Proof. Point (a) is given by generalised associativity in the binary group G. Point (b) is a straightforward exercise and is left to the reader.

Proof. Let $u\circ v\in \mathbf {D}$ . Then Definition 1.1(3) applies to $(\emptyset )\circ u\circ v$ and yields $(\Pi (u))\circ v\in \mathbf {D}$ with $\Pi (u\circ v)=\Pi ((\Pi (u))\circ v)$ . Now apply Definition 1.1(3) to $(\Pi (u))\circ v\circ (\emptyset )$ to obtain (a).

Let $u\circ v\circ w\in \mathbf {D}$ . Then $u\circ v$ and w are in $\mathbf {D}$ by Definition 1.1(1), and $\mathbf {D}$ -multiplicativity yields $\Pi (u\circ v\circ w)=\Pi (u\circ v)\Pi (w)$ . Similarly, $\Pi (u\circ v\circ w)=\Pi (u)\Pi (v\circ w)$ , and (b) holds.

Since $\mathbf {1}=\Pi (\emptyset )$ , it is immediate from Definition 1.1(3) that if $w=u\circ v\in \mathbf {D}$ , then $w'=u\circ (\mathbf {1})\circ v\in \mathbf {D}$ . On the other hand, suppose that $w'\in \mathbf {D}$ . If u and v are empty, then $w=(\emptyset )\in \mathbf {D}$ . Suppose u is not the empty word, and write $u=u_0\circ (x)$ . Then $w'=u_0\circ (x,\mathbf {1})\circ v$ , and since $\Pi (x,\mathbf {1})=x$ , we obtain $w\in \mathbf {D}$ via $\mathbf {D}$ -associativity. Similarly, $w\in \mathbf {D}$ if v is nonempty, so (c) holds.

Let $u\circ v\in \mathbf {D}$ . Then $v^{-1}\circ u^{-1}\circ u\circ v\in \mathbf {D}$ by Definition 1.1(4), and then $u^{-1}\circ u\circ v\in \mathbf {D}$ by Definition 1.1(1). Multiplicativity then yields

$$ \begin{align*} \Pi(u^{-1}\circ u\circ v)=\Pi(u^{-1}\circ u)\Pi(v)=\mathbf{1}\Pi(v)= \Pi(\emptyset)\Pi(v)=\Pi(\emptyset\circ v)=\Pi(v). \end{align*} $$

As $(w^{-1})^{-1}=w$ for any $w\in \mathbf {W}$ , one obtains $w\circ w^{-1}\in \mathbf {D}$ for any $w\in \mathbf {D}$ , and $\Pi (w\circ w^{-1})=\mathbf {1}$ . From this one easily completes the proof of (d).

Now let $u\circ v$ and $u\circ w$ be in $\mathbf {D}$ , with $\Pi (u\circ v)=\Pi (u\circ w)$ . Then (d) (together with multiplicativity and associativity, which will not be explicitly mentioned hereafter) yield

$$ \begin{align*}\Pi(v)=\Pi(u^{-1}\circ u\circ v)=\Pi(u^{-1})\Pi(u)\Pi(v)=\Pi(u^{-1})\Pi(u)\Pi(w)= \Pi(u^{-1}\circ u\circ w)=\Pi(w), \end{align*} $$

and (e) holds.

Let $u\in \mathbf {D}$ . Then $u\circ u^{-1}\in \mathbf {D}$ , and then $\Pi (u)\Pi (u^{-1})=\mathbf {1}$ . But also $(\Pi (u),\Pi (u)^{-1})\in \mathbf {D}$ , and $\Pi (u)\Pi (u)^{-1}=\mathbf {1}$ . Now (f) follows by Definition 1.1(2) and cancellation.

Let $u,v,w$ be as in (g). Then $u^{-1}\circ u\circ v$ and $u^{-1}\circ u\circ w$ are in $\mathbf {D}$ by (d). By two applications of (d), $\Pi (u^{-1}\circ u\circ v)=\Pi (v)=\Pi (w)=\Pi (u^{-1}\circ u\circ w)$ , so $\Pi (u\circ v)=\Pi (u\circ w)$ by (e), and (g) holds.

Proof.

1. (a) We are given $(f,g)\in \mathbf {D}$ , so $(f^{-1},f,g)\in \mathbf {D}$ and $\Pi (f^{-1},f,g)=g$ , by Lemma 1.4(d) and $\mathbf {D}$ -associativity. We are also given $f\in \mathbf {D}(g)$ and $fg=gf$ , so

   $$ \begin{align*}f^g=\Pi(g^{-1},f,g)=\Pi((g^{-1},fg)=\Pi(g^{-1},gf)=\Pi(g^{-1},g,f)=f. \end{align*} $$

2. (b) Set $v=(f^{-1},g,g^{-1},f,g)$ . As $(g^{-1},f,g)\in \mathbf {D}$ , it follows from Lemma 1.4(d) that also $v\in \mathbf {D}$ and $\Pi (v)=g$ . Then $(f^{-1},g,f)=(f^{-1},g,f^g)\in \mathbf {D}$ by $\mathbf {D}$ -associativity and $\Pi (v)=g^f=g$ . Also, from $v\in \mathbf {D}$ , we obtain $(f,g)\in \mathbf {D}$ from Definition 1.1(1), so $(g,g^{-1},f,g)\in \mathbf {D}$ by Lemma 1.4(d). Then $\mathbf {D}$ -associativity yields $fg=\Pi (g,g^{-1},f,g)=gf^g=gf$ .

Notation. From now on, in any given partial group $\mathcal {L}$ , usage of the symbol ‘ $x^g$ ’ shall be taken to imply $x\in \mathbf {D}(g)$ . More generally, for X a subset of $\mathcal {L}$ and $g\in \mathcal {L}$ , usage of ‘ $X^g$ ’ shall be taken to mean that $X\subseteq \mathbf {D}(g)$ , whereupon $X^g$ is by definition the set of all $x^g$ with $x\in X$ .

Proof. By Definition 1.1(4), $g\circ \emptyset \circ g^{-1}=g\circ g^{-1}\in \mathbf {D}$ , so $\mathbf {1}\in \mathbf {D}(g)$ and then $\mathbf {1}^g=\mathbf {1}$ by Lemma 1.4(c). Thus (a) holds. Now let $x\in \mathbf {D}(g)$ , and set $w=(g^{-1},x,g)$ . Then $w\in \mathbf {D}$ , and $w^{-1}=(g^{-1},x^{-1},g)$ by definition in Definition 1.1. Then Definition 1.1(4) yields $w^{-1}\circ w\in \mathbf {D}$ , so $w^{-1}\in \mathbf {D}$ by Definition 1.1(1). This shows that $\mathbf {D}(g)$ is closed under inversion. Also, Definition 1.1(4) yields $\mathbf {1}=\Pi (w^{-1}\circ w)=(x^{-1})^g x^g$ , and then $(x^{-1})^g=(x^g)^{^{-1}}$ by Lemma 1.4(f). This completes the proof of (b).

As $w\in \mathbf {D}$ , Lemma 1.4(d) implies that $g\circ w$ and then $g\circ w\circ g^{-1}$ are in $\mathbf {D}$ . Now Definition 1.1(3) and two applications of Lemma 1.4(d) yield

$$ \begin{align*} gx^gg^{-1}=\Pi(g,g^{-1},x,g,g^{-1})=\Pi((g,g^{-1},x)\circ g\circ g^{-1})=\Pi(g,g^{-1},x)=x. \end{align*} $$

Thus $x^g\in \mathbf {D}(g^{-1})$ with $(x^g)^{g^{-1}}=x$ , and thus (c) holds.

Finally, $\mathbf {1}=\mathbf {1}^{-1}$ by Lemma 1.4(f), and $\emptyset \circ x\circ \emptyset =x\in \mathbf {D}$ for any $x\in \mathcal {L}$ , proving (d).

Proof. One observes that in all the points (a) through (e), the requisite closure with respect to inversion obtains. Thus, we need only be concerned with products.

1. (a) Let $\mathcal {E}\leq \mathcal {H}$ be a partial subgroup of $\mathcal {H}$ . Then

   $$ \begin{align*}\mathbf{D}(\mathcal{E})=\mathbf{W}(E)\cap\mathbf{D}(\mathcal{H})=\mathbf{W}(E)\cap(\mathbf{W}(\mathcal{H})\cap\mathbf{D}(\mathcal{L})= \mathbf{W}(E)\cap\mathbf{D}(\mathcal{L}), \end{align*} $$
   and (a) follows.

2. (b) Suppose $\mathcal {K}\subseteq \mathcal {H}$ , and let $w\in \mathbf {W}(\mathcal {K})\cap \mathbf {D}(\mathcal {H})$ . As $\mathbf {D}(\mathcal {H})\leq \mathbf {D}(\mathcal {L})$ , and since $\mathcal {K}\leq \mathcal {L}$ by hypothesis, we obtain $\Pi (w)\in \mathcal {K}$ .

3. (c) Assuming now that $\mathcal {H}$ is a subgroup of $\mathcal {L}$ , we have $\mathbf {W}(\mathcal {H})\subseteq \mathbf {D}(\mathcal {L})$ , and then $\mathbf {D}(\mathcal {H}\cap \mathcal {K})\subseteq \mathbf {D}(\mathcal {H})\cap \mathbf {D}(\mathcal {K})$ , so that $\mathcal {H}\cap \mathcal {K}$ is a subgroup of both $\mathcal {H}$ and $\mathcal {K}$ .

4. (d) Let $\mathcal {K}\trianglelefteq \mathcal {L}$ , and let $x\in \mathcal {H}\cap \mathcal {K}$ and $h\in \mathcal {H}$ with $(h^{-1},x,h)\in \mathbf {D}(\mathcal {H})$ . Then $(h^{-1},x,h)\in \mathbf {D}(\mathcal {L})$ , and $x^h\in \mathcal {K}$ . As $\mathcal {H}\leq \mathcal {L}$ , we also have $x^h\in \mathcal {H}$ , so $\mathcal {H}\cap \mathcal {K}\trianglelefteq \mathcal {H}$ . Now suppose further that $\mathcal {H}$ is a subgroup of $\mathcal {L}$ . That is, assume that $\mathbf {W}(\mathcal {H})\subseteq \mathbf {D}(\mathcal {L})$ . Then $\mathbf {W}(\mathcal {H}\cap \mathcal {K})\subseteq \mathbf {D}(\mathcal {L})$ , hence $\mathcal {H}\cap \mathcal {K}$ is a subgroup of $\mathcal {H}$ and evidently a normal subgroup.

5. (e) Set $\mathcal {X}=\bigcap \{\mathcal {H}_i\}_{i\in I}$ . Then $\Pi (w)\in \mathcal {X}$ for all $w\in \mathbf {W}(\mathcal {X})\cap \mathbf {D}(\mathcal {L})$ , so $\mathcal {X}\leq \mathcal {L}$ . The last part of (e) may be left to the reader.

Proof. Let Y be the union of the sets $X_i$ . Each $X_i$ is closed under inversion by Lemma 1.4(f), and $Y\neq \emptyset $ since $\mathbf {1}=\Pi (\emptyset )$ . Since Y is closed under products, by construction, we get $Y\leq \langle X\rangle $ , and then $Y=\langle X\rangle $ by the definition of $\langle X\rangle $ .

Proof. The proof is identical to the proof for binary groups and is left to the reader.

Proof. Since $\mathbf {1} \mathbf {1}=\mathbf {1}$ , (H1) and (H2) yield $\mathbf {1}\beta =(\mathbf {1}\mathbf {1})\beta =(\mathbf {1}\beta )(\mathbf {1}\beta )$ , and then $\mathbf {1}\beta =\mathbf {1}'$ by left or right cancellation. Since $(g,g^{-1})\in \mathbf {D}$ for any $g\in \mathcal {L}$ by Lemma 1.4(d), (H1) yields $(g\beta ,(g^{-1})\beta )\in \mathbf {D}'$ , and then $\mathbf {1}\beta =(gg^{-1})\beta =(g\beta )((g^{-1})\beta )$ by (H2). As $\mathbf {1}\beta =\mathbf {1}'=(g\beta )(g\beta )^{-1}$ , left cancellation yields $(g^{-1})\beta =(g\beta )^{-1}$ .

Proof. By Lemma 1.13, $\mathcal {N}$ is closed under inversion. For w in $\mathbf {W}(\mathcal {N})\cap \mathbf {D}$ , the map $\beta ^*:\mathbf {W}\to \mathbf {W}'$ sends w to a word of the form $(\mathbf {1}',\cdots ,\mathbf {1}')$ . Then $\Pi '(w\beta ^*)=\mathbf {1}'$ by Lemma 1.4(c), and thus $\Pi (w)\in \mathcal {N}$ and $\mathcal {N}$ is a partial subgroup of $\mathcal {L}$ . Now let $f\in \mathcal {L}$ , and let $g\in \mathcal {N}\cap \mathbf {D}(f)$ . Then

$$ \begin{align*} (f^{-1},g,f)\beta^*=((f\beta)^{-1},{\mathbf{1}}',f\beta) \quad\text{(by Lemma 1.13)}, \end{align*} $$

so that

$$ \begin{align*} (g^f)\beta=\Pi'((f^{-1},g,f)\beta^*)=\Pi'((f\beta)^{-1},\mathbf{1}',f\beta)=\mathbf{1}' \end{align*} $$

(again using Lemma 1.4(c)). Thus $\mathcal {N}\trianglelefteq \mathcal {L}$ .

Proof. We are given $\mathbf {W}(M)\subseteq \mathbf {D}(\mathcal {L})$ , so $\beta ^*$ maps $\mathbf {W}(M)$ into $\mathbf {D}(\mathcal {L}')$ .

Proof. We leave to the reader the proof that if $\alpha $ is a homomorphism of partial groups, then $\alpha $ is a homomorphism of binary groups. Now suppose that $\alpha $ is a homomorphism of binary groups. As $\mathbf {W}(G)=\mathbf {D}(G)$ (and similarly for $G'$ ), it is immediate that $\alpha ^*$ maps $\mathbf {D}(G)$ into $\mathbf {D}(G)$ . Assume that $\alpha $ is not a homomorphism of partial groups, and let $w\in \mathbf {D}(G)$ be of minimal length subject to $\Pi '(w\alpha ^*)\neq (\Pi (w))\alpha $ . Then $n>1$ , and we can write $w=u\circ v$ with both u and v nonempty. Then

$$ \begin{align*}\Pi'(w\alpha^*)=\Pi'(u\alpha^*\circ v\alpha^*)=\Pi'(u\alpha^*)\Pi'(v\alpha^*)=((\Pi(u))\alpha)((\Pi(v))\alpha)=(\Pi(u)\Pi(v))\alpha, \end{align*} $$

as $\alpha $ is a homomorphism of binary groups. Since $(\Pi (u)\Pi (v))\alpha =(\Pi (w))\alpha $ , the proof is complete.

Proof.

1. (a) Let $X\in \Delta $ , and let $u\in \mathbf {W}(N_{\mathcal {L}}(X))$ . Then $u\in \mathbf {D}$ via X, $\mathbf {1}\in N_{\mathcal {L}}(X)$ (Lemma 1.6(d)), and $N_{\mathcal {L}}(X)^{-1}=N_{\mathcal {L}}(X)$ (Lemma 1.6(c)).

2. (b) Let $x,y\in N_{\mathcal {L}}(X)$ , and set $v=(g^{-1},x,g,g^{-1},y,g)$ . Then (with assistance from Lemma 1.6(c)) $v\in \mathbf {D}$ via Y, and then $\Pi (v)=(xy)^g=x^gy^g$ (using points (a) and (b) of Lemma 1.4). Thus, the conjugation map $c_g:N_{\mathcal {L}}(X)\to N_{\mathcal {L}}(Y)$ is a homomorphism of binary groups (see Lemma 1.3). Since $c_{g^{-1}}=c_g^{-1}$ by Lemma 1.6(c), $c_g$ is an isomorphism of groups.

3. (c) Let $x\in N_{\mathcal {L}}(X_0)$ , set $u_x=w^{-1}\circ (x)\circ w$ and observe (using Lemma 1.6(c)) that $u_x\in \mathbf {D}$ via $X_n$ . Then $\Pi (u_x)$ can be written as $(\cdots (x)^{g_1}\cdots )^{g_n}$ , and this yields (c).

Proof. Point (a) is a fact concerning partial groups in general and is immediate from Lemma 1.4(c). Now consider the setup in (b). As $(f,g)\in \Delta $ , we also have $(f^{-1},f,g)\in \Delta $ and $g=\Pi (f^{-1},f,g)=f^{-1}(fg)$ . Now observe that $(f^{-1},fg)\in \mathbf {D}$ via $P^f$ , and apply Lemma 2.3(c) to obtain $X^{fg}=((X^f)^{f^{-1}})^{fg}=(X^f)^g$ .

Proof. Fix $g\in \mathcal {L}$ . Then the word $(g)$ of length 1 is in $\mathbf {D}$ by Definition 1.1(2), and since $\mathbf {D}=\mathbf {D}_\Delta $ by (O1), there exists $X\in \Delta $ such that $Y:=X^g\in \Delta $ . Let $a\in S_g$ , and set $b=a^g$ . Then $X^a$ and $Y^b$ are subgroups of S (as $a,b\in S$ ), so $X^a$ and $Y^b$ are in $\Delta $ by (O2). Then $(a^{-1},g,b)\in \mathbf {D}$ via $X^a$ , so also $(g,b)\in \mathbf {D}$ . Also $(a,g)\in \mathbf {D}$ via $X^{a^{-1}}$ . Since $g^{-1} ag=b$ , we get $ag=gb$ by Lemma 2.4(a), and hence

$$ \begin{align*}a^{-1} gb=a^{-1}(gb)=a^{-1}(ag)=(a^{-1} a)g=g \end{align*} $$

by $\mathbf {D}$ -associativity. Since $a^{-1} gb$ conjugates $X^a$ to $Y^b$ by Lemma 2.3(c), we draw the following conclusion.

1. (1) $X^a\leq S_g$ and $(X^a)^g\in \Delta $ for all $a\in S_g$ .

Now let $c,d\in S_g$ . Then (1) shows that both $X^c$ and are members of $\Delta $ that are conjugated to members of $\Delta $ by g. Setting $w=(g^{-1},c,g,g^{-1},d,g)$ , we conclude (by following $X^g$ along the chain of conjugations given by w) that $w\in \mathbf {D}$ via $X^g$ . Then $\mathbf {D}$ -associativity yields

(2) $$ \begin{align} \Pi(w)=(cd)^g=c^g d^g. \end{align} $$

Since $c^g$ and $d^g$ are in S, we conclude that $cd\in S_g$ . Since $S_g$ is closed under inversion by Lemma 1.6(b), $S_g$ is a subgroup of S. As $X\leq S_g\leq S$ , where X and S are in $\Delta $ , (O2) now yields $S_g\in \Delta $ . Thus (a) holds.

Since $c_{g^{-1}}=(c_g)^{-1}$ by Lemma 1.6(c), it follows that $S_{g^{-1}}=(S_g)^g$ . Points (b) and (c) are then immediate from (a) and Lemma 2.3(b).

Proof. Let $x_0,y_0\in S_w$ , and let $\sigma =(x_0,\cdots ,x_n)$ and $\tau =(y_0,\cdots ,y_n)$ be the corresponding sequences of elements of S, obtained from $x_0$ and $y_0$ via the sequence of maps $c_{g_1}\circ \cdots \circ c_{g_k}$ . Set $S_j=S_{g_j} (1\leq j\leq n)$ . Then $x_{i-1}$ and $y_{i-1}$ are elements of $S_{i-1}$ , so $x_{i-1}y_{i-1}\in S_{i-1}$ by Proposition 2.5(a). As $c_{g_i}$ restricts to a homomorphism on $S_{i-1}$ (see Lemma 2.3(b)), it follows that $x_iy_i\in S_i$ . Thus $S_w$ is closed under the binary product in S. That $S_w$ is closed under inversion is given by Lemma 1.6(c), so $S_w$ is a subgroup of S.

Suppose that $S_w\in \Delta $ , set $P_0=S_w$ , and recursively define $P_k$ for $0\langle k\leq n$ by $P_k=(P_{k-1})^{g_k}$ . Then $P_k$ is a subgroup of S by Proposition 2.5(c) and induction on k. Then (O2) yields $P_k\in \Delta $ , so $w\in \mathbf {D}$ by (O1). Conversely, if $w\in \mathbf {D}$ , then (O1) shows that $P\leq S_w$ for some $P\in \Delta $ , and then $S_w\in \Delta $ by (O2).

Proof. Set $P=(S_w)^{g_0^{-1}}$ . One then observes that the word displayed in (*) is in $\mathbf {D}$ via P. Now fix $g\in N_{\mathcal {L}}(S)$ . Applying (*) with $\ell (w)=1$ and with $g_0=g^{-1}$ and $g=g_1$ , we conclude that there is a well-defined mapping $\alpha :\mathcal {L}\to \mathcal {L}$ given by conjugation by g. To show that $\alpha $ is a homomorphism, we need to check that the induced map $\alpha _*:\mathbf {W}(\mathcal {L})\to \mathbf {W}(\mathcal {L})$ sends $\mathbf {D}$ into $\mathbf {D}$ and that

(**) $$ \begin{align} \Pi(w\alpha^*)=\Pi(w)\alpha. \end{align} $$

Set

$$ \begin{align*}u=(g^{-1},x_1,g,g^{-1},x_2,\cdots,x_{n-1},g,g^{-1},x_n,g). \end{align*} $$

Applying (*) with $g^{-1}$ in the role of $g_i$ for all i then yields $u\in \mathbf {D}$ via $P^g$ . Now set $v=((x_1)^g,\cdots ,(x_n)^g)$ . Then $\Pi (v)=\Pi (u)=\Pi (w)^g$ by $\mathbf {D}$ -associativity, and this yields (**). Replacing g with $g^{-1}$ yields the homomorphism $\alpha ^{-1}$ by Lemma 1.6(c), and thus $\alpha $ is an automorphism of $\mathcal {L}$ .

Proof. Observe first of all that by (L2), the lemma holds for $P=S$ and $g=\mathbf {1}$ . Among all P for which the lemma fails to hold, choose P so that first $|P|$ and then $|N_S(P)|$ is as large as possible. Set $R=N_S(P)$ , and let $R^*$ be a Sylow p-subgroup of $N_{\mathcal {L}}(P)$ containing R. Then $R<R^*$ (proper subgroup), and then also $R<N_{R^*}(R)$ . We have $P\neq R$ as $P\neq S$ , and then the maximality of $|P|$ yields the existence of an element $f\in N_{\mathcal {L}}(R,S)$ with $N_S(R^f)\in Syl_p(N_{\mathcal {L}}(R^f))$ .

By Lemma 2.3(b), f-conjugation induces an isomorphism

$$ \begin{align*}N_{\mathcal{L}}(R)\xrightarrow{c_f} N_{\mathcal{L}}(R^f). \end{align*} $$

By Sylow’s theorem, there exists $x\in N_{\mathcal {L}}(R^f)$ such that $(N_{R^*}(R)^f)^x\leq N_S(R^f)$ . Here $(f,x)\in \mathbf {D}$ via R, so Lemma 2.3(c) yields $(N_{R^*}(R)^f)^x=N_{R^*}(R)^{fx}$ . Thus, by replacing f with $fx$ , we may assume that f was chosen to begin with so that $N_{R^*}(R)^f\leq N_S(R^f)$ . Since $R^*$ normalises P and $c_f$ is an isomorphism, it follows that $N_{R^*}(R)^f$ normalises $P^f$ , and thus $|N_S(P^f)|>|N_S(P)|$ . The maximality of $|N_S(P)|$ in the choice of P then implies that $P^f$ is not a counterexample to the lemma. Set $Q=P^f$ . Thus there exists $h\in N_{\mathcal {L}}(N_S(Q),S)$ such that $N_S(Q^h)$ is a Sylow subgroup of $N_{\mathcal {L}}(Q^h)$ . Here $(f,h)\in \mathbf {D}$ via R, so $Q^h=P^g$ , where $g=fh$ , so P is not a counterexample to the lemma, and therefore no counterexample exists.

Proof. For any $w=(h_1,\cdots ,h_n)\in \mathbf {W}(H)$ , let $w'$ be the word $(g_1,\cdots ,g_n)$ defined by $g_i=h_1\cdots h_i$ . As H is finite, we may choose w so as to maximise the cardinality of the set $X=\{g_1,\cdots ,g_n\}$ . Suppose that $X\neq H$ , let $g\in H-X$ , and set $h=\Pi (w)^{-1} g$ . Then the set of entries of $(w\circ (h))'$ is $X\cup \{g\}$ , contrary to the maximality of X. Thus $X=H$ .

We have $\mathbf {W}(H)\subseteq \mathbf {D}$ as H is a subgroup of $\mathcal {L}$ , and thus $w\in \mathbf {D}$ via some $P\in \Delta $ . Then $P^{g_i}=P^{h_1\cdots h_i}\leq S$ for all i, so $P^h\leq S$ for all $h\in H$ . Set $U=\langle P^h\mid h\in H\rangle $ (the subgroup of S generated by the union of all $P^h$ for $h\in H$ ). Then $U\in \Delta $ by (O2). To complete the proof of (a) and (b) it now suffices to show that $H\leq N_{\mathcal {L}}(U)$ , and for this it is enough to observe that, by Lemma 2.4(b), $(P^f)^g$ is defined and is equal to $P^{fg}$ for all $f,g\in H$ .

Assume now that H is a p-group. By Lemma 2.9, there exist $V\in \Delta $ and $g\in \mathcal {L}$ such that $V=U^g$ and $N_S(V)\in Syl_p(N_{\mathcal {L}}(V))$ . Let $c_g:N_{\mathcal {L}}(U)\to N_{\mathcal {L}}(V)$ be the isomorphism given by Lemma 2.3(b). Thus $H^g$ is a p-subgroup of $N_{\mathcal {L}}(V)$ , so there exists $x\in N_{\mathcal {L}}(V)$ with $(H^g)^x\leq N_S(V)$ . Since $(g,x)\in \mathbf {D}$ via U, we may apply Lemma 2.3(c), obtaining $(H^g)^x=H^{gx}$ . Thus (c) holds with $gx$ in the role of g.

Proof. Set $\mathcal {L}_R=N_{\mathcal {L}}(R)$ . Then Lemma 2.3(c) shows that $\Pi $ maps $\mathbf {W}(\mathcal {L}_R)\cap \mathbf {D}$ into $\mathcal {L}_R$ , while Lemma 1.6(c) shows that $\mathcal {L}_R$ is closed under the inversion in $\mathcal {L}$ . Thus $\mathcal {L}_R$ is a partial subgroup of $\mathcal {L}$ .

Let $\mathbf {D}(\mathcal {L}_R)$ be the domain of the product in the partial group $\mathcal {L}_R$ . Then $\mathbf {D}(\mathcal {L}_R)=\mathbf {W}(\mathcal {L}_R)\cap \mathbf {D}$ , by definition. Assume now that $\Gamma \subseteq \Delta $ , and let $w\in \mathbf {D}(\mathcal {L}_R)$ via some $P\in \Delta $ . Then $w\in \mathbf {D}(\mathcal {L}_R)$ via $N_P(R)$ , and thus $\mathcal {L}_R$ satisfies the condition (O1) for objectivity in Definition 2.1. Also, as $\Gamma \subseteq \Delta $ , we have $\Gamma =\{P\in \Delta \mid P\leq N_S(R)\}$ , so $\Gamma $ is closed in the sense of condition (O2) in Definition 2.1. This yields (b). Point (c) is then immediate from Definition 2.7.

Proof. Set $Y=O_p(\mathcal {L})$ , let $g\in \mathcal {L}$ , and let $w\in \mathbf {W}(\mathcal {L})$ . Then $Y\leq S_g$ and $Y\leq S_{(g)\circ w}$ , so $Y^g\leq S_w$ . Thus $Y^g\leq Y$ , so $Y\trianglelefteq \mathcal {L}$ .

Now let $X\leq S$ with $X\trianglelefteq \mathcal {L}$ . Then $X\trianglelefteq S$ . Let $g\in \mathcal {L}$ , and set $P=S_g$ . Then conjugation by g is defined on $N_{\mathcal {L}}(P)$ by Lemma 2.3(b), so $N_X(P)^g$ is defined, and then $N_X(P)^g\leq X$ as $X\trianglelefteq \mathcal {L}$ . Thus $N_X(P)^g\leq S$ , so $N_X(P)\leq S_g$ . As $S_gX=PX$ is a subgroup of S, it follows that $X\leq P$ , so $X^g=X$ , and $X\leq O_p(\mathcal {L})$ .

Proof. Write $\mathcal {S}$ for $Syl_p(\mathcal {L})$ . For $S'\in \mathcal {S}$ , note that $N_{\mathcal {L}}(S')$ is a subgroup of $\mathcal {L}$ . By Lemma 2.8, the condition (*) holds with $N_{\mathcal {L}}(S')$ in the role of K. Let $\mathcal {U}$ be the union (taken over all $S'\in \mathcal {S}$ ) of the groups $N_{\mathcal {L}}(S')$ , and let H be the partial subgroup of $\mathcal {L}$ generated by $\mathcal {U}$ . The above observation concerning Lemma 2.8 together with a straightforward argument by induction on word length then yields $\mathbf {W}(\mathcal {U})\subseteq \mathbf {D}$ , and thus H is a subgroup of $\mathcal {L}$ . Moreover, $\mathbf {W}(\mathcal {U})$ has the following property: For any $(g_1,\cdots ,g_n)\in \mathbf {D}(\mathcal {L})$ and any $(u_0,\cdots ,u_n)$ with each $u_i\in \mathbf {W}(\mathcal {U})$ , the word

$$ \begin{align*}u_0\circ g_1\circ u_1^{-1}\circ u_1\circ g_2\circ\cdots\circ u_{n-1}^{-1}\circ u_{n-1}\circ g_n\circ u_n \end{align*} $$

is in $\mathbf {D}$ by induction on the sum of the lengths of the words $u_i$ . The condition (*) then holds with H in the role of K, by $\mathbf {D}$ -associativity. In particular, conjugation by $g\in H$ is an automorphism of $\mathcal {L}$ , and thus H is a subset of K.

Now let $g\in K$ . Then $(\mathcal {L},\Delta ^g,S^g)$ is a locality, so $S^g\in \mathcal {S}$ . Then $S^g\in Syl_p(H)$ , and there exists $h\in H$ with $S^g=S^h$ . The product $gh^{-1}$ is defined, and then $gh^{-1}\in N_{\mathcal {L}}(S)$ . Thus $gh^{-1}\in H$ , so $g\in H$ , and we conclude that $H=K$ . This completes the proof of (a), (b) and (d). Point (e) is immediate from (d), so it remains only to prove (c).

By Proposition 2.10(a), K normalises a member of $\Delta $ . Then, since $S\in Syl_p(K)$ , we obtain $Q:=O_p(K)\in \Delta $ . By Proposition 2.10(b), there exists a word $u\in \mathbf {W}(K)$ such that $Q=S_u$ . Then also $Q=S_{u^{-1}}$ . Let $w\in \mathbf {D}$ , and let $f\in N_{\mathcal {L}}(Q)$ . The word $v:=u\circ w\circ u$ is in $\mathbf {D}$ by (*), and clearly $S_v\leq Q$ and $S_{v^{-1}}\leq Q$ . Set $P=(S_v)^{\Pi (u)}$ . Then $P\leq S_w\cap Q$ and $P^{\Pi (w)}\leq Q$ . As $S_v\in \Delta $ , we have $P\in \Delta $ . Write $w=(x_1,\cdots ,x_n)$ . Then

$$ \begin{align*} (f^{-1},x_1,f,f^{-1},\cdots,f,f^{-1},x_n,f)\in\mathbf{D}\ \ \text{via } P^f, \end{align*} $$

and this shows that conjugation by f is an inner automorphism of $\mathcal {L}$ . Thus $K=N_{\mathcal {L}}(Q)$ , and the proof is complete.

Proof. Take $\Delta _0$ to be the overgroup-closure in S of the set of all $S_w$ for $w\in \mathbf {D}$ . Take $\Delta _1$ to be the union of all the sets $\Gamma $ of subgroups of S that fulfil (L1) and (L2).

Proof.

1. (a) Let $g\in \mathcal {L}$ , let $P\leq S_g$ , and let $t\in T\cap P$ . Then $t^g\in S$ , and $t^g\in \mathcal {N}$ as $\mathcal {N}\trianglelefteq \mathcal {L}$ . Thus $t^g\in T$ , so

   (*) $$ \begin{align} (T\cap P)^g\leq T\cap P^g. \end{align} $$
   Now replace g with $g^{-1}$ and replace P with the subgroup $P^g$ of $S_{g^{-1}}$ . Then (*) becomes the statement
   (**) $$ \begin{align} (T\cap P^g)^{g^{-1}}\leq T\cap P. \end{align} $$
   Conjugating (**) by g yields $T\cap P^g\leq (T\cap P)^g$ , and thus (*) is an equality, as required.

2. (b) Let $a\in P$ . Then $(P^x)^a\leq S$ and $P^a=P$ . Setting $w=(a^{-1},x^{-1},a,x)$ , we then have $w\in \mathbf {D}$ via $P^{xa}$ . Now $\Pi (w)=a^{-1} a^x\in S$ , while also $\Pi (w)=(x^{-1})^ax\in \mathcal {N}$ , so $\Pi (w)\in T$ . Then $a^x\in aT$ , and we have thus shown that $P^x\leq PT$ . Then $P^x T\leq PT$ . The reverse inequality is given (as in the proof of (a)) by replacing x with $x^{-1}$ and replacing P with $P^x$ .

3. (c) Let R be a p-subgroup of $\mathcal {N}$ containing T. By Proposition 2.10(c), there exists $g\in \mathcal {L}$ with $R^g\leq S$ , and then $R^g\leq S\cap \mathcal {N}=T$ . As $T^g=T$ and conjugation by g is injective, we conclude that $R=T$ .

Proof. Set $Q=S_{(x,f)}$ and $P=S_x\cap S_f$ . As $x\in \mathcal {N}$ , we have $Q^xT=QT$ by Lemma 3.1(b). Then since $f\in N_{\mathcal {L}}(T)$ , we obtain $QT\leq S_f$ . Thus $Q\leq P$ . But also $P^xT=PT$ , so $P=Q$ . Now $(f,f^{-1},x,f)\in \mathbf {D}$ via Q, and $\Pi (f,f^{-1},x,f)=xf=fx^f$ . We have

$$ \begin{align*}Q=S_f\cap S_x = S_{(f,f^{-1},x,f)}\leq S_{(f,x^f)}, \end{align*} $$

so in order to complete the proof of (a), it remains to show that $S_{(f,x^f)}\leq Q$ .

On the other hand, in addressing (b), set $R=S_{(f,y)}$ . Then Lemma 3.1(b) yields $R^{fy}T=R^fT$ , so that $R^{fy}T\leq \leq S_{f^{-1}}$ . Set $v=(f,y,f^{-1},f)$ . Then $v\in \mathbf {D}$ via R, and we have $R=S_v$ . Now $\mathbf {D}$ -associativity yields $fy=\Pi (v)=y^{f^{-1}}f$ . In the special case that $y=x^f$ , we obtain

$$ \begin{align*}R=S_{(f,x^f)}=S_{(f,x^f,f^{-1},f)}\leq S_{(x,f)}=Q, \end{align*} $$

completing the proof of (a). The remainder of (b) now follows as an application of (a) to $(y^{f^{-1}},f)$ , and the remainder of (b) as an application of (a) to $(y^{f^{-1}},f)$ .

Proof. We prove only (a), leaving it to the reader to supply a similar argument for (b). As $P^xT=PT$ by Lemma 3.1(b), and since both $P^x$ and T are contained in $S_w$ , we have $P\leq S_w$ . Then $P^g\leq S$ by Lemma 2.3(c), and $P^g\leq S_u$ . In particular, $u\in \mathbf {D}$ via $P^g$ . As $S_u\leq S_{w^{-1}}$ , and $(S_u)^{g^{-1}}\leq P$ , we obtain $S_u\leq P^g$ . Thus $S_u=P^g$ .

Proof. The case $n=1$ is given by Lemma 3.2. For the general case, write

$$ \begin{align*}w=(f_1,g_1)\circ w_1, \end{align*} $$

and set $u_1=(f_2,\cdots ,f_n)$ . Induction on n implies that there exists $x_1\in \mathcal {N}$ such that $S_{w_1}\leq S_{u_1\circ (x_1)}$ and $\Pi (w_1)=\Pi (u_1\circ (x_1))$ . Set

$$ \begin{align*}w'=(f_1,g_1)\circ u_1\circ(x_1). \end{align*} $$

Thus $S_w\leq S_{w'}$ , and we have $\Pi (w)=\Pi (w')$ by $\mathbf {D}$ -associativity. Apply Lemma 3.3(a) to the word $(g_1)\circ u_1$ to obtain (*). A similar argument in which one begins by writing $w=w_n\circ (f_n,g_n)$ yields (**).

Proof. Let $g\in N_{\mathcal {N}}(T)$ , and set $P=S_g$ . Then $P^g=P$ by Lemma 3.1(b). Set $M=N_{\mathcal {L}}(P)$ , $K=\mathcal {N}\cap M$ and $D=N_{C_S(T)}(P)$ . Here M is a subgroup of $\mathcal {L}$ as $P\in \Delta $ , and then K is a normal subgroup of M by Definition 1.7(d). We have $T\trianglelefteq M$ by Lemma 3.1(a), so $D\leq C_M(T)$ , and then $[K,D]\leq C_K(T)$ . Notice that $T\in Syl_p(K)$ by Lemma 3.1(c), so that $K/T$ is a $p'$ -group. Then $C_K(T)=Z(T)\times Y$ , where Y is a $p'$ -group; and thus Y is a normal $p'$ -subgroup of M. Then $[O_p(M),Y]=1$ , so $Y\leq O_p(M)$ by hypothesis, and thus $Y=1$ . We then have $[K,D]\leq T$ , so $[g,D]\leq T$ , and $D\leq P$ . As $C_S(T)P$ is a p-group we thereby obtain $C_S(T)\leq P$ . This shows that $C_S(T)$ is g-invariant for all $g\in \mathcal {N}_{\mathcal {N}}(T)$ and yields the lemma.

Proof. Point (a) is immediate from Definition 3.6. Now suppose that f is $\uparrow $ -maximal, and let $g\in \mathcal {L}$ with $(f^{-1},S_{f^{-1}})\uparrow (g^{-1},S_{g^{-1}})$ . Since $S_{f^{-1}}=(S_f)^f$ and $S_{g^{-1}}=(S_g)^g$ , one obtains a diagram

as in Definition 3.6, from which it is easy to read off the relation $(f,S_f)\uparrow (g,S_g)$ . Then $|S_f|=|S_g|$ as f is $\uparrow $ -maximal, and then also $|S_{f^{-1}}|=|S_{g^{-1}}|$ . Thus $f^{-1}$ is $\uparrow $ -maximal, and (b) holds. Point (c) is immediate from the transitivity of $\uparrow $ .

Proof. By definition of the relation $\uparrow $ , there exist elements $u\in N_{\mathcal {N}}(Q,R)$ and $v\in N_{\mathcal {N}}(Q^g,R^h)$ such that $(u,h,v^{-1},g^{-1})\in \mathbf {D}$ via Q and $\Pi (w)=\mathbf {1}$ .

In particular, $uh=gv$ . Since $T\leq R$ , points (a) and (b) of Lemma 3.1 yield

$$ \begin{align*} T=T^h, Q^uT=QT\leq R, \text{ and } Q^gT=Q^{gv}T\leq R^h. \end{align*} $$

Then

$$ \begin{align*} w:=(u,h,v^{-1},h^{-1})\in\mathbf{D}\quad{\text via}\quad (Q,Q^u,Q^{uh},Q^{uhv^{-1}}=Q^g,Q^{gh^{-1}}). \end{align*} $$

Set $y=\Pi (w)$ . Then $y=u(v^{-1})^{h^{-1}}\in N_{\mathcal {N}}(Q,R)$ . Since $(u,h,v^{-1},h^{-1},h)$ and $(g,v,v^{-1})$ are in $\mathbf {D}$ (as $\mathcal {L}$ is a partial group), we get $yh=uhv^{-1}=g$ . This yields (a). The uniqueness of y is given by right cancellation.

Suppose now that $N_T(Q^g)\in Syl_p(N_{\mathcal {N}}(Q^g))$ . As $N_T(Q^y)^h=N_T(Q^g)$ , it follows from Lemma 2.3(b) that $N_T(Q^y)\in Syl_p(N_{\mathcal {N}}(Q^y))$ .

Proof. Set $P=S_g$ and $Q=P^g$ . We first show

1. (1) Let $y\in N_{\mathcal {N}}(P,S)$ . Then $|T\cap P|=|T\cap P^y|$ and $(g,P)\uparrow (y^{-1} g,P^y)$ . In particular, $y^{-1} g$ is $\uparrow $ -maximal.

We have $|T\cap P|=|T\cap P^y|$ by Lemma 3.1(a). The following diagram

(*)

shows that $(g,P)\uparrow (y^{-1} g,P^y)$ , and then $y^{-1} g$ is $\uparrow $ -maximal by Lemma 3.7(c). Thus (1) holds.

Suppose next that $N_T(P)\in Syl_p(N_{\mathcal {N}}(P))$ . Then $N_T(P)^g\in Syl_p(N_{\mathcal {N}}(Q))$ , by Lemma 2.3(b), and there exists $x\in N_{\mathcal {N}}(Q)$ such that $N_T(Q)^x\leq N_T(P)^g$ . Here $(x,g^{-1})\in \mathbf {D}$ via Q, and we get $N_T(Q)^{xg^{-1}}\leq N_T(P)\leq S$ . Thus $(g^{-1},Q)\uparrow (xg^{-1},N_T(Q)Q)$ . As $g^{-1}$ is $\uparrow $ -maximal by Lemma 3.7, it follows that $N_T(Q)\leq Q$ . As $TQ$ is a p-group, we then have $T\leq Q$ , so $T\leq P$ . We have thus shown

1. (2) If $T\nleq P$ , then $N_T(P)\notin Syl_p(N_{\mathcal {N}}(P))$ .

We next show

1. (3) Suppose that there exists $y\in \mathcal {N}$ such that $P\leq S_y$ and such that $N_T(P^y)\in Syl_p(N_{\mathcal {N}}(P^y))$ . Then $T\leq P$ .

Indeed, under the hypothesis of (3), we have $(g,P)\uparrow (y^{-1} g,P^y)$ by (1). Then $(y^{-1} g,P^y)$ is maximal in $\mathcal {L}\circ \Delta $ , and $P^y=S_{y^{-1} g}$ . If $T\nleq P$ , then $T\nleq P^y$ , and then (2) applies to $(y^{-1} g,P^y)$ in the role of $(g,P)$ and yields a contradiction. So, (3) holds.

Among all counterexamples, let g be chosen so that $|P|$ is as large as possible. By Lemma 2.9, there exists $f\in \mathcal {L}$ so that $Q^f\leq S$ and $N_S(Q^f)\in Syl_p(N_{\mathcal {L}}(Q^f))$ . Set $h=gf$ (where the product is defined via P) and set $R=P^h$ . Let $(h',P')$ be maximal in $\mathcal {L}\circ \Delta $ with $(h,P)\uparrow (h',P')$ . We note that since $R=P^h=Q^f$ , we have $N_T(R)\in Syl_p(N_{\mathcal {N}}(R))$ . So, if $T\leq P'$ , then Lemma 3.8(b) applies and yields $h=yh'$ for some $y\in \mathcal {N}$ such that $P\leq S_y$ and such that $N_T(P^y)\in Syl_p(N_{\mathcal {N}}(P^y))$ . The existence of such an element y contradicts (3), so we conclude that $T\nleq P'$ . Then $(h',P')$ is a counterexample to the proposition, and the maximality of $|P|$ yields $|P|=|P'|$ . Then $(h,P)$ is maximal in $\mathcal {L}\circ \Delta $ , as is $(h^{-1},R)$ by Lemma 3.7(b), so $T\nleq R$ . But $N_T(R)\in Syl_p(N_{\mathcal {N}}(R))$ , so (2) applies with $(h^{-1},R)$ in the role of $(g,P)$ and yields $T\leq R$ . Then $T\leq P$ , and the proof is complete.