$\def\puz{\mathop{\strut\mathrm{puz}}}$

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 $\rho\tau^{-1}\rho\pi^{-1}\tau\pi^2\rho\pi^2\rho^{-1}=(z,a_1,c_1)$.