\def\puz{\mathop{\strut\mathrm{puz}}}
We are now prepared to prove the main lemma, 2.2.6. As we remarked on page 2.2, it suffices to show that A(V(G)\backslash\{v\})\subseteq \Gamma(v) for a single vertex v in G. The proof is by induction on \beta(G), with base case \beta(G)=2, that is, when G is a \theta-graph other than \theta_0.
Let G be a simple \theta-graph with vertices of degree three y and z. Denote the three paths between y and z by p_1, p_2, and p_3, with \eqalign{ p_1&=(z,a_n,a_{n-1},\ldots,a_1,y)\cr p_2&=(y,b_m,b_{m-1}\ldots,b_1,z)\cr q&=(v,c_1,c_2,\ldots,c_s,y)\cr r&=(v,d_1,d_2,\ldots,d_t,z)\cr p_3&=\overline q r,\cr } where v is an internal vertex of p_3.
Now let \eqalign{ \pi&=\sigma_{q p_2 p_1\overline q}=(a_1,\ldots,a_n,z,b_1,\ldots,b_m)\cr \rho&=\sigma_{r p_1 p_2 \overline r}=(b_1,\ldots,b_m,y,a_1,\ldots,a_n).\cr } These are in \Gamma(v).
Assume first that n\ge 1 and that \{m,n\} is not \{1,2\}, \{2,2\}, or \{2,4\}, so G is not of type (4,2,2), (2,2,2), or (2,2,1). By lemma 2.3.9, A(\{a_1,\ldots,a_n,b_1,\ldots,b_m,y,z\})\subseteq\langle\pi,\rho\rangle. This implies that every 3-cycle with support in \{a_1,\ldots,a_n,b_1,\ldots,b_m,y,z\} is in \Gamma(v). Let \tau=\sigma_{q p_2 \overline r} =(d_1,\ldots,d_t,z,b_1,\ldots,b_m,y,c_s\ldots c_1)\in\Gamma(v). We can produce a 3-cycle in \Gamma(v) containing a_1 and any vertex in \{c_1,\ldots,c_s,d_1,\ldots,d_t\} in the form \tau^p(a_1,y,z)\tau^{-p}. Thus, the 3-cycles in \Gamma(v) generate a subgroup transitive on V(G)\backslash\{v\}, and so by lemma 2.3.5, A(V(G)\backslash\{v\})\subseteq\Gamma(v).
Now suppose that G is of type (4,2,2). Let n=4, m=2, s=1, and t=0. Then \tau^2\rho\pi^{-1}\tau^{-2}\pi^{-1}\rho=(z,a_4,a_1). We can then produce a 3-cycle mapping z to any vertex in V(G)\backslash\{v\} in the form \rho^p(z,a_4,a_1)\rho^{-p} or \tau^{4}(z,a_4,a_1)\tau^{-4}(z,a_4,a_1)^2=(z,c_1,a_4). Once again the result follows from lemma 2.3.5.
If G is of type (2,2,2), let m=n=2, s=1, and t=0. Then \rho\tau^{-1}\rho\pi^{-1}\tau\pi^2\rho\pi^2\rho^{-1}=(z,a_1,c_1). We can then produce a 3-cycle mapping z to any vertex in V(G)\backslash\{v\} in the form \rho^p(z,a_1,c_1)\rho^{-p}
Now let G be a simple 2-connected graph with \beta(G)\ge 3. Write G=H\cup A as in lemma 2.3.10; by lemma 2.3.12, we may do this so that H is not \theta_0 when \beta(G)=3. Let v and w be the endpoints of the handle A. Since \Gamma_G(v)\supseteq\Gamma_H(v), \Gamma_G(v) contains a 3-cycle, and so by lemma 2.3.8 it suffices to show that \Gamma_G(v) is doubly transitive on V(G)\backslash\{v\}.
Let A be the path p=(v,a_t,a_{t-1},\ldots,a_1,w). Let u be a vertex of H other than v and w. Since H is 2-connected, there is a simple path q=(w,b_s,\ldots,b_1,v) in H that does not include u. Then let \sigma=\sigma_{pq}=(w,a_1,\ldots,a_t,b_1,\ldots,b_s), so \sigma^i(w)=a_i and \sigma(u)=u. By the induction hypothesis, A(V(H)\backslash\{v\}\subseteq\Gamma_H(v), so for all vertices x of H, other than u and v, \Gamma_H(v) contains a permutation mapping w to x and fixing u (see exercise 6 in section 2.3). Thus, for all vertices u in H, other than v and w, and for all vertices y and z in G, other than v and u, there is a permutation in \Gamma_G(v) mapping y to z and fixing u.
Finally, we need to show that for all u in \{w,a_1,\ldots,a_t\} and vertices y and z in V(G)\backslash\{v,u\}, there is a permutation in \Gamma_G(v) mapping y to z and fixing u.
Let x\in V(H)\backslash\{v,w\}; there is a permutation \tau\in\Gamma_G(v) such that \tau(u)=x. Then as we have already seen, there is a permutation \mu mapping \tau(y) to \tau(z) and fixing x. Thus \tau^{-1}\mu\tau maps y to z and fixes u.
Exercises 2.4
Ex 2.4.1 In the case that G is of type (2,2,2), Verify that \rho\tau^{-1}\rho\pi^{-1}\tau\pi^2\rho\pi^2\rho^{-1}=(z,a_1,c_1).