Proof. A 3-cycle is an even permutation, so the group $G$ generated by $C$ is contained in $A(X)$. Suppose $\sigma\in G$, so $\sigma$ is a product of an even number of transpositions. Any transposition $(a,b)$ can be written $(a,b)=(y,a)(y,b)(y,a)$, so $\sigma$ can be written as $\sigma=(y,a_1)(y,a_2)\cdots(y,a_m)$, $m$ even, and hence $\sigma=(y,a_2,a_1)(y,a_4,a_3)\cdots(y,a_m,a_{m-1})$, since $(y,a)(y,b)=(y,b,a)$. If $a_{2i-1}=x$, $(y,a_{2i},a_{2i-1})=(x,y,a_{2i})$. Otherwise, $(y,a_{2i},a_{2i-1})=(x,y,a_{2i})(x,y,a_{2i-1})(x,y,a_{2i})^{-1}$. Thus, $\sigma$ can be written as a product of 3-cycles of the form $(x,y,z)$ and $(x,y,z)^{-1}$, so $A(X)\subseteq G$. $\qed$

Proof. If $H\le S(X)$ and $Y\subseteq X$, let $H|_Y=\{\sigma\in H | \|\sigma\|\subseteq Y\}$. $H|_Y$ is a subgroup of $S(X)$ but we interpret it as a subgroup of $S(Y)$.

It is easy to see that (1) implies (2). Suppose that $\Sigma$ has property 2. Let $Y\subseteq X$ such that $|Y|\ge 3$, $\langle\Sigma\rangle|_Y=A(Y)$, and $Y$ is maximal with these properties. We show that $Y=X$, which finishes the proof.

There is a set $Y$ with the first two properties: Let $\sigma=(a,b,c)\in\Sigma$, and let $Y=\{a,b,c\}$. Then $\langle\Sigma\rangle|_Y=\{e,(a,b,c),(a,c,b)\}=A(Y)$. Now let $Y$ be a maximal set with these properties, and suppose that $Y\not=X$.

Since $\Sigma$ has property 2, $\langle\Sigma\rangle$ contains a permutation $\sigma$ that maps an element of $Y$ to an element of $X\backslash Y$. Therefore, $\Sigma$ contains a 3-cycle that maps an element of $Y$ to an element of $X\backslash Y$. This 3-cycle has one of two forms: $(x,y,z)$ with $x,y\in Y$ and $z\notin Y$, or $(x,y,z)$ with $z\in Y$ and $x,y\notin Y$.

In the first case, since $\langle\Sigma\rangle|_Y=A(Y)$, $\langle\Sigma\rangle$ contains all 3-cycles $(x,y,t)$, $t\in Y$. Thus $\langle\Sigma\rangle$ contains all 3-cycles $(x,y,t)$, $t\in Y\cup \{z\}$. By lemma 2.3.1, $\langle\Sigma\rangle|_{Y\cup\{z\}}=A(Y\cup\{z\})$ contradicting the maximality of $Y$.

In the second case, for each $t\in Y$, $\langle\Sigma\rangle|_Y$ contains a permutation $\sigma$ such that $\sigma(z)=t$, because $\langle\Sigma\rangle|_Y=A(Y)$. Hence $\sigma(xyz)\sigma^{-1}=(xyt)\in\langle\Sigma\rangle$. Since this is true for every $t\in Y$, $\langle\Sigma\rangle|_{Y\cup\{x,y\}}=A(Y\cup\{x,y\})$, again by lemma 2.3.1, which is again a contradiction. $\qed$

Proof. Let $\Sigma$ be the set of 3-cycles in $G$. We claim that $G$ preserves the orbits of $\langle\Sigma\rangle$.

First, we show that $\langle\Sigma\rangle\triangleleft G$. Suppose $\sigma\in\langle\Sigma\rangle$, that is, $\sigma$ is a product of 3-cycles. Let $\tau\in G$; we need to show that $\tau\sigma\tau^{-1}\in \langle\Sigma\rangle$. Write $$ \tau\sigma\tau^{-1}=\tau c_1 c_2 c_3\cdots c_k\tau^{-1} =\tau c_1\tau\tau^{-1} c_2\tau\tau^{-1} c_3\tau\tau^{-1} \cdots \tau\tau^{-1}c_k\tau^{-1}, $$ where the $c_i$ are 3-cycles. Now it suffices to show that $\tau c_i\tau^{-1}$ is a 3-cycle. It is easy to verify that $\tau (a,b,c)\tau^{-1}=(\tau(a),\tau(b),\tau(c))$.

Now suppose that $\tau\in G$, that $X_1$ and $X_2$ are orbits of $\langle\Sigma\rangle$, that $x\in X_1$, and that $\tau(x)\in X_2$. We need to show that $\tau(X_1)=X_2$. Suppose that $y\in X_1$, so there is a $\sigma\in\langle\Sigma\rangle$ such that $\sigma(x)=y$. Then $\tau(y)=\tau\sigma\tau^{-1}(\tau(x))$. Since $\tau\sigma\tau^{-1}\in\langle\Sigma\rangle$, $\tau(y)\in X_2$, so $\tau(X_1)\subseteq X_2$.

If $z\in X_2$, there is a $\sigma\in\langle\Sigma\rangle$ such that $\sigma\tau(x)=z$. Then $z=\sigma\tau(x)=\tau(\tau^{-1}\sigma\tau(x))$, and since $\tau^{-1}\sigma\tau\in \langle\Sigma\rangle$, $\tau^{-1}\sigma\tau(x)\in X_1$, and so $z\in \tau(X_1)$. Thus, $\tau(X_1)\supseteq X_2$. Since $G$ is primitive and preserves the orbits of $\langle\Sigma\rangle$, the orbits must form a trivial partition of $X$. Since $\langle\Sigma\rangle$ contains a 3-cycle, the orbits cannot be the singletons, so there must be a single orbit, all of $X$. Thus, $\langle\Sigma\rangle$ is transitive and by lemma 2.3.5, $A(X)=\langle\Sigma\rangle\subseteq G$. $\qed$

Proof. Without loss of generality, we may assume $n\ge m$.

It suffices to show that $A(X)\subseteq\langle\pi,\rho\rangle$, since $\pi$ and $\rho$ have the same parity as $m+n$.

The group $\langle \pi,\rho\rangle=S(X)$ is doubly transitive on $X$: it is clearly transitive on $X$, and it is then not hard, but somewhat tedious, to see that it is doubly transitive. For example, suppose we seek a $\sigma$ that fixes $a_i$ and such that $\sigma(b_k)=a_j$, $i\not=j$. First, note that $\pi^{n-i+1}(a_i)=z$ and $\pi^{n-i+1}(b_k)=x$, where $x\in \{a_1,…,a_n,b_1,\ldots,b_m\}$. Then there is a $p$ such that $\rho^p(x)=\pi^{n-i+1}(a_j)\in \{a_1,…,a_n,b_1,\ldots,b_m\}$. Now $\pi^{i-n-1}\circ\rho^p\circ\pi^{n-i+1}$ fixes $a_i$ and maps $b_k$ to $a_j$.

If $n=m=0$, $\langle \pi,\rho\rangle=\{()\}=A(\{y,z\}$. Otherwise, by lemma 2.3.8, it suffices to show that $\langle\pi,\rho\rangle$ contains a 3-cycle.

Suppose first that $n\ge m\ge 3$. Then $\langle\pi,\rho\rangle$ contains the following permutations: $$\eqalign{ \sigma_1&=\rho\pi^{-1}=(y,a_1)(z,b_1)\cr \sigma_2&=\pi\sigma_1\pi^{-1}=(y,a_2,b_1,b_2)\cr \sigma_3&=\pi\sigma_2\pi^{-1}=(y,a_3,b_2,b_3)\cr \sigma_4&=\rho^{-m}\sigma_1\rho^m=(b_1,b_2)(z,a_{k+1}), k=n-m\le n-3\cr \sigma_5&=\sigma_2\sigma_4\cr \sigma_6&=\pi\sigma_5\pi^{-1}\cr \sigma_7&=\rho^{1-m}\sigma_1\rho^{m-1}\cr \sigma_8&=\sigma_7\sigma_3=(y,a_3)(z,a_{k+2})\cr \sigma_9&=\sigma_8\sigma_6=(z,a_k+2,b_1).\cr }$$

Next, suppose $n>m\ge 0$, $n\not=2m$, $m< 3$. If $m=0$, $\rho\pi^{-1}=(a_1,z,y)$. If $m>0$, $$\eqalign{ \tau_1&=\rho\pi^{-1}=(y,a_1)(z,b_1)\cr \tau_2&=\rho^m\tau_1\rho^{-m}=(a_m,a_{m+1})(zy)\cr \tau_3&=\pi^{2m}\tau_1\pi^{-2m}=(y,c)(a_m,a_{m+1}),\cr }$$ where $c=\pi^{2m}(a_1)\notin\{y,z,a_m,a_{m+1}\}$. Then $\tau_3\tau_2=(y,z,c)$.

The only remaining case is $n=m=1$, in which case $\pi$ and $\rho$ are 3-cycles. $\qed$

Proof. Given $G$ and $K$, let $H$ be a maximal proper subgraph of $G$ containing $K$. If $V(H)=V(G)$, let $e$ be an edge not in $H$. Since $H$ plus the edge $e$ is 2-connected, it must be $G$, by the maximality of $H$. Hence $A$ is the path consisting of $e$ and its endpoints.

Suppose that $v$ is in $V(G)$ but not $V(H)$. Let $u$ be a vertex of $H$. Since $G$ is 2-connected, there is a cycle $C$ containing $v$ and $u$. Following this cycle from $v$ to $u$, Let $w$ be the first vertex in $H$. Continuing on the cycle from $u$ to $v$, let $x$ be the last vertex in $H$. If $x\not=w$, let $A$ be the path $(x,v_1,v_2,\ldots,v_k,v=v_{k+1},v_{k+2},\ldots,v_m,w)$, that is, the portion of the cycle between $x$ and $w$ containing no vertices of $H$ except $x$ and $w$. Since $H$ together with $A$ is 2-connected, it is $G$, as desired.

If $x=w$ then $x=w=u$. Let $y$ be a vertex of $H$ other than $u$. Since $G$ is 2-connected, there is a path $P$ from $v$ to $y$ that does not include $u$. Let $v_j$ be the last vertex on $P$ that is in $\{v_1,\ldots,v,\ldots,v_m\}$; without loss of generality, suppose $j\ge k+1$. Let $z$ be the first vertex on $P$ after $v_j$ that is in $H$. Then let $A$ be the path $(u,v_1,\ldots,v=v_{k+1},\ldots, v_j,\ldots,z)$, where from $v_j$ to $z$ we follow path $P$. Now $H\cup A$ is a 2-connected subgraph of $G$, but it is not $G$, as it does not contain the edge $\{u,v_m\}$, contradicting the maximality of $H$. Thus $x\not=w$. $\qed$

Proof. There are eight distinct ways to add a handle to $\theta_0$:

In the first case, we may remove the handle with interior node $v$ to form a $\theta$-graph of type $(x,0,4)$. The rest are similar. $\qed$