction to (2.2) in the first case, and a contradiction to (2.1) in the second case. This proves the assertion.

Proof. It is sufficient to prove the first part of the assertion. Indeed, this part is a special case of Stellmacher’s Splitting Lemma for localities [Reference Chermak3, Lemma 3.12], but we give an elementary proof.

Let $g\in G$ . By the Frattini Argument, there exist $m\in N$ and $h\in N_G(T)$ with $g=mh$ . Set $P:=S_g$ . As $(P^m)^h=P^g\leq S\leq N_G(T)$ and $h\in N_G(T)$ , we have $P^m\leq N_G(T)$ . Moreover, $P^m\leq SN$ , so by a Dedekind Argument $P^m\leq N_G(T)\cap SN=SN_N(T)$ . By Sylow’s theorem, there exists thus $x\in N_N(T)$ with $P^{mx}=(P^m)^x\leq S$ . Notice that $n:=mx\in N$ , $f:=x^{-1}h\in N_G(T)$ and $g=mh=nf$ . In particular, $S_{(n,f)}\leq S_{nf}=P$ by (2.4). As $P^n\leq S$ and $P^{nf}=P^g\leq S$ , we have also $P\leq S_{(n,f)}$ . This proves the assertion.

Proof. This is a special case of [Reference Henke8, Theorem 1], but again, we give an elementary proof. Let $g\in MN$ and write $g=m'n'$ for some $m'\in M$ and $n'\in N$ . Notice that $Q:=S_g^{m'}\leq S^{m'}\leq {\langle }S,M{\rangle }=SM$ . Moreover, $Q^{n'}=S_g^g\leq S$ and thus $Q\leq S^{(n')^{-1}}\leq {\langle }S,N{\rangle }=SN$ . Hence, $Q\leq SM\cap SN$ . Notice that $O^p(SM\cap SN)\leq O^p(SM)\cap O^p(SN)\leq M\cap N$ and thus, as S is a Sylow p-subgroup of $SM\cap SN$ , we have $SM\cap SN=SO^p(SM\cap SN)=S(M\cap N)$ . It follows now from Sylow’s theorem that there exists $x\in M\cap N$ with $S_g^{m'x}=Q^x\leq S$ . Now $m:=m'x\in M$ , $n:=x^{-1}n'\in N$ , $g=mn$ , $S_g^m\leq S$ and $S_g^{mn}=S_g^g\leq S$ . In particular, $S_g\leq S_{(m,n)}$ . Together with (2.4), we obtain $S_g=S_{(m,n)}$ .

Proof. Assume $G=MN_G(H)$ and consider $K:={\langle }H^M{\rangle }$ . Notice that G acts on the set of normal subgroups of M, and thus $N_G(H)$ acts on the set of normal subgroups of M containing H. As K is the smallest normal subgroup of M containing H, it follows that $N_G(H)\leq N_G(K)$ . By construction, $K\,{\trianglelefteq }\, M$ , and so K is normal in $G=MN_G(H)$ . This implies that $M={\langle }H^G{\rangle }=K$ . Using the definition of K and the fact that H is subnormal in M, we can therefore conclude that $H=M\,{\trianglelefteq }\, G$ .

Proof. It is sufficient to show that, for every $g\in NH$ , there exist $n\in N$ and $h\in H$ with $g=nh$ and $S_g=S_{(n,h)}$ . Let $(G,N,H,g)$ be a counterexample to that assertion such that $|G|+|N|$ is minimal. Set $M:={\langle }H^G{\rangle }$ .

Notice that $g\in NH\subseteq NM$ . Hence, by Lemma 2.3, there exist $n'\in N$ and $m\in M$ such that $g=n'm$ and $S_g=S_{(n',m)}$ . In particular, as $(G,N,H,g)$ is a counterexample, it follows

(2.5) $$ \begin{align} H\neq M. \end{align} $$

Write $g=ab$ with $a\in N$ and $b\in H$ . Then $ab=n'm$ and so $mb^{-1}=(n')^{-1}a\in M\cap N$ . We obtain thus $m\in (M\cap N)b\subseteq (M\cap N)H$ .

Assume now that $N\not \leq M$ . Then $M\cap N$ is a proper subgroup of N, which is normal in G. The minimality of $|G|+|N|$ yields thus that $(G,M\cap N,H,m)$ is not a counterexample. Hence, there exist $y\in M\cap N$ and $h\in H$ with $m=yh$ and $S_m=S_{(y,h)}$ . This implies $g=n'm=n'yh=(n'y)h$ and

$$\begin{align*}S_g=S_{(n',m)}=S_{(n',y,h)}\leq S_{(n'y,h)}\leq S_{(n'y)h}=S_g,\end{align*}$$

where the second equality uses $S_m=S_{(y,h)}$ and the inclusions use (2.4). So $n:=n'y\in N$ with $g=nh$ and $S_{(n,h)}=S_g$ . This contradicts the assumption that $(G,N,H,g)$ is a counterexample. Hence, we have shown that

$$\begin{align*}N\leq M\leq MS.\end{align*}$$

By (2.5) and Lemma 2.4, we have $G\neq MN_G(H)$ . In particular, as $S\leq N_G(H)$ , it follows $MS\neq G$ . Thus, the minimality of $|G|+|N|$ yields that $(MS,N,H,g)$ is not a counterexample. However, this implies that $(G,N,H,g)$ is not a counterexample, contradicting our assumption.

Proof. If H is properly contained in a subgroup $H_i$ of G, then ${\langle }H^{H_i}{\rangle }$ is a proper subgroup of $H_i$ , as H is subnormal in $H_i$ . It follows thus by induction on $|G:H|$ that we can construct a subnormal series as above.

If $i>1$ and $H_i$ is $N_G(H)$ -invariant, then $N_G(H)$ acts via conjugation on the set of normal subgroups of $H_i$ containing H, and thus also on $H_{i-1}={\langle }H^{H_i}{\rangle }$ , which is the smallest normal subgroup of $H_i$ containing H. Thus, by induction on $n-i$ , one proves that $H_i$ is $N_G(H)$ -invariant for $i=1,2,\dots ,n$ .

Proof of Theorem C.

Let $(G,H_1,H_2)$ be a counterexample to Theorem C such that first $|G|$ and then $|G:H_1|+|G:H_2|$ is minimal. Set

$$\begin{align*}S_1:=S\cap H_1\text{ and }S_2:=S\cap H_2.\end{align*}$$

The minimality of $|G|$ together with Meierfrankenfeld’s Lemma 2.1 yields that

(2.6) $$ \begin{align} G={\langle}H_1,H_2{\rangle}\text{ and }S={\langle}S_1,S_2{\rangle}. \end{align} $$

If $H_1\neq H_1^x$ for some $x\in S$ , then $H_1<H_1^*:={\langle }H_1,H_1^x{\rangle }{\,\trianglelefteq \trianglelefteq \,} G$ by Wielandt’s Join Theorem. Hence, it follows from the minimality of $|G:H_1|+|G:H_2|$ that

$$\begin{align*}\mathcal{F}_S(G)={\langle}\;\mathcal{F}_S({\langle}H_1^*,S{\rangle}),\;\mathcal{F}_S({\langle}H_2,S{\rangle})\;{\rangle}.\end{align*}$$

As ${\langle }H_1^*,S{\rangle }={\langle }H_1,S{\rangle }$ , this contradicts the assumption that $(G,H_1,H_2)$ is a counterexample. So $H_1$ is S-invariant. As the situation is symmetric in $H_1$ and $H_2$ , we have thus shown that

(2.7) $$ \begin{align} S\leq N_G(H_1)\cap N_G(H_2). \end{align} $$

In particular, we need to show that $\mathcal {F}_S(G)={\langle }\mathcal {F}_S(H_1S),\mathcal {F}_S(H_2S){\rangle }$ to obtain a contradiction to the assumption that $(G,H_1,H_2)$ is a counterexample. We will use (2.6) and (2.7) throughout this proof, most of the time without further reference.

Set

$$\begin{align*}N_i:={\langle}H_i^G{\rangle}\text{ for }i=1,2.\end{align*}$$

As $(G,H_1,H_2)$ is a counterexample, it follows from Lemma 2.5 (applied with $(H_i,H_{3-i})$ in place of $(N,H)$ ) that $H_i$ is not normal in G for each $i=1,2$ . In particular, Lemma 2.4 together with (2.7) yields

(2.8) $$ \begin{align} N_1S\neq G\neq N_2S. \end{align} $$

We prove now that

(2.9) $$ \begin{align} H_1\cap N_2=H_1\cap H_2=H_2\cap N_1\,{\trianglelefteq}\, G. \end{align} $$

For the proof assume first that $H_1\cap N_2\not \leq H_2$ . Then

$$\begin{align*}H_2<\tilde{H}_2:={\langle}H_2,H_1\cap N_2{\rangle}. \end{align*}$$

Notice that $H_1\cap N_2$ is subnormal in G and hence, by Wielandt’s Join Theorem, $\tilde {H}_2$ is subnormal in G. The minimality of $|G:H_1|+|G:H_2|$ yields thus

$$\begin{align*}\mathcal{F}_S(G)={\langle}\mathcal{F}_S(H_1S),\mathcal{F}_S(\tilde{H}_2S){\rangle}.\end{align*}$$

Notice that $K_1:=H_1\cap (N_2S){\,\trianglelefteq \trianglelefteq \,} N_2S$ and $K_1\cap S=S_1$ . Using (2.6) and a Dedekind argument, observe moreover that

$$\begin{align*}K_1=H_1\cap (N_2S)=H_1\cap (N_2S_1)=(H_1\cap N_2)S_1.\end{align*}$$

In particular, we can conclude that $\tilde {H}_2S={\langle }K_1,H_2{\rangle }$ and $K_1S=(H_1\cap N_2)S$ . By (2.8) and the minimality of $|G|$ , $(N_2S,K_1,H_2)$ is not a counterexample. Hence,

$$\begin{align*}\mathcal{F}_S(\tilde{H}_2S)=\mathcal{F}_S({\langle}K_1,H_2{\rangle})={\langle}\mathcal{F}_S(K_1S),\mathcal{F}_S(H_2S){\rangle}\end{align*}$$

where $\mathcal {F}_S(K_1S)=\mathcal {F}_S((H_1\cap N_2)S)\subseteq \mathcal {F}_S(H_1S)$ . Hence,

$$\begin{align*}\mathcal{F}_S(G)={\langle}\mathcal{F}_S(H_1S),\mathcal{F}_S(\tilde{H}_2S){\rangle}={\langle}\mathcal{F}_S(H_1S),\mathcal{F}_S(H_2S){\rangle},\end{align*}$$

which contradicts the assumption that $(G,H_1,H_2)$ is a counterexample. This shows that $H_1\cap N_2\leq H_2$ and so $H_1\cap N_2=H_1\cap H_2$ . A symmetric argument gives $H_2\cap N_1=H_1\cap H_2$ . Thus,

$$\begin{align*}H_1\cap N_2=H_1\cap H_2=H_2\cap N_1\,{\trianglelefteq}\, {\langle}H_1,H_2{\rangle}=G\end{align*}$$

and so (2.9) holds. We argue next that

(2.10) $$ \begin{align} H_1=(H_1\cap H_2)N_{H_1}(S_2). \end{align} $$

For the proof recall that $H_1$ is S-invariant by (2.7) and $S_2\leq N_2\,{\trianglelefteq }\, G$ . Hence (2.9) yields

$$\begin{align*}[H_1,S_2]\leq H_1\cap N_2=H_1\cap H_2\,{\trianglelefteq}\, G.\end{align*}$$

Thus, $H_1$ normalizes $(H_1\cap H_2)S_2$ . In particular, $H_1$ acts on $\operatorname {Syl}_p((H_1\cap H_2)S_2)$ , and $H_1\cap H_2$ acts transitively on this set as $S_2\in \operatorname {Syl}_p((H_1\cap H_2)S_2)$ . The general Frattini Argument [Reference Kurzweil and Stellmacher14, 3.1.4] implies thus (2.10). As a next step we prove

(2.11) $$ \begin{align} O^{p^\prime}(H_2)=O^{p^\prime}(N_2)\,{\trianglelefteq}\, G. \end{align} $$

For the proof, Lemma 2.6 allows us to pick a subnormal series

$$\begin{align*}H_2=M_0\,{\trianglelefteq}\, M_1\,{\trianglelefteq}\,\cdots\,{\trianglelefteq}\, M_k=N_2\,{\trianglelefteq}\, M_{k+1}=G\end{align*}$$

for $H_2$ in G with $M_i={\langle }H_2^{M_{i+1}}{\rangle }$ for $i=0,1,2,\dots ,k$ . Moreover, $M_0,M_1,\dots ,M_k$ are $N_G(H_2)$ -invariant. In particular, $M_0,M_1,\dots ,M_k$ are S-invariant by (2.7). Fix now $i\in \{0,1,2,\dots ,k\}$ minimal with $O^{p^\prime }(M_i)=O^{p^\prime }(N_2)$ . Notice that such i exists as $M_k=N_2$ . Assume (2.11) fails. Then $i>0$ and $H_2<M_i$ . In particular, the minimality of $|G:H_1|+|G:H_2|$ yields that $(G,H_1,M_i)$ is not a counterexample and thus

$$\begin{align*}\mathcal{F}_S(G)={\langle}\mathcal{F}_S(H_1S),\mathcal{F}_S(M_iS){\rangle}.\end{align*}$$

Note that $S\cap N_2$ is strongly closed in S as $N_2\,{\trianglelefteq }\, G$ . So the above equality implies

$$\begin{align*}\operatorname{Aut}_G(S\cap N_2)={\langle}\operatorname{Aut}_{H_1}(S\cap N_2),\operatorname{Aut}_{M_i}(S\cap N_2),\operatorname{Aut}_S(S\cap N_2){\rangle}.\end{align*}$$

Recall that $M_{i-1}$ is normalized by $M_i$ and by S. Thus, $M_{i-1}\cap S$ is $\operatorname {Aut}_{M_i}(S\cap N_2)$ -invariant and $\operatorname {Aut}_S(S\cap N_2)$ -invariant. Moreover, using (2.9) and that fact that $H_1$ is S-invariant, we see that

$$\begin{align*}[N_{H_1}(S\cap N_2),S\cap M_{i-1}]\leq [N_{H_1}(S\cap N_2),S\cap N_2]\leq S\cap N_2\cap H_1 =S\cap H_1\cap H_2\leq S_2\leq S\cap M_{i-1}. \end{align*}$$

This implies that $S\cap M_{i-1}$ is $\operatorname {Aut}_{H_1}(S\cap N_2)$ -invariant. Altogether, we have seen that $S\cap M_{i-1}$ is $\operatorname {Aut}_G(S\cap N_2)$ -invariant and thus

$$\begin{align*}N_G(S\cap N_2)\leq N_G(S\cap M_{i-1}).\end{align*}$$

As $O^{p^\prime }(N_2)$ is normal in G and by assumption contained in $M_i$ , a Frattini Argument yields now

$$\begin{align*}G=O^{p^\prime}(N_2)N_G(S\cap N_2)=M_iN_G(S\cap M_{i-1}).\end{align*}$$

Hence, by a Dedekind Argument, setting $X:=N_{M_{i+1}}(S\cap M_{i-1})$ , we have $M_{i+1}=M_iX$ . As $H_2\leq M_{i-1}\,{\trianglelefteq }\, M_i\,{\trianglelefteq }\, M_{i+1}$ , it follows that

$$ \begin{align*} M_i&={\langle}H_2^{M_{i+1}}{\rangle}={\langle}M_{i-1}^{M_{i+1}}{\rangle}={\langle}M_{i-1}^X{\rangle}\\ &= \prod_{x\in X}M_{i-1}^x. \end{align*} $$

Notice that, for every $x\in X$ , $M_{i-1}^x$ is a normal subgroup of $M_i$ with $M_{i-1}^x\cap S=(M_{i-1}\cap S)^x=M_{i-1}\cap S$ . Hence, using Lemma 2.1 (or an elementary argument for this special case), one sees that

$$\begin{align*}M_i\cap S=\prod_{x\in X}(M_{i-1}^x\cap S)=M_{i-1}\cap S.\end{align*}$$

As $M_{i-1}\,{\trianglelefteq }\, M_i$ , it follows that $O^{p^\prime }(M_{i-1})=O^{p^\prime }(M_i)=O^{p^\prime }(N_2)$ , contradicting the minimality of i. Thus, (2.11) holds. We show next that

(2.12) $$ \begin{align} S_2\,{\trianglelefteq}\, G. \end{align} $$

For the proof we use that, by (2.9) and (2.11), $H_1\cap H_2$ and $O^{p^\prime }(H_2)$ are normal in G. In particular,

$$\begin{align*}\hat{H}_2:=(H_1\cap H_2)O^{p^\prime}(H_2)\,{\trianglelefteq}\, G.\end{align*}$$

A Frattini Argument yields $H_2=O^{p^\prime }(H_2)N_{H_2}(S_2)$ and (2.10) gives $H_1=(H_1\cap H_2)N_{H_1}(S_2)$ . Hence,

$$\begin{align*}G={\langle}H_1,H_2{\rangle}=\hat{H}_2{\langle}N_{H_1}(S_2),N_{H_2}(S_2){\rangle}.\end{align*}$$

Notice that $\hat {H}_2\leq H_2$ and so $N_{\hat {H}_2}(S_2)\leq N_{H_2}(S_2)$ . Hence, a Dedekind Argument gives that

$$\begin{align*}N_G(S_2)=N_{\hat{H}_2}(S_2){\langle}N_{H_1}(S_2),N_{H_2}(S_2){\rangle}={\langle}N_{H_1}(S_2),N_{H_2}(S_2){\rangle}.\end{align*}$$

Observe that $S_2\,{\trianglelefteq }\, S$ by (2.7). Assume now that $S_2$ is not normal in G. Then the minimality of $|G|$ yields that $(N_G(S_2),N_{H_1}(S_2),N_{H_2}(S_2))$ is not a counterexample, that is,

$$\begin{align*}\mathcal{F}_S(N_G(S_2))={\langle}\;\mathcal{F}_S(N_{H_1}(S_2)S),\;\mathcal{F}_S(N_{H_2}(S_2)S)\;{\rangle}.\end{align*}$$

As $O^{p^\prime }(H_2)$ is normal in G, Lemma 2.2 yields thus that

$$\begin{align*}\mathcal{F}_S(G)={\langle}\mathcal{F}_S(O^{p^\prime}(H_2)S),\mathcal{F}_S(N_G(S_2)){\rangle}={\langle}\mathcal{F}_S(H_1S),\mathcal{F}_S(H_2S){\rangle},\end{align*}$$

which contradicts the assumption that $(G,H_1,H_2)$ is a counterexample. Hence, (2.12) holds.

The fact that $S_2$ is normal in G implies by Wielandt’s Join Theorem in particular that $H_1S_2$ is subnormal in G. If $S_2\not \leq H_1$ , then the minimality of $|G:H_1|+|G:H_2|$ yields that $(G,H_1S_2,H_2)$ is not a counterexample. As $(H_1S_2)S=H_1S$ , this would imply that $(G,H_1,H_2)$ is not a counterexample, a contradiction. Hence, $S_2\leq H_1\cap S=S_1$ . A symmetric argument yields $S_1\leq S_2$ and thus

$$\begin{align*}S=S_1=S_2\,{\trianglelefteq}\, G.\end{align*}$$

It follows now easily that $(G,H_1,H_2)$ is not a counterexample contradicting our assumption.

Proof. (a) Let $g\in {\mathcal {L}}$ . By the Frattini Lemma [Reference Chermak3, Corollary 3.11], there exist $n\in \mathcal {N}$ and $h\in {\mathcal {L}}$ such that $(n,h)\in \mathbf {D}$ and h is $\uparrow $ -maximal with respect to $\mathcal {N}$ in the sense of [Reference Chermak3, Definition 3.6]. Then $S_g=S_{(n,h)}$ by the Splitting Lemma [Reference Chermak3, Lemma 3.12]. Using first [Reference Chermak3, Proposition 3.9] and then [Reference Chermak3, Lemma 3.1(a)], it follows that $h\in N_{\mathcal {L}}(T)$ .

(b) As $\mathcal {F}_S({\mathcal {L}})$ is generated by maps of the form $c_g\colon S_g\rightarrow S$ , it is sufficient to prove that such a map is in ${\langle }\mathcal {F}_S(\mathcal {N} S),\mathcal {F}_S(N_{\mathcal {L}}(T)){\rangle }$ . Fixing $g\in {\mathcal {L}}$ , it follows from (a) that there exist $n\in \mathcal {N}$ and $h\in N_{\mathcal {L}}(T)$ such that $(n,h)\in \mathbf {D}$ , $g=nh$ and $S_g=S_{(n,h)}$ . So $c_g\colon S_g\rightarrow S$ can be written as a composite of restrictions of the conjugation map $c_n\colon S_n\rightarrow S$ (which is a morphism in $\mathcal {F}_S(\mathcal {N} S)$ ) and of $c_h\colon S_h\rightarrow S$ (which is a morphism in $\mathcal {F}_S(N_{\mathcal {L}}(T))$ ). This implies (b).

Proof. It is a special case of (3.4) that $S_g\cap T^*\in \Delta $ . Hence, as $T^*$ is strongly closed, $(g,f)$ , $u:=(f,f^{-1},g,f)$ and $(g,f,f^{-1})$ are in $\mathbf {D}$ via $S_g\cap T^*$ . In particular, by the axioms of a partial group, $gf=\Pi (u)=fg^f$ and $(gf)f^{-1}=\Pi (g,f,f^{-1})=g$ . So

$$\begin{align*}S_{(g,f)}\cap T^*\leq S_{gf}\cap T^*\leq S_{(gf,f^{-1},f)}\cap T^*\leq S_{((gf)f^{-1},f)}\cap T^*=S_{(g,f)}\cap T^*,\end{align*}$$

where the first and the third inclusion use (3.2) and the second inclusion uses $f\in N_{\mathcal {L}}(T^*)$ . Thus, $S_{(g,f)}\cap T^*=S_{gf}\cap T^*$ . Observe also that $(f^{-1},f,g)\in \mathbf {D}$ via $S_g\cap T^*$ . In particular, $(f,g)\in \mathbf {D}$ and $g=\Pi (f^{-1},f,g)=f^{-1}(fg)$ . It follows that

$$\begin{align*}S_{(f,g)}\cap T^*\leq S_{fg}\cap T^*\leq S_{(f,f^{-1},fg)}\cap T^*\leq S_{(f,f^{-1}(fg))}\cap T^*= S_{(f,g)}\cap T^*,\end{align*}$$

where again the first and the third inclusion use (3.2) and the second inclusion uses $f\in N_{\mathcal {L}}(T^*)$ . Hence, $S_{(f,g)}\cap T^*=S_{fg}\cap T^*$ .

Proof. By [Reference Henke10, Lemma 11.9], we have

$$\begin{align*}T^*=S\cap F^*({\mathcal{L}})=(S\cap E({\mathcal{L}}))O_p({\mathcal{L}}).\end{align*}$$

As $E({\mathcal {L}})\,{\trianglelefteq }\, {\mathcal {L}}$ , $S\cap E({\mathcal {L}})=T^*\cap E({\mathcal {L}})$ and ${\mathcal {L}}=N_{\mathcal {L}}(O_p({\mathcal {L}}))$ , the assertion follows.

Proof. It is a consequence of Theorem 2 and Lemma 10.11(c) in [Reference Henke10] that, for every $f\in N_{\mathcal {L}}(T^*)$ , we have ${\mathcal {L}}=\mathbf {D}(f)$ and the conjugation map $c_f$ is an automorphism of ${\mathcal {L}}$ . Moreover, by [Reference Henke12, Lemma 2.13], $N_{\mathcal {L}}(T^*)$ acts on the set ${\mathcal {L}}$ by conjugation. Thus, by [Reference Henke10, Lemma 11.12], $N_{\mathcal {L}}(T^*)$ acts also on the set of components of ${\mathcal {L}}$ .

Proof. By [Reference Henke10, Proposition 11.7], $\mathcal {N}_1$ , $\mathcal {N}_2$ and $\mathcal {N}$ are well-defined and partial normal subgroups of $F^*({\mathcal {L}})$ . As $F^*({\mathcal {L}})$ forms by Theorem 3.2 a regular locality with Sylow subgroup $T^*$ , it follows in particular from [Reference Henke9, Theorem 1] applied with $F^*({\mathcal {L}})$ in place of ${\mathcal {L}}$ that $(\mathcal {N}_1\mathcal {N}_2)\cap S=(\mathcal {N}_1\mathcal {N}_2)\cap T^*=(\mathcal {N}_1\cap T^*)(\mathcal {N}_2\cap T^*)=(\mathcal {N}_1\cap S)(\mathcal {N}_2\cap S)$ . Thus, for (a), it remains only to prove that $\mathcal {N}=\mathcal {N}_1\mathcal {N}_2$ . In fact, as $\mathcal {N}_i\subseteq \mathcal {N}$ for $i=1,2$ and $\mathcal {N}$ is a partial subgroup, we have $\mathcal {N}_1\mathcal {N}_2\subseteq \mathcal {N}$ . So for (a) it is sufficient to prove that

(3.5) $$ \begin{align} \mathcal{N}\subseteq \mathcal{N}_1\mathcal{N}_2. \end{align} $$

As $N_{\mathcal {L}}(T^*)$ is a subgroup, we have also $N_{\mathcal {N}_1}(T^*) N_{\mathcal {N}_2}(T^*)\subseteq N_{\mathcal {N}}(T^*)$ . So for (c) it remains only to show that

(3.6) $$ \begin{align} N_{\mathcal{N}}(T^*)\subseteq N_{\mathcal{N}_1}(T^*) N_{\mathcal{N}_2}(T^*). \end{align} $$

Observe now that, replacing $\mathfrak {C}_2$ by $\mathfrak {C}_2\backslash \mathfrak {C}_1$ , we may assume for the proof of (3.5) and (3.6) that $\mathfrak {C}_1\cap \mathfrak {C}_2=\emptyset $ . So we will assume this property from now on throughout.

Applying first [Reference Henke10, Theorem 11.18(a)] and then [Reference Henke10, Lemma 4.5, Lemma 4.8], one sees that ${\mathcal {N}=\mathcal {N}_1\mathcal {N}_2}$ , and that $\mathcal {N}_i\subseteq C_{\mathcal {L}}(\mathcal {N}_{3-i})$ for $i=1,2$ . In particular, (3.5) holds and thus (a).

For the proof of the remaining statement in (b) let now $n\in \mathcal {N}_1$ and $m\in \mathcal {N}_2$ . As $\mathcal {N}_1\subseteq C_{\mathcal {L}}(\mathcal {N}_2)$ , it follows from [Reference Henke10, Lemma 3.5] that $(n,m)$ and $(m,n)$ are in $\mathbf {D}$ and that $nm=mn$ . In particular, $\mathcal {N}_1$ and $\mathcal {N}_2$ commute in the sense of [Reference Henke10, Definition 2], that is, they are commuting partial normal subgroups of $F^*({\mathcal {L}})$ . Hence, it follows from [Reference Henke10, Theorem 1(d)] applied with $F^*({\mathcal {L}})$ in place of ${\mathcal {L}}$ that $S_{nm}\cap T^*=S_{(n,m)}\cap T^*$ . This proves (b).

For the proof of (3.6) let $f\in N_{\mathcal {N}}(T^*)$ . By (a), we may pick $n\in \mathcal {N}_1$ and $m\in \mathcal {N}_2$ with $(n,m)\in \mathbf {D}$ and $f=nm$ . It follows then from (b) that $T^*=T^*\cap S_f=T^*\cap S_{nm}\leq S_{(n,m)}$ . As $T^*$ is strongly closed in $\mathcal {F}_S({\mathcal {L}})$ , it follows that $n\in N_{\mathcal {N}_1}(T^*)$ and $m\in N_{\mathcal {N}_2}(T^*)$ . This proves (3.6) and thus (c).

Proof. Since $T^*\,{\trianglelefteq }\, S$ , we have $S\cap {\mathcal {H}}\subseteq N_{\mathcal {H}}(T^*)$ and thus $S\cap {\mathcal {H}}=S\cap N_{\mathcal {H}}(T^*)$ . Since ${\mathcal {H}}{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ , it follows moreover easily that $N_{\mathcal {H}}(T^*)={\mathcal {H}}\cap N_{\mathcal {L}}(T^*)$ is subnormal in $N_{\mathcal {L}}(T^*)$ .

As mentioned before, it follows from [Reference Henke10, Remark 11.2] that $E({\mathcal {H}})\subseteq E({\mathcal {L}})\subseteq F^*({\mathcal {L}})$ . Thus, $E({\mathcal {H}})\cap S=E({\mathcal {H}})\cap T^*$ . Since $E({\mathcal {H}})\,{\trianglelefteq }\, {\mathcal {H}}$ by [Reference Henke10, Lemma 11.13], it follows that $N_{\mathcal {H}}(T^*)\subseteq N_{\mathcal {H}}(T^*\cap E({\mathcal {H}}))=N_{\mathcal {H}}(S\cap E({\mathcal {H}}))$ . It remains thus only to show that $N_{\mathcal {H}}(S\cap E({\mathcal {H}}))\subseteq N_{\mathcal {H}}(T^*)$ .

Applying first [Reference Henke10, Theorem 11.18(c)] and then [Reference Henke10, Lemma 4.8], one sees that ${\mathcal {H}}$ is in the centralizer of $\mathcal {M}:=\prod _{\mathcal {K}\in \operatorname {Comp}({\mathcal {L}})\backslash \operatorname {Comp}({\mathcal {H}})}\mathcal {K}$ . In particular, ${\mathcal {H}}\subseteq C_{\mathcal {L}}(\mathcal {M}\cap S)$ . Moreover, by Lemma 3.6(a), $E({\mathcal {L}})\cap S=(E({\mathcal {H}})\cap S)(\mathcal {M}\cap S)$ . Using Lemma 3.4, we can thus conclude that $N_{{\mathcal {H}}}(S\cap E({\mathcal {H}}))\subseteq N_{\mathcal {L}}(S\cap E({\mathcal {L}}))=N_{\mathcal {L}}(T^*)$ . This proves the assertion.

Proof. It is shown in [Reference Grazian and Henke7, Theorem 6.1(a)] that $\mathcal {N} H=H\mathcal {N}$ is a partial subgroup; essentially, the argument uses the property stated in Lemma 3.3. Suppose now $H\cap S\in \operatorname {Syl}_p(H)$ . As $N_{\mathcal {N}}(T^*)\,{\trianglelefteq }\, N_{\mathcal {L}}(T^*)$ with $\mathcal {N}\cap S=N_{\mathcal {N}}(T^*)\cap S\in \operatorname {Syl}_p(N_{\mathcal {N}}(T^*))$ , it follows that

$$\begin{align*}S_0:=(\mathcal{N}\cap S)(H\cap S)\in\operatorname{Syl}_p(N_{\mathcal{N}}(T^*)H).\end{align*}$$

Thus, it is a consequence of [Reference Grazian and Henke7, Theorem 6.1(c)] that $S_0$ is a maximal p-subgroup of $\mathcal {N} H$ . Since $S_0\leq S\cap (\mathcal {N} H)$ and $S\cap (\mathcal {N} H)$ is a p-subgroup of $\mathcal {N} H$ , we can conclude that $S\cap \mathcal {N} H=S_0$ is a maximal p-subgroup of $\mathcal {N} H$ .

Proof. It is a special case of [Reference Henke13, Theorem 2] that $\hat {{\mathcal {H}}}$ is subnormal in ${\mathcal {L}}$ , but we supply a shorter direct argument here: Let $H=H_0\,{\trianglelefteq }\, H_1\,{\trianglelefteq }\,\cdots \,{\trianglelefteq }\, H_n=N_{\mathcal {L}}(T^*)$ be a subnormal series for H in $N_{\mathcal {L}}(T^*)$ . Then by Lemma 4.1, $E({\mathcal {L}}) H_i=H_i E({\mathcal {L}})$ is a partial subgroup of ${\mathcal {L}}$ for all $i=0,1,2,\dots ,n$ . By the Frattini Lemma [Reference Chermak3, Corollary 3.11] and Lemma 3.4, we have ${\mathcal {L}}=N_{\mathcal {L}}(S\cap E({\mathcal {L}}))E({\mathcal {L}})=N_{\mathcal {L}}(T^*)E({\mathcal {L}})=H_nE({\mathcal {L}})$ . Thus, it is sufficient to prove that $H_{i-1}E({\mathcal {L}})\,{\trianglelefteq }\, H_iE({\mathcal {L}})$ for $i=1,2,\dots ,n$ . So fix $i\in \{1,2,\dots ,n\}$ , $x,y\in E({\mathcal {L}})$ , $h\in H_{i-1}$ and $f\in H_i$ with

$$\begin{align*}w:=((fy)^{-1},hx,fy)\in\mathbf{D}.\end{align*}$$

By [Reference Chermak3, Lemma 1.4(f)], $(fy)^{-1}=y^{-1}f^{-1}$ . Moreover, Lemma 3.3 gives

$$\begin{align*}S_{hx}\cap T^*=S_{(h,x)}\cap T^*,\end{align*}$$

$$\begin{align*}S_{fy}\cap T^*=S_{(f,y)}\cap T^*\text{ and }S_{(fy)^{-1}}\cap T^*=S_{y^{-1}f^{-1}}\cap T^*=S_{(y^{-1},f^{-1})}\cap T^*.\end{align*}$$

Hence, setting $u:=(y^{-1},f^{-1},h,x,f,y)\in \mathbf {D}$ , it follows that $S_u\cap T^*=S_w\cap T^*$ . As $w\in \mathbf {D}$ , the property (3.4) yields now first that $S_u\cap T^*=S_w\cap T^*\in \Delta $ and then $u\in \mathbf {D}$ . Using $f\in N_{\mathcal {L}}(T^*)$ , we see also that

$$\begin{align*}v:=(y^{-1},f^{-1},h,f,f^{-1},x,f,y)\in\mathbf{D}\end{align*}$$

via $S_w\cap T^*$ . Hence,

$$\begin{align*}(hx)^{fy}=\Pi(w)=\Pi(u)=\Pi(v)=(h^fx^f)^y.\end{align*}$$

As $h\in H_{i-1}\,{\trianglelefteq }\, H_i$ and $f\in H_i$ , we have $h^f\in H_{i-1}$ . Moreover, $x\in E({\mathcal {L}})\,{\trianglelefteq }\,{\mathcal {L}}$ implies $x^f\in E({\mathcal {L}})$ . Hence, $h^fx^f\in H_{i-1}E({\mathcal {L}})$ . Since $y\in E({\mathcal {L}})\subseteq H_{i-1}E({\mathcal {L}})$ and $H_{i-1}E({\mathcal {L}})$ is a partial subgroup, it follows $(hx)^{fy}=(h^fx^f)^y\in H_{i-1}E({\mathcal {L}})$ . This proves $H_{i-1}E({\mathcal {L}})\,{\trianglelefteq }\, H_iE({\mathcal {L}})$ and thus $\hat {{\mathcal {H}}}:=E({\mathcal {L}}) H=H E({\mathcal {L}}){\,\trianglelefteq \trianglelefteq \,} {\mathcal {L}}$ . It follows from [Reference Henke10, Remark 11.2] that $\operatorname {Comp}(\hat {{\mathcal {H}}})=\operatorname {Comp}({\mathcal {L}})$ and thus $E(\hat {{\mathcal {H}}})=E({\mathcal {L}})$ .

Proof. By Lemma 4.2, $\hat {{\mathcal {H}}}:=E({\mathcal {L}})H=HE({\mathcal {L}})$ is a partial subnormal subgroup of ${\mathcal {L}}$ with $E({\mathcal {L}})=E(\hat {{\mathcal {H}}})$ . We argue now that

(4.1) $$ \begin{align} \mathcal{N}\,{\trianglelefteq}\, \hat{{\mathcal{H}}}. \end{align} $$

For that let $x\in E({\mathcal {L}})$ , $h\in H$ and $n\in \mathcal {N}$ with $u:=((xh)^{-1},n,(xh))\in \mathbf {D}$ . By [Reference Chermak3, Lemma 1.4(f)], $(xh)^{-1}=h^{-1}x^{-1}$ . Set $v:=(h^{-1},x^{-1},n,x,h)$ . It follows from Lemma 3.3 that $S_v\cap T^*=S_u\cap T^*$ . Now (3.4) yields first $S_v\cap T^*=S_u\cap T^*\in \Delta $ and then $v\in \mathbf {D}$ . Hence,

$$\begin{align*}n^{xh}=\Pi(u)=\Pi(v)=(n^x)^h.\end{align*}$$

By [Reference Henke10, Proposition 11.7], $\mathcal {N}\,{\trianglelefteq }\, F^*({\mathcal {L}})\supseteq E({\mathcal {L}})$ and thus $n^x\in \mathcal {N}$ . As $\mathfrak {C}$ is by assumption H-invariant and since H induces automorphisms of ${\mathcal {L}}$ via conjugation (cf. Lemma 3.5), it follows that $\mathcal {N}$ is invariant under conjugation by H. Thus, $n^{xh}=(n^x)^h\in \mathcal {N}$ . This proves (4.1).

Since $\hat {{\mathcal {H}}}$ is a subnormal, it is a regular locality with Sylow subgroup $S_0:=S\cap \hat {{\mathcal {H}}}$ . Note that $H\cap S_0=H\cap S\in \operatorname {Syl}_p(H)$ , as $H{\,\trianglelefteq \trianglelefteq \,} N_{\mathcal {L}}(T^*)$ and $S\in \operatorname {Syl}_p(N_{\mathcal {L}}(T^*))$ . Moreover, since $E(\hat {{\mathcal {H}}})=E({\mathcal {L}})$ , it follows from Lemma 3.4 that $N_{\hat {{\mathcal {H}}}}(F^*(\hat {{\mathcal {H}}})\cap S)=N_{\hat {{\mathcal {H}}}}(E(\hat {{\mathcal {H}}})\cap S)=N_{\hat {{\mathcal {H}}}}(E({\mathcal {L}})\cap S)=N_{\hat {{\mathcal {H}}}}(T^*)$ and thus $H{\,\trianglelefteq \trianglelefteq \,} N_{\hat {{\mathcal {H}}}}(F^*(\hat {{\mathcal {H}}})\cap S)$ . The property (4.1) allows us now to apply Lemma 4.1 with $\hat {{\mathcal {H}}}$ in place of ${\mathcal {L}}$ to obtain that

$$\begin{align*}{\mathcal{H}}:=\mathcal{N} H=H\mathcal{N}\end{align*}$$

is a partial subgroup of ${\mathcal {L}}$ with $\mathcal {N} H\cap S=\mathcal {N} H\cap S_0=(\mathcal {N}\cap S_0)(H\cap S_0)=(\mathcal {N}\cap S)(H\cap S)$ . We show next that

(4.2) $$ \begin{align} {\mathcal{H}}\,{\trianglelefteq}\, \hat{{\mathcal{H}}}\quad\text{and}\quad{\mathcal{H}}{\,\trianglelefteq\trianglelefteq\,}{\mathcal{L}}. \end{align} $$

As $\hat {\mathcal {H}}{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ , it is indeed sufficient to prove that ${\mathcal {H}}\,{\trianglelefteq }\,\hat {\mathcal {H}}$ . For the proof fix $a\in {\mathcal {H}}$ and $f\in \hat {{\mathcal {H}}}$ such that $w:=(f^{-1},a,f)\in \mathbf {D}$ . As ${\mathcal {H}}$ is a partial subgroup, we only need to prove that $a^f\in {\mathcal {H}}$ . Write $f=yh$ with $y\in E({\mathcal {L}})$ and $h\in H$ . Then [Reference Chermak3, Lemma 1.4(f)] gives $f^{-1}=h^{-1} y^{-1}$ . Setting $w':=(h^{-1},y^{-1},a,y,h)$ , it follows from Lemma 3.3 that $S_{w'}\cap T^*=S_w\cap T^*$ . Hence, (3.4) gives first $S_w\cap T^*\in \Delta $ and then $w'\in \mathbf {D}$ . Using the axioms of a partial group, we can thus conclude that

$$\begin{align*}a^f=\Pi(w)=\Pi(w')=(a^y)^h.\end{align*}$$

By part (a) of Lemma 3.6, there exist $n\in \mathcal {N}$ and $m\in \mathcal {M}$ with $y=nm$ . Moreover, part (b) of that lemma yields then $S_y\cap T^*=S_{nm}\cap T^*=S_{(n,m)}\cap T^*$ . So by (3.4), we have $(m^{-1},n^{-1},a,n,m)\in \mathbf {D}$ via $S_{(y^{-1},a,y)}\cap T^*$ and thus

$$\begin{align*}a^y=(a^n)^m.\end{align*}$$

Notice that $a,n\in \mathcal {N} H={\mathcal {H}}$ and thus $a^n\in {\mathcal {H}}$ . Lemma 3.6(b) gives that $\mathcal {N}\subseteq C_{\mathcal {L}}(\mathcal {M})$ . By assumption we have moreover $H\subseteq C_{\mathcal {L}}(\mathcal {M})$ . As $\mathcal {M}\,{\trianglelefteq }\, F^*({\mathcal {L}})$ is subnormal in ${\mathcal {L}}$ by [Reference Henke10, Proposition 11.7], it follows from [Reference Henke12, Theorem A(f)] that $C_{\mathcal {L}}(\mathcal {M})$ is a partial subgroup of ${\mathcal {L}}$ . Hence, ${\mathcal {H}}=\mathcal {N} H\subseteq C_{\mathcal {L}}(\mathcal {M})$ and thus $\mathcal {M}\subseteq C_{\mathcal {L}}({\mathcal {H}})$ by [Reference Henke10, Lemma 3.5]. It follows that $a^y=(a^n)^m=a^n\in {\mathcal {H}}$ . Recall that $h\in H\subseteq {\mathcal {H}}$ . Hence, $a^f=(a^y)^h\in {\mathcal {H}}$ . This completes the proof of (4.2). We prove next that

(4.3) $$ \begin{align} N_{\mathcal{H}}(S\cap \mathcal{N})=N_{\mathcal{N}}(S\cap \mathcal{N})H\quad\text{and}\quad N_{\mathcal{H}}(S\cap\mathcal{N})=N_{\mathcal{H}}(T^*){\,\trianglelefteq\trianglelefteq\,} N_{\mathcal{L}}(T^*) \end{align} $$

To see this notice that $H\subseteq N_{\mathcal {H}}(T^*)\subseteq N_{\mathcal {H}}(T^*\cap \mathcal {N})=N_{\mathcal {H}}(S\cap \mathcal {N})$ as $\mathcal {N}\,{\trianglelefteq }\,\hat {\mathcal {H}}\supseteq {\mathcal {H}}$ by (4.1). Hence, by the Dedekind Lemma [Reference Chermak3, Lemma 1.10], we have $N_{\mathcal {H}}(S\cap \mathcal {N})=N_{\mathcal {N}}(S\cap \mathcal {N})H$ . Applying Lemma 3.7 with $\mathcal {N}$ in place of ${\mathcal {H}}$ and noting that $\mathcal {N}=E(\mathcal {N})$ , we obtain $N_{\mathcal {N}}(S\cap \mathcal {N})=N_{\mathcal {N}}(T^*)$ . Hence, $N_{\mathcal {H}}(S\cap \mathcal {N})\leq N_{\mathcal {H}}(T^*)$ . This shows $N_{\mathcal {H}}(S\cap \mathcal {N})=N_{\mathcal {H}}(T^*)$ . As ${\mathcal {H}}{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ , one sees easily that $N_{\mathcal {H}}(T^*){\,\trianglelefteq \trianglelefteq \,} N_{\mathcal {L}}(T^*)$ . This completes the proof of (4.3). It remains now only to prove that

(4.4) $$ \begin{align} \operatorname{Comp}({\mathcal{H}})=\mathfrak{C}. \end{align} $$

Note first that, by [Reference Henke10, Remark 11.2(a)], we have $\mathfrak {C}\subseteq \operatorname {Comp}({\mathcal {H}})$ . Assuming that (4.4) is false, there exists thus $\mathcal {K}\in \operatorname {Comp}({\mathcal {H}})\backslash \mathfrak {C}$ . Then $\mathcal {K}$ centralizes $\mathcal {N}$ by [Reference Henke10, Theorem 11.18(e)]. In particular, $\mathcal {K}\subseteq C_{\mathcal {H}}(S\cap \mathcal {N})\subseteq N_{\mathcal {H}}(S\cap \mathcal {N})$ . Using (4.3) and the fact that $\mathcal {K}$ is subnormal in ${\mathcal {H}}$ , we see then that $\mathcal {K}{\,\trianglelefteq \trianglelefteq \,} N_{\mathcal {H}}(S\cap \mathcal {N}){\,\trianglelefteq \trianglelefteq \,} N_{\mathcal {L}}(T^*)$ and so $\mathcal {K}$ is a subnormal subgroup of the group $N_{\mathcal {L}}(T^*)$ . As $N_{\mathcal {L}}(T^*)$ is a group of characteristic p, it follows from [Reference Meierfrankenfeld and Stellmacher15, Lemma 1.2(a)] that $\mathcal {K}$ is a group of characteristic p. As $\mathcal {K}$ is quasisimple, [Reference Henke10, Lemma 7.10] gives $Z(\mathcal {K})=O_p(\mathcal {K})\neq \mathcal {K}$ . This yields a contradiction. Hence, (4.4) holds. In particular, $E({\mathcal {H}})=\mathcal {N}$ and the proof is complete.

Hypothesis 4.4. Let $({\mathcal {L}},\Delta ,S)$ be a regular locality and set $T^*:=F^*({\mathcal {L}})\cap S$ . Let ${\mathcal {H}}_1$ and ${\mathcal {H}}_2$ be partial subnormal subgroups of ${\mathcal {L}}$ . Set

$$\begin{align*}{\mathcal{H}}:={\langle}{\mathcal{H}}_1,{\mathcal{H}}_2{\rangle},\end{align*}$$

$$\begin{align*}S_i:={\mathcal{H}}_i\cap S,\;T_i=E({\mathcal{H}}_i)\cap S\text{ and }H_i:=N_{{\mathcal{H}}_i}(T_i)\text{ for }i=1,2.\end{align*}$$

Set moreover

$$\begin{align*}H:={\langle}H_1,H_2{\rangle},\;\mathfrak{C}:=\operatorname{Comp}({\mathcal{H}}_1)\cup\operatorname{Comp}({\mathcal{H}}_2),\;\mathcal{N}:=\prod_{\mathcal{K}\in\mathfrak{C}}\mathcal{K}\text{ and }\mathcal{M}:=\prod_{\mathcal{K}\in\operatorname{Comp}({\mathcal{L}})\backslash\mathfrak{C}}\mathcal{K}.\end{align*}$$

Proof. Lemma 3.7 implies that $H_i=N_{{\mathcal {H}}_i}(T^*){\,\trianglelefteq \trianglelefteq \,} N_{\mathcal {L}}(T^*)$ and $S_i=S\cap H_i$ for each $i=1,2$ . In particular, it follows from Wielandt’s Join Theorem for groups that $H={\langle }H_1,H_2{\rangle }$ is a subnormal subgroup of $N_{\mathcal {L}}(T^*)$ and from Meierfrankenfeld’s Lemma 2.1 that $H\cap S={\langle }S_1,S_2{\rangle }$ .

Proof. It is sufficient to argue that $\mathfrak {C}$ is $H_i$ -invariant for each $i=1,2$ . For the proof fix $i\in \{1,2\}$ . By Lemma 3.5 (applied with ${\mathcal {H}}_i$ in place of ${\mathcal {L}}$ ), $\operatorname {Comp}({\mathcal {H}}_i)$ is $H_i$ -invariant. Moreover, applying first [Reference Henke10, 11.17] and then [Reference Henke10, Lemma 3.5], one sees that ${\mathcal {H}}_i\subseteq C_{\mathcal {L}}(\mathcal {K})$ for every $\mathcal {K}\in \operatorname {Comp}({\mathcal {L}})\backslash \operatorname {Comp}({\mathcal {H}}_i)$ . In particular, $H_i$ centralizes every component in $\mathfrak {C}\backslash \operatorname {Comp}({\mathcal {H}}_i)$ . Thus $\mathfrak {C}$ is $H_i$ -invariant.

Proof. By [Reference Henke10, Proposition 11.7], $\mathcal {M}\,{\trianglelefteq }\, F^*({\mathcal {L}})$ is subnormal in ${\mathcal {L}}$ . Hence, it follows from [Reference Henke12, Theorem A(f)] that $C_{\mathcal {L}}(\mathcal {M})$ is a partial subgroup of ${\mathcal {L}}$ . Therefore, it is sufficient to show that ${\mathcal {H}}_i\subseteq C_{\mathcal {L}}(\mathcal {M})$ for each $i=1,2$ .

Fix now $i\in \{1,2\}$ and notice that $\mathfrak {C}':=\operatorname {Comp}({\mathcal {L}})\backslash \mathfrak {C}\subseteq \operatorname {Comp}({\mathcal {L}})\backslash \operatorname {Comp}({\mathcal {H}}_i)$ . Hence, it is a special case of [Reference Henke10, Theorem 11.18(c)] that ${\mathcal {H}}_i\mathcal {M}$ is an internal central product of the elements of $\{{\mathcal {H}}_i\}\cup \mathfrak {C}'$ (in the sense defined in [Reference Henke10, Definition 4.1]). In particular, by [Reference Henke10, Lemma 4.8], we have ${\mathcal {H}}_i\subseteq C_{\mathcal {L}}(\mathcal {M})$ as required.

Proof. Recall that $H{\,\trianglelefteq \trianglelefteq \,} N_{\mathcal {L}}(T^*)$ by Lemma 4.5, that $\mathfrak {C}$ is H-invariant by Lemma 4.6 and that ${H\subseteq C_{\mathcal {L}}(\mathcal {M})}$ by Lemma 4.7. Hence, it follows from Lemma 4.3 that $\mathcal {N} H=H\mathcal {N}$ is a partial subnormal subgroup of ${\mathcal {L}}$ ,

$$\begin{align*}\operatorname{Comp}(\mathcal{N} H)=\mathfrak{C},\;E(\mathcal{N} H)=\mathcal{N}\,{\trianglelefteq}\,\mathcal{N} H,\;(\mathcal{N} H)\cap S=(\mathcal{N}\cap S)(H\cap S)\text{ and }N_{\mathcal{N} H}(\mathcal{N}\cap S)=N_{\mathcal{N}}(T^*)H.\end{align*}$$

It follows from the Frattini Lemma [Reference Chermak3, Corollary 3.11] that ${\mathcal {H}}_i=E({\mathcal {H}}_i)H_i\subseteq \mathcal {N} H$ for every $i\in \{1,2\}$ . So the fact that $\mathcal {N} H$ is a partial subgroup implies ${\mathcal {H}}:={\langle }{\mathcal {H}}_1,{\mathcal {H}}_2{\rangle }\subseteq \mathcal {N} H$ . As $\mathcal {N}$ and H are both contained in ${\mathcal {H}}$ , we can conclude that ${\mathcal {H}}=\mathcal {N} H=H\mathcal {N}$ . In particular, ${\mathcal {H}}$ is subnormal in ${\mathcal {L}}$ ,

$$\begin{align*}\operatorname{Comp}({\mathcal{H}})=\mathfrak{C},\;E({\mathcal{H}})=\mathcal{N}\,{\trianglelefteq}\,{\mathcal{H}},\;{\mathcal{H}}\cap S=(\mathcal{N}\cap S)(H\cap S)\text{ and }N_{\mathcal{H}}(\mathcal{N}\cap S)=N_{\mathcal{N}}(T^*)H.\end{align*}$$

By Lemma 4.5, $H\cap S={\langle }S_1,S_2{\rangle }$ . It follows moreover from Lemma 3.6(a) that

$$\begin{align*}\mathcal{N}\cap S=(E({\mathcal{H}}_1)\cap S)(E({\mathcal{H}}_2)\cap S)\leq {\langle}S_1,S_2{\rangle}=H\cap S.\end{align*}$$

Hence, ${\mathcal {H}}\cap S=(\mathcal {N}\cap S)(H\cap S)=H\cap S={\langle }S_1,S_2{\rangle }$ . Notice that Lemma 3.6(a) yields also that $\mathcal {N}=E({\mathcal {H}}_1)E({\mathcal {H}}_2)$ , that is, $E({\mathcal {H}})=\mathcal {N}=E({\mathcal {H}}_1)E({\mathcal {H}}_2)$ .

As $\mathcal {N}=E({\mathcal {H}})$ , it follows from Lemma 3.4 that $N_{\mathcal {H}}(S\cap F^*({\mathcal {H}}))=N_{\mathcal {H}}(S\cap \mathcal {N})$ and from Lemma 3.7 that $N_{\mathcal {H}}(\mathcal {N}\cap S)=N_{\mathcal {H}}(T^*)$ . It remains thus only to show that $N_{\mathcal {H}}(T^*)=H$ . By Lemma 3.6(c), we have $N_{\mathcal {N}}(T^*)=N_{E({\mathcal {H}}_1)}(T^*)N_{E({\mathcal {H}}_2)}(T^*)$ . Note that $N_{E({\mathcal {H}}_i)}(T^*)\leq N_{{\mathcal {H}}_i}(T^*)=H_i$ for each $i=1,2$ by Lemma 4.5. Thus, it follows that $N_{\mathcal {N}}(T^*)\leq {\langle }H_1,H_2{\rangle }=H$ . Using the properties above, we obtain therefore that $N_{\mathcal {H}}(T^*)=N_{\mathcal {H}}(\mathcal {N}\cap S)=N_{\mathcal {N}}(T^*)H=H$ . This completes the proof.

Proof. Recall that $\mathcal {N}\,{\trianglelefteq }\, {\mathcal {H}}{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ and $N_{\mathcal {H}}(S\cap \mathcal {N})=H$ by Theorem 4.8. In particular, by Theorem 3.2, ${\mathcal {H}}$ supports the structure of a regular locality. Hence, the assertion follows from Lemma 3.1(b) applied with ${\mathcal {H}}$ in place of ${\mathcal {L}}$ .

Proof. As $T\leq N_S({\mathcal {H}}_i)$ and $E({\mathcal {H}}_i)$ is by [Reference Henke10, Lemma 11.12] invariant under automorphisms of ${\mathcal {H}}_i$ , we have

$$ \begin{align*} T\leq N_S(E({\mathcal{H}}_i))\text{ and }T_i\,{\trianglelefteq}\, T\text{ for }i=1,2. \end{align*} $$

Let $i\in \{1,2\}$ . By [Reference Henke12, Lemma 3.18(a)], ${\mathcal {H}}_i T$ is a partial subgroup of ${\mathcal {L}}$ and $({\mathcal {H}}_iT,\delta (\mathcal {F}_T({\mathcal {H}}_iT)),T)$ is a regular locality with $E({\mathcal {H}}_i)=E({\mathcal {H}}_iT)$ . In particular, $E({\mathcal {H}}_i)=E({\mathcal {H}}_iT)\,{\trianglelefteq }\,{\mathcal {H}}_iT$ by [Reference Henke10, Lemma 11.13]. It follows thus from Lemma 3.1(b) that

$$\begin{align*}\mathcal{F}_T({\mathcal{H}}_iT)={\langle}\mathcal{F}_T(E({\mathcal{H}}_i)T),\mathcal{F}_T(N_{{\mathcal{H}}_iT}(T_i)){\rangle}\text{ for }i=1,2. \end{align*}$$

By the Dedekind Lemma [Reference Chermak3, Lemma 1.10], $N_{{\mathcal {H}}_iT}(T_i)=N_{{\mathcal {H}}_i}(T_i)T=H_iT$ . Hence,

(4.5) $$ \begin{align} \mathcal{F}_T({\mathcal{H}}_iT)={\langle}\mathcal{F}_T(E({\mathcal{H}}_i)T),\mathcal{F}_T(H_iT){\rangle}\text{ for }i=1,2. \end{align} $$

As $\mathcal {N}\,{\trianglelefteq }\, {\mathcal {H}}$ by Theorem 4.8, the product $\mathcal {N} T$ is a partial subgroup of ${\mathcal {L}}$ by [Reference Henke10, Lemma 3.15]. We show next that

(4.6) $$ \begin{align} E({\mathcal{H}}_i)\,{\trianglelefteq}\, \mathcal{N} T \text{ for each }i=1,2. \end{align} $$

For the proof let $i\in \{1,2\}$ , $x\in E({\mathcal {H}}_i)$ and $f\in \mathcal {N} T$ with $u:=(f^{-1},x,f)\in \mathbf {D}$ . Then there exist $n\in \mathcal {N}$ and $s\in T$ with $f=ns$ . By [Reference Chermak3, Lemma 1.4(f)], $f^{-1}=s^{-1}n^{-1}$ . Moreover, by [Reference Henke11, Lemma 2.8], we have $S_f=S_{(n,s)}$ and $S_{f^{-1}}=S_{(s^{-1},n^{-1})}$ . Hence, for $v:=(s^{-1},n^{-1},x,n,s)$ , we have $S_v=S_u$ . Applying (3.1) twice, it follows first that $S_v=S_u\in \Delta $ and then that $v\in \mathbf {D}$ . Hence, by the axioms of a partial group, we have $x^f=\Pi (u)=\Pi (v)=(x^n)^s$ . As $E({\mathcal {H}}_i)\,{\trianglelefteq }\, F^*({\mathcal {L}})\supseteq \mathcal {N}$ by [Reference Henke10, Proposition 11.7], we have $x^n\in E({\mathcal {H}}_i)$ . So $s\in T\leq N_S(E({\mathcal {H}}_i))$ implies that $x^f=(x^n)^s\in E({\mathcal {H}}_i)$ . This shows (4.6). We show next:

(4.7) $$ \begin{align} \mathcal{F}_T(\mathcal{N} T)={\langle}\mathcal{F}_T(E({\mathcal{H}}_1)T),\mathcal{F}_T(E({\mathcal{H}}_2)T){\rangle} \end{align} $$

As $E({\mathcal {H}}_i)T\subseteq \mathcal {N} T$ for $i=1,2$ , we have clearly $\mathcal {F}_0:={\langle }\mathcal {F}_T(E({\mathcal {H}}_1)T),\mathcal {F}_T(E({\mathcal {H}}_2)T){\rangle }\subseteq \mathcal {F}_T(\mathcal {N} T)$ and so it remains to show the opposite inclusion. Thus, fixing $f\in \mathcal {N} T$ , we need to show that $c_f|_{S_f\cap T}$ is a morphism in $\mathcal {F}_0$ . Write $f=nt$ for some $n\in \mathcal {N}$ and $t\in T$ . By [Reference Henke11, Lemma 2.8], we have $S_f=S_{(n,t)}$ , which implies that $c_f\colon S_f\cap T\rightarrow T$ is a composite of restrictions of $c_n|_{S_n\cap T}$ and $c_t|_T$ . As $c_t|_T$ is a morphism in $\mathcal {F}_0$ , it is thus sufficient to show that $c_n|_{S_n\cap T}$ is a morphism in $\mathcal {F}_0$ . By [Reference Henke12, Lemma 3.18(a)] $(\mathcal {N} T,\delta (\mathcal {F}_T(\mathcal {N} T)),T)$ is a regular locality. Moreover, by Theorem 4.8, $\mathcal {N}=E({\mathcal {H}}_1)E({\mathcal {H}}_2)$ and by (4.6), $E({\mathcal {H}}_i)\,{\trianglelefteq }\, \mathcal {N} T$ for each $i=1,2$ . Therefore, [Reference Henke9, Theorem 1] applied with $\mathcal {N} T$ in place of ${\mathcal {L}}$ yields the existence of elements $x\in E({\mathcal {H}}_1)$ and $y\in E({\mathcal {H}}_2)$ with $(x,y)\in \mathbf {D}$ , $n=xy$ and $S_n\cap T=S_{(x,y)}\cap T$ . So $c_n|_{S_n\cap T}$ is the composite of restrictions of $c_x|_{S_x\cap T}$ (which is a morphism in $\mathcal {F}_T(E({\mathcal {H}}_1)T)$ ) and of $c_y|_{S_y\cap T}$ (which is a morphism in $\mathcal {F}_T(E({\mathcal {H}}_2)T)$ ). So $c_n|_{S_n\cap T}$ is a morphism in $\mathcal {F}_0$ and (4.7) holds.

We use now that $T={\mathcal {H}}\cap S=H\cap S$ by Theorem 4.8. Recall also that $H_1$ and $H_2$ are subnormal in $N_{\mathcal {L}}(T^*)$ by Lemma 4.5. Our assumption yields moreover that $H_1$ and $H_2$ are T-invariant and so ${\langle }H_i,T{\rangle }=H_iT$ for $i=1,2$ . Hence, it follows from Theorem C that

$$\begin{align*}\mathcal{F}_T(H)={\langle}\mathcal{F}_T(H_1 T),\mathcal{F}_T(H_2 T){\rangle}.\end{align*}$$

So Lemma 4.9 implies

$$\begin{align*}\mathcal{F}_T({\mathcal{H}})={\langle}\mathcal{F}_T(\mathcal{N} T),\mathcal{F}_T(H){\rangle}={\langle}\mathcal{F}_T(\mathcal{N} T),\mathcal{F}_T(H_1 T),\mathcal{F}_T(H_2 T){\rangle}.\end{align*}$$

Hence, the assertion follows from (4.5) and (4.7).

Proof. This follows from Theorem A using induction on n.

Proof. (a) It was first observed by Puig [Reference Puig16, $\S$ 1.1] that $S\cap O^p(G)=\mathfrak {hyp}(\mathcal {F})$ ; a detailed proof can be found in [Reference David6, Theorem 1.33]. Thus $\mathcal {F}_{S\cap O^p(\mathcal {F})}(O^p(\mathcal {F}))$ is a saturated subsystem of $\mathcal {F}$ over $\mathfrak {hyp}(\mathcal {F})$ . As $O^p(\mathcal {F})$ is characterized in [Reference Aschbacher, Kessar and Oliver1, Theorem I.7.4] as the unique saturated subsystem of $\mathcal {F}$ over $\mathfrak {hyp}(\mathcal {F})$ , part (a) follows.

(b) It follows from [Reference Aschbacher, Kessar and Oliver1, Proposition I.6.2] that $\mathcal {E}$ is subnormal in $\mathcal {F}$ , and one observes easily that $N_S(H)\leq N_S(\mathcal {E})$ . For $P\leq N_S(H)$ notice that $O^p(HP)=O^p(H)$ and so (a) yields $O^p(\mathcal {F}_{TP}(HP))=\mathcal {F}_{O^p(H)\cap S}(O^p(H))=O^p(\mathcal {E})$ . As $\mathcal {E} P=(\mathcal {E} P)_{\mathcal {F}}$ is by [Reference Henke12, Theorem B(a)] the unique saturated subsystem of $\mathcal {F}$ over $TP$ with $O^p(\mathcal {E} P)=O^p(\mathcal {E})$ , statement (b) follows.

Hypothesis 5.3. Let $\mathcal {F}$ be a saturated fusion system and $n\geq 1$ . For $i=1,2,\dots ,n$ let $\mathcal {E}_i$ be a subnormal subsystem of $\mathcal {F}$ over $S_i\leq S$ .

Proof. (a) By Corollary 4.11, we have ${\mathcal {H}}\cap S=T$ and ${\mathcal {H}}{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ . In particular, $\mathcal {E}=\mathcal {F}_{{\mathcal {H}}\cap S}({\mathcal {H}})=\mathcal {F}_T({\mathcal {H}})$ is a subsystem over T. Moreover, it follows from [Reference Chermak and Henke5, Proposition 7.1(a)] that $\mathcal {E}$ is subnormal in $\mathcal {F}$ . As ${\mathcal {H}}_i\subseteq {\mathcal {H}}$ and ${\mathcal {H}}_i{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ , it is a consequence of [Reference Chermak and Henke5, Proposition 7.1(c)] that $\mathcal {E}_i=\mathcal {F}_{S\cap {\mathcal {H}}_i}({\mathcal {H}}_i){\,\trianglelefteq \trianglelefteq \,}\mathcal {F}_T({\mathcal {H}})=\mathcal {E}$ for all $i=1,2,\dots ,n$ .

(b) This follows since

$$\begin{align*}{\mathcal{H}}:={\langle}{\mathcal{H}}_1,{\mathcal{H}}_2,\dots,{\mathcal{H}}_n{\rangle}={\langle}\;{\langle}{\mathcal{H}}_1,\dots{\mathcal{H}}_{i_1}{\rangle},{\langle}{\mathcal{H}}_{i_1+1},\dots,{\mathcal{H}}_{i_2}{\rangle},\dots,{\langle} {\mathcal{H}}_{i_{k-1}+1},\dots ,{\mathcal{H}}_{i_k}{\rangle}\;{\rangle}.\\[-35pt]\end{align*}$$

Proof. We use throughout that $T={\mathcal {H}}\cap S$ by Lemma 5.5(a). By [Reference Henke12, Lemma 3.8], $N_S(\mathcal {E}_i)=N_S({\mathcal {H}}_i)$ for $i=1,2$ . Hence, $T\leq N_S({\mathcal {H}}_1)\cap N_S({\mathcal {H}}_2)$ and so Lemma 4.10 gives

$$\begin{align*}\mathcal{E}={\langle}\mathcal{F}_T({\mathcal{H}}_1T),\mathcal{F}_T({\mathcal{H}}_2T){\rangle}.\end{align*}$$

By [Reference Henke12, Theorem B(b)], $\mathcal {F}_T({\mathcal {H}}_iT)=(\mathcal {E}_iT)_{\mathcal {F}}$ for $i=1,2$ . Hence the assertion holds.

Proof. Since $c_x\in \operatorname {Aut}(S)$ induces an automorphism of $\mathcal {F}$ , we have $\mathcal {E}_1^x{\,\trianglelefteq \trianglelefteq \,}\mathcal {F}$ . Indeed, it is shown in [Reference Henke12, Lemma 3.26(a)] that ${\mathcal {H}}_1^x{\,\trianglelefteq \trianglelefteq \,}{\mathcal {L}}$ and $\mathcal {E}_1^x=\mathcal {F}_{S\cap {\mathcal {H}}_1^x}({\mathcal {H}}_1^x)$ (which implies by [Reference Chermak and Henke5, Theorem 7.1(a)] also that $\mathcal {E}_1^x{\,\trianglelefteq \trianglelefteq \,}\mathcal {F}$ ). As $x\in T\subseteq {\mathcal {H}}={\langle }{\mathcal {H}}_1,{\mathcal {H}}_2,\dots ,{\mathcal {H}}_n{\rangle }$ , we have moreover that ${\mathcal {H}}={\langle }{\mathcal {H}}_1,{\mathcal {H}}_1^x,{\mathcal {H}}_2,\dots ,{\mathcal {H}}_n{\rangle }$ . It follows now from the definitions of $\mathcal {E}$ and of ${\langle }\!{\langle } \mathcal {E}_1,\mathcal {E}_1^x,\mathcal {E}_2,\dots ,\mathcal {E}_n{\rangle }\!{\rangle }$ that

$$\begin{align*}\mathcal{E}=\mathcal{F}_{{\mathcal{H}}\cap S}({\mathcal{H}})={\langle}\!{\langle}\mathcal{E}_1,\mathcal{E}_1^x,\mathcal{E}_2,\dots,\mathcal{E}_n{\rangle}\!{\rangle}\\[-42pt]\end{align*}$$

Proof. If $n=1$ , then $\mathcal {E}=\mathcal {E}_1{\,\trianglelefteq \trianglelefteq \,}\mathcal {G}$ . Hence, using Lemma 5.5(b) and induction on n, we can reduce to the case $n=2$ . Thus, we assume from now on that

$$\begin{align*}n=2.\end{align*}$$

Suppose moreover that $(\mathcal {F},\mathcal {G},\mathcal {E}_1,\mathcal {E}_2,{\mathcal {L}},\Delta ,S)$ is a counterexample such that first $|S:T|$ and then $|\mathcal {E}_1|+|\mathcal {E}_2|$ is maximal, where $T={\langle }S_1,S_2{\rangle }$ as before and $|\mathcal {E}_i|$ denotes the number of morphisms in $\mathcal {E}_i$ for $i=1,2$ .

Let $S'\leq S$ such that $\mathcal {G}$ is a subsystem over $S'$ . As $\mathcal {E}_1$ and $\mathcal {E}_2$ are contained in $\mathcal {G}$ , we have $T\leq S'$ . Fix now a regular locality $({\mathcal {L}}',\Delta ',S')$ over $\mathcal {G}$ (which exists by [Reference Henke10, Lemma 10.4]). By [Reference Chermak and Henke5, Theorem E(a)], there exist partial subnormal subgroups ${\mathcal {H}}_1',{\mathcal {H}}_2'$ of ${\mathcal {L}}'$ such that $S_i={\mathcal {H}}_i'\cap S'$ and $\mathcal {E}_i=\mathcal {F}_{S_i}({\mathcal {H}}_i')$ for $i=1,2$ . So we may define

$$\begin{align*}{\langle}\!{\langle} \mathcal{E}_1,\mathcal{E}_2{\rangle}\!{\rangle}_{\mathcal{G}}:=\mathcal{F}_{S\cap {\mathcal{H}}'}({\mathcal{H}}')\text{ where }{\mathcal{H}}':={\langle}{\mathcal{H}}_1',{\mathcal{H}}_2'{\rangle}.\end{align*}$$

Notice that ${\langle }\!{\langle } \mathcal {E}_1,\mathcal {E}_2{\rangle }\!{\rangle }_{\mathcal {G}}{\,\trianglelefteq \trianglelefteq \,}\mathcal {G}$ by Lemma 5.5(a) applied with $(\mathcal {G},{\mathcal {L}}',\Delta ',S')$ in place of $(\mathcal {F},{\mathcal {L}},\Delta ,S)$ .

Assume now first that $T\leq N_S(\mathcal {E}_1)\cap N_S(\mathcal {E}_2)$ . Then Lemma 5.6 yields $\mathcal {E}={\langle } (\mathcal {E}_1T)_{\mathcal {F}},(\mathcal {E}_2T)_{\mathcal {F}}{\rangle }$ . As $T\leq S'$ , we have $T\leq N_{S'}(\mathcal {E}_1)\cap N_{S'}(\mathcal {E}_2)$ . Hence, Lemma 5.6 applied with $(\mathcal {G},{\mathcal {L}}',\Delta ',S')$ in place of $(\mathcal {F},{\mathcal {L}},\Delta ,S)$ yields also that ${\langle }\!{\langle } \mathcal {E}_1,\mathcal {E}_2{\rangle }\!{\rangle }_{\mathcal {G}}={\langle }(\mathcal {E}_1 T)_{\mathcal {G}},(\mathcal {E}_2 T)_{\mathcal {G}}{\rangle }$ . It follows now from Remark 5.1 that $(\mathcal {E}_i T)_{\mathcal {G}}=(\mathcal {E}_i T)_{\mathcal {F}}$ for $i=1,2$ and thus

$$\begin{align*}\mathcal{E}={\langle}\!{\langle} \mathcal{E}_1,\mathcal{E}_2{\rangle}\!{\rangle}_{\mathcal{G}}{\,\trianglelefteq\trianglelefteq\,}\mathcal{G}.\end{align*}$$

This contradicts the assumption that $(\mathcal {F},\mathcal {G},\mathcal {E}_1,\mathcal {E}_2,{\mathcal {L}},\Delta ,S)$ is a counterexample. Hence, $T\not \leq N_S(\mathcal {E}_i)$ for some $i=1,2$ . Since the situation is symmetric in $\mathcal {E}_1$ and $\mathcal {E}_2$ , we may assume without loss of generality that

$$\begin{align*}T\not\leq N_S(\mathcal{E}_1).\end{align*}$$

This means that $T_0:=T\cap N_S(\mathcal {E}_1)<T$ and thus $T_0<N_T(T_0)$ . Fix $x\in N_T(T_0)\backslash T_0$ . By Lemma 5.7, we have $\mathcal {E}_1^x{\,\trianglelefteq \trianglelefteq \,}\mathcal {F}$ and $\mathcal {E}_1^x{\,\trianglelefteq \trianglelefteq \,}\mathcal {G}$ . Set now

$$\begin{align*}\tilde{\mathcal{E}}_1:={\langle}\!{\langle} \mathcal{E}_1,\mathcal{E}_1^x{\rangle}\!{\rangle}.\end{align*}$$

Notice that $\mathcal {E}_1^x$ is a subsystem over $S_1^x$ and $\tilde {S}_1:={\langle }S_1,S_1^x{\rangle }\leq T_0<T$ as $S_1\leq T_0$ . Thus, $|S:\tilde {S}_1|>|S:T|$ , and so the maximality of $|S:T|$ yields that $(\mathcal {F},\mathcal {G},\mathcal {E}_1,\mathcal {E}_1^x,{\mathcal {L}},\Delta ,S)$ is not a counterexample. Hence,