Notation Let $t\in \operatorname {Sym}(\mathbb {Z})$ be defined as $t: z\mapsto z+1$ for all $z \in \mathbb {Z}$.

Proof. Note that $g=\sigma t^{c}$ for some $\sigma \in \operatorname {FSym}(\mathbb {Z})$ and $c\in \mathbb {Z}$. Thus, if $x\not \in \operatorname {supp}(\sigma )$, then the result follows by considering $\{(x)g^{m}\;:\;m\in \mathbb {N}\}$ and $\{(x)g^{-m}\;:\;m\in \mathbb {N}\}$. Then $d_1$ and $d_2$ are both equal to $c$ in this case. Now, if $y\in \operatorname {supp}(\sigma )$, either $y$ lies on a finite orbit or one of the infinite orbits already described.

Proof. Let $N$ be a non-trivial normal subgroup of $G_k$. Then $\operatorname {Alt}(\mathbb {Z})\leqslant N$, and so

\[G_{k}/N\cong(G_{k}/\operatorname{Alt}(\mathbb{Z}))/(N/\operatorname{Alt}(\mathbb{Z}))\]

i.e. $G_{k}/N$ is a quotient of $G_{k}/\operatorname{Alt}(\mathbb{Z})\cong \langle t^{k}\rangle$, and hence cyclic.

Notation For any $k\in \mathbb {N}$, let $\Omega _k=\{1,\, 2,\, \ldots,\, k\}$.

Proof. We will show that all $3$-cycles in $\operatorname {Alt}(\mathbb {Z})$ lie in $\langle \operatorname {Alt}(\Omega _{2k}),\, t^{k}\rangle$. We only need to construct those $3$-cycles with support in $\mathbb {N}$, since for any $\sigma \in \operatorname {Alt}(\mathbb {Z})$ there is an $n \in \mathbb {N}$ such that $\operatorname {supp}(t^{-kn}\sigma t^{kn})\subset \mathbb {N}$. Moreover, $\{(1\;2\;m)\;:\;m\geqslant 3\}$ generates $\operatorname {Alt}(\mathbb {N})$ since for any distinct $m_1,\, m_2,\, m_3\in \mathbb {N}$ we have $(1\;2\;m_1)(1\;2\;m_2)^{-1}=(2\;m_1\;m_2)$ and $(1\;2\;m_3)^{-1}(2\;m_1\;m_2)(1\;2\;m_3)=(m_1\;m_2\;m_3)$.

By definition, $(1\;2\;m)\in \operatorname {Alt}(\Omega _{2k})$ for $m=3,\, \ldots,\, 2k$. To use induction on $n$, let us assume, for each $r\in \{1,\, \ldots,\, k\}$, that $(1,\, 2,\, nk+r)\in \langle \operatorname {Alt}(\Omega _{2k}),\, t^{k}\rangle$. Thus, we want to show that $\langle \operatorname {Alt}(\Omega _{2k}),\, t^{k}\rangle$ also contains $(1,\, 2,\, (n+1)k+r)$. Clearly $(r,\, r+1,\, r+k)\in \langle \Omega _{2k},\, t^{k}\rangle$. Let

\[ \omega:=t^{{-}nk}(r, r+1, r+k)t^{nk}=(nk+r, nk+r+1, (n+1)k+r). \]

Conjugating $(1,\, 2,\, nk+r)$ by $\omega ^{-1}$ yields the required element $(1,\, 2,\, (n+1)k+r)$.

Proof. Our aim is to reduce to the previous lemma. From the assumption that $k\geqslant 3$, all $3$-cycles in $\operatorname {Alt}(\Omega _{2k})$ are conjugate. Our generating set will consist of $t^{k}$ and an element $\alpha \in \operatorname {Alt}(\mathbb {Z})$.

Let $n=\binom {2k}{3}$ and let $\omega _1,\, \ldots,\, \omega _n$ be a choice of distinct $3$-cycles in $\operatorname {Alt}(\Omega _{2k})$ with $\omega _i\ne \omega _j^{-1}$ for every $i,\,j \in \{1,\, \ldots,\, n\}$. Thus $\langle \omega _1,\, \ldots,\, \omega _n\rangle =\operatorname {Alt}(\Omega _{2k})$. Set $\sigma _0=(1\;3)$ and $\sigma _{n+1}=(2\;3)$ and observe that $(\sigma _{n+1}^{-1}\sigma _0^{-1}\sigma _{n+1})\sigma _0=(1\;2\;3)$. Now choose $\sigma _1,\, \ldots,\, \sigma _{n}\in \operatorname {Alt}(\Omega _{2k})$ such that for each $m\in \{1,\,\ldots,\, n\}$ we have $\sigma _m^{-1}(1\;2\;3)\sigma _m=\omega _m$.

Let $\alpha :=\prod _{i=0}^{n+1}t^{-2ik}\sigma _it^{2ik}$ and, for each $m\in \{1,\,\ldots,\, n+1\}$, let $\beta _m:=t^{2mk}\alpha t^{-2mk}$. Then $\beta _{n+1}^{-1}\alpha ^{-1}\beta _{n+1}\alpha =\sigma _{n+1}^{-1}\sigma _0^{-1}\sigma _{n+1}\sigma _0=(1\;2\;3)$ and, for $m\in \{1,\, \ldots,\, n\}$, we have that $\beta _m^{-1}(1\;2\;3)\beta _m=\omega _m$. Hence $\langle \alpha,\, t^{k}\rangle =G_k$.

Proof. We will show, given any $h \in G_k$, that $\langle (1\;2\;3),\, h\rangle$ does not contain $\operatorname {Alt}(\mathbb {Z})$ and so cannot equal $G_k$.

If $h\in \operatorname {Alt}(\mathbb {Z})$, then $\langle (1\;2\;3),\, h\rangle$ is finite. Thus, without loss of generality, $h=\omega t^{n}$ where $\omega \in \operatorname {Alt}(\mathbb {Z})$ and $n \in k\mathbb {N}$. From our description of the orbits of $H_2$, we know that $h$ has at least $k$ infinite orbits. Thus, $h$ has an infinite orbit $\mathcal {O}_4$ which does not intersect $\{1,\, 2,\, 3\}$. Let $x\in \mathcal {O}_4$. Then the orbit of $x$ under $\langle (1\;2\;3),\, h\rangle$ is $\mathcal {O}_4$.

Proof. With a similar approach to the previous lemma, we will show, given any $h \in G_3$, that $\langle (1\;2\;3),\, h\rangle \ne G_3$.

If $h\in \operatorname {Alt}(\mathbb {Z})$, then $\langle (1\;2\;3),\, h\rangle$ is finite. Similarly, if $h=\omega t^{n}$ where $\omega \in \operatorname {Alt}(\mathbb {Z})$ and $n\not \in \{3,\, -3\}$, then $|n|>3$ and the argument in the previous lemma proves the result. Thus, without loss of generality, $h=\omega t^{3}$ where $\omega \in \operatorname {Alt}(\mathbb {Z})$.

Let $\mathcal {O}_1,\, \mathcal {O}_2$, and $\mathcal {O}_3$ denote the three infinite orbits of $h$. After relabelling, we can assume that $i \in \mathcal {O}_i$ for $i=1,\, 2,\, 3$ (for if, say, $\mathcal {O}_3$ did not contain a point from $\{1,\, 2,\, 3\}$, then we could run the argument from the previous lemma). Moreover, if $\mathbb {Z}\setminus (\mathcal {O}_1\cup \mathcal {O}_2\cup \mathcal {O}_3)=Y\ne \emptyset$, then the orbit of points in $Y$ under $\langle (1\;2\;3),\, h\rangle$ is the same as the orbit of points in $Y$ under $\langle h\rangle$. By construction, this orbit cannot be $\mathbb {Z}$, which contradicts that $\operatorname {Alt}(\mathbb {Z})\leqslant G_3$ acts transitively on $\mathbb {Z}$. Then $\mathcal {O}_1\cup \mathcal {O}_2\cup \mathcal {O}_3=\mathbb {Z}$, and by relabelling $\mathbb {Z}$ we can assume that

\[ \mathcal{O}_i=\{n \in \mathbb{Z}\;:\;n\equiv i\mod 3\}\text{ for }i=1, 2, 3 \]

and $(1\;2\;3)$ has been relabelled as $\sigma =(a_1\;a_2\;a_3)$ where $a_i \in \mathcal {O}_i$ for $i=1,\, 2,\, 3$. But then $\langle (a_1\;a_2\;a_3),\, h\rangle \cong C_3\wr \mathbb {Z}$, since for every $m\in \mathbb {Z}\setminus \{0\}$ we have that $h^{-m}(a_1\;a_2\;a_3)h^{m}$ and $(a_1\;a_2\;a_3)$ are disjoint. Therefore $\langle (a_1\;a_2\;a_3),\, h\rangle$ has an imprimitive action on $\mathbb {Z}$, and so cannot contain $\operatorname {Alt}(\mathbb {Z})$.

Notation For any $\omega \in \operatorname {FSym}(\mathbb {Z})$, let $s_\omega :=\omega ^{-1}t\omega$.

Proof. Fix a $k \in \mathbb {N}$. We start by taking powers of the $g_1,\, \ldots,\, g_n$ so to obtain

\[ \omega_1t^{{-}c}, \ldots, \omega_nt^{{-}c} \]

for some $c\in k\mathbb {N}$ and $\omega _1,\, \ldots,\, \omega _n \in \operatorname {Alt}(\mathbb {Z})$. Our aim is to show that there exists a $\sigma \in \operatorname {Alt}(\mathbb {Z})$ so that $s_{\sigma }^{k}$ is a suitable choice for our element $h\in G_k$.

From Remark 2.4, there exists an $m\in \mathbb {N}$ such that for every $z\in \mathbb {Z}\setminus \{-m,\, \ldots,\, m\}$ and every $i\in \{1,\, \ldots,\, n\}$ we have $\omega _it^{-c}(z)=z-c$. Fix an $i\in \{1,\, \ldots,\, n\}$. Let $\omega _i^{*}:=(\omega _it^{-c})^{2}t^{2c}$ so that $(\omega _it^{-c})^{2}=\omega _i^{*}t^{-2c}$. Then, for any $\sigma \in \operatorname {Alt}(\mathbb {Z})$,

\[ w_i^{*}t^{{-}2c}s_\sigma^{2c}=\omega_i^{*}(t^{{-}2c}\sigma^{{-}1}t^{2c})\sigma \]

and we can choose $\sigma$ so that $\operatorname {supp}(\sigma )\cap \operatorname {supp}(t^{-2c}\sigma ^{-1}t^{2c})=\emptyset$. Recall that $\Omega _{2c}=\{1,\, \ldots,\, 2c\}$. We will choose $\sigma$ so that $\operatorname {supp}(\sigma )\cap \{-m,\, \ldots,\, m\}=\emptyset$ and

\[ \operatorname{supp}(\sigma)\subset \bigcup_{j\in \mathbb{Z}}\Omega_{2c}t^{4jc}. \]

We now mimick the proof of Lemma 3.7 (with $\omega _it^{-c}$ taking the role of $t^{k}$) and show that there exists $\sigma \in \operatorname {FSym}(\mathbb {Z})$, with the conditions above, such that $\langle \omega _i^{*}(t^{-2c}\sigma ^{-1}t^{2c})\sigma,\, \omega _it^{-c}\rangle$ contains $\Lambda :=\{\alpha \in \operatorname {Alt}(\mathbb {Z})\;:\;\min (\operatorname {supp}(\alpha ))\geqslant m\}$. Let $n=\binom {2c}{3}$, pick a set of distinct $3$-cycles $\tau _1,\, \ldots,\, \tau _n$ with support in $\Omega _{2c}$ such that $\tau _i\ne \tau _j^{-1}$ for any $i,\, j\in \{1,\, \ldots,\, n\}$, and choose $\sigma _1,\, \ldots,\, \sigma _n$ with support in $\Omega _{2c}$ such that $\sigma _i^{-1}(1\;2\;3)\sigma _i=\tau _i$ for $i=1,\, \ldots,\, n$. Choose $\sigma$ so that $t^{4c(m+i)}\sigma t^{-4c(m+i)}$ acts on $\Omega _{2c}$ in the same way as $\sigma _i$ for $i=1,\, \ldots,\, n$, and moreover make $\sigma$ act on $\Omega _{2c}t^{-4cm}$ and $\Omega _{2c}t^{4c(m+n+1)}$ in such a way that there exists $d\in \mathbb {N}$ so that the commutator of $(\omega _it^{-c})^{d}(\omega _i^{*}(t^{-2c}\sigma ^{-1}t^{2c})\sigma )(\omega _it^{-c})^{-d}$ and $\omega _i^{*}(t^{-2c}\sigma ^{-1}t^{2c})\sigma$ produces $t^{-4c(m+n+1)}(1\;2\;3)t^{4c(m+n+1)}$. We can assume that $\sigma \in \operatorname {Alt}(\mathbb {Z})$ by potentially composing it with $(x\;x+1)$ for some suitably large $x\in \mathbb {N}$. Then, for every $i\in \{1,\, \ldots,\, n\}$, we have that $\langle s_{\sigma }^{k},\, g_i\rangle \geqslant \langle s_{\sigma }^{k},\, \Lambda \rangle \geqslant \langle s_\sigma ^{k},\, \operatorname {Alt}(\mathbb {Z})\rangle =G_k$.

Proof. We change our perspective by relabelling $\mathbb {Z}$ so to replace $s$ with $t$ and $(a\;0\;b)$ with $(-d_a\;0\;d_b)$. If $\gcd (d_a,\, d_b)=1$, then apply the Euclidean algorithm to $(-d_a\;0\;d_b)$ using $t$ to obtain $(-1\;0\;m)$ or $(-m\;0\;1)$ for some $m \in \mathbb {N}$, which both then reduce to $(-1\;0\;1)$.

On the other hand, if $\gcd (d_a,\, d_b)=k>1$, then we claim that the set of torsion elements in $\langle (-d_a\;0\;d_b),\, t\rangle$ preserve a non-trivial block structure of $\mathbb {Z}$. We work with $\langle (-k\;0\;k),\, t\rangle$, since this contains $\langle (-d_a\;0\;d_b),\, t\rangle$. A torsion element in $\langle (-k\;0\;k),\, t\rangle$ will be a product of $t$-conjugates of $(-k\;0\;k)$ and $(-k\;0\;k)^{-1}$. Then both $(-k\;0\;k)$ and $(-k\;0\;k)^{-1}$ fix all but one of the sets

\[ X_i:=\{z\in \mathbb{Z}\;:\;z\equiv i \mod{k}\}\text{ where }i=1, \ldots, k \]

and every $t$-conjugate of $(-k\;0\;k)^{\pm 1}$ also has this property.

Proof. As with the previous lemma, we start by reordering $\mathbb {Z}$ to consider $t$ rather than $s_\omega$, at the cost of then working with $\sigma _*=\gamma ^{-1}\sigma \gamma$ rather than $\sigma$. With this perspective, the claims regarding $\tau$ follow immediately.

Let $\sigma _*=\sigma _1\ldots \sigma _n$ be written in disjoint cycle notation and let $\max \{\operatorname {supp}(\sigma _*)\}=\max \{\operatorname {supp}(\sigma _n)\}=y$. We can choose $\gamma$ so that $(y)\sigma _*^{-1}=y-1$.

Conjugate $\sigma _*$ by a power of $t$ to obtain $\alpha$ where $\min \{\operatorname {supp}(\alpha )\}=y$. We can then conjugate $\sigma _*$ by $\alpha$ to obtain $\sigma '$ where, by construction, $(x)\sigma _*=(x)\sigma '$ for all $x< y$.

We will now consider the element $\sigma _*(\sigma ')^{-1}$. Recall that $\sigma _*=\sigma _1\ldots \sigma _n$, and, from our construction, $\sigma '=\sigma _1\ldots \sigma _{n-1}\sigma _n'$ where

\[ \sigma_n=(x_1^{(n)}\ldots x_{r_n-1}^{(n)}\;x_{r_n}^{(n)})\text{ and }\sigma_n'=(x_1^{(n)}\ldots x_{r_n-1}^{(n)}\;m)\text{ for some }m>x_{r_n}^{(n)}. \]

Therefore $\sigma _*(\sigma ')^{-1}=\sigma _n(\sigma '_n)^{-1}=(x_{r_n-1}^{(n)}\;x_{r_n}^{(n)}\;m)=(y-1\;y\;m)$, which then, together with $t$, generates $G_1$ by Lemma 5.1.

Proof. Starting with an arbitrary odd $d\in \mathbb {Z}$, if we conjugate $(d\;0\;2)$ by $t^{-2}(d\;0\;2)t^{2}$ then we obtain $(d\;0\;4)$. Then $(d\;0\;2)(d\;4\;0)=(0\;2\;4)$ which conjugates $(d\;0\;2)$ to $(d\;2\;4)$. Now $t^{2}(d\;2\;4)t^{-2}=(e\;0\;2)$ where $e=d-2$, and hence $\langle (d\;0\;2),\, t^{2}\rangle$ contains $(1\;0\;2)$ and so $\operatorname {Alt}(\mathbb {Z})$. For the other case, we note that $tG_2t^{-1}=G_2$.

Proof. As with the proof of Lemma 5.2, we change our perspective and so work with $t^{2}$ rather than $s_\omega ^{2}$, at the cost of then working with $\sigma _*=\gamma ^{-1}\sigma \gamma$ rather than $\sigma$. This again resolves the claims regarding $\tau$.

If $\sigma$ is a $3$-cycle, then we choose $\gamma$ so that $\operatorname {supp}(\sigma _*)=\{0,\, 1,\, 2\}$. Otherwise we recall the notation from Lemma 5.2, and write $\sigma _*=\sigma _1\ldots \sigma _n$ where

\[ \sigma_i=(x_1^{(i)}\ldots x_{r_i}^{(i)})\text{ for each } i\in\{1, \ldots, n\} \]

and choose $\gamma \in \operatorname {Alt}(\mathbb {Z})$ so that:

• • $x_{r_n}^{(n)}$ is even and maximal in $\operatorname {supp}(\sigma _*)$;

• • $x_{r_n-1}^{(n)}$ is odd;

• • $x_1^{(1)}=0$ and is minimal in $\operatorname {supp}(\sigma _*)$; and

• • $x_2^{(1)}=2$.

Now conjugate $\sigma _*$ by a suitable power of $t^{2}$ to produce an element $\alpha \in \operatorname {Alt}(\mathbb {Z})$ with $\min \{\operatorname {supp}(\alpha )\}=x_{r_n}^{(n)}$ and conjugate $\sigma _*$ by $\alpha$ to obtain $\sigma '$. With this construction $\sigma _*(\sigma ')^{-1}=(x_{r_n-1}^{(n)}\;x_{r_n}^{(n)}\;x_{r_n}^{(n)}+2)$. Then $\langle \sigma _*(\sigma ')^{-1},\, t^{2}\rangle =G_2$ by Lemma 5.3.

Proof. Consider the elements in $L=\{(k\;k+2\;k+4)\;:\;k\in \mathbb {N}\}$. Given any finite sets $S_1\subset G_1$ and $S_2\subset G_2$, first produce $S_1^{*}$ and $S_2^{*}$ such that $S_i^{*}$ contains elements with exactly $i$ infinite orbits. We do this because no element in $S_i\setminus S_i^{*}$ can generate $G_i$ with any element in $L$.

Recall, by Remark 2.4, that there exists an $m\in \mathbb {N}$ such that each $g\in S_1^{*}\cup S_2^{*}$ translates all points $z>m$. Now $(m\;m+2\;m+4)$ cannot generate $G_1$ with any $s \in S_1^{*}$ by Lemma 5.1, and $(m\;m+2\;m+4)$ cannot generate $G_2$ with any $s'\in S_2^{*}$ since $m$, $m+2$, and $m+4$ all lie on the same orbit of each $s'\in S_2^{*}$.

Notation Given $g,\, h \in G$, let $g^{h}:=h^{-1}gh$.

Proof. By reordering $\mathbb {Z}$, we can work with $t$ rather than $h$ and aim to find an $\omega \in \operatorname {Alt}(\mathbb {Z})$ such that $\langle g_1^{\omega },\, t\rangle =\langle g_2^{\omega },\, t\rangle =G_1$.

First assume that $\operatorname {supp}(g_1)=\operatorname {supp}(g_2)$. If $|\operatorname {supp}(g_1)|=3$, then $g_1\in \{g_2,\, g_2^{-1}\}$ and so we are done. Otherwise, let $y=\max \{\operatorname {supp}(g_1)\}$, and define $x_1:=(y)g_1^{-1}$, $x_2:=(y)g_2^{-1}$, and fix some $a\in \operatorname {supp}(g_1)\setminus \{x_1,\, x_2,\, y\}$. Choose $\omega \in \operatorname {Alt}(\mathbb {Z})$ so that:

• • $(y)\omega =:y'$ is still maximal in $\operatorname {supp}(g_1^{\omega })$;

• • $a$ is minimal in $\operatorname {supp}(g_1^{\omega })$; and

• • $(x_1)\omega =y'-1$.

Applying the steps in the proof of Lemma 5.2 to $g_1^{\omega }$ then yields $(y'-1\;y'\;m)$ for some $m>y'$, and so Lemma 5.1 implies that $\langle g_1^{\omega },\, t\rangle =G_1$. We note that there is also some $d\in \mathbb {N}$ such that $(y')t^{-d}=(x_2)\omega$. Given $b_2:=(a)g_2^{\omega }$, we can therefore adjust $\omega$ so that it moves $a$ arbitrarily far away from the other points in $\operatorname {supp}(g_1^{\omega })$, i.e. we can assume that $(a)t^{p}=b_2$, where $p$ is a suitably large prime number so that $\gcd (d,\, p)=1$. Again, applying the steps from the proof of Lemma 5.2 will yield $(y'-d\;y'\;y'+p)$ and so $\langle g_2^{\omega },\, t\rangle \geqslant \langle (-d\;0\;p),\, t\rangle =G_1$ by Lemma 5.1.

If $\operatorname {supp}(g_1)\ne \operatorname {supp}(g_2)$ then, without loss of generality, $\operatorname {supp}(g_1)\setminus \operatorname {supp}(g_2)\ne \emptyset$ and we can fix some $\alpha$ in this set. Imagine that $\{\alpha g_1^{k}\;:\;k\in \mathbb {N}\}\cap \operatorname {supp}(g_2)=\emptyset$. By choosing $\omega \in \operatorname {FSym}(\mathbb {Z})$ such that $(\alpha )g_1^{\omega }=\alpha +1$ and $\alpha =\min (\operatorname {supp}(g_1^{\omega })\cup \operatorname {supp}(g_2^{\omega }))$, we have that $\langle t,\, g_1^{\omega }\rangle =G_1$. Thus, choose $\omega$ so that some point $y$ in $\operatorname {supp}(g_2^{\omega })$ is maximal in $\operatorname {supp}(g_1^{\omega })\cup \operatorname {supp}(g_2^{\omega })$, and set $(y)(g_2^{\omega })^{-1}=y-1$.

Now let $\alpha \in \operatorname {supp}(g_1)\setminus \operatorname {supp}(g_2)$ but assume that $(\alpha )g_1^{\omega }=\beta \in \operatorname {supp}(g_2^{\omega })$ where $\alpha$ is minimal in $\operatorname {supp}(g_1^{\omega })\cup \operatorname {supp}(g_2^{\omega })$ and $\beta =\alpha +1$. Thus, $\langle t,\, g_1^{\omega }\rangle =G_1$ but also, by construction, $\beta$ is minimal in $\operatorname {supp}(g_2^{\omega })$. We can therefore choose $\omega$ so that $(\beta )g_2^{\omega }=\beta +1$, so that $\langle t,\, g_2^{\omega }\rangle =G_1$.

Proof. Let $g_1,\, g_2 \in G_1$. We will show that there exists $\alpha \in \operatorname {FSym}(\mathbb {Z})$ such that $\langle s_\alpha,\, g_1\rangle = \langle s_\alpha,\, g_2\rangle =G_1$. The case where $g_1,\, g_2 \in \operatorname {Alt}(\mathbb {Z})$ is dealt with in the previous lemma. If $g_1,\, g_2 \in G_1\setminus \operatorname {Alt}(\mathbb {Z})$, then Theorem 4.4 implies the result. Finally, if $g_1 \in \operatorname {Alt}(\mathbb {Z})$ and $g_2\in G_1\setminus \operatorname {Alt}(\mathbb {Z})$, then we can apply Lemma 5.2 so to find $s_\omega$ which generates $G_1$ with $g_1$. This lemma also states that choosing any $\tau \in \operatorname {FSym}(\mathbb {Z})$ such that $\operatorname {supp}(\tau )$ is disjoint from $\operatorname {supp}(\omega )\cup \operatorname {supp}(g_1)$ will result in $\langle s_{\tau \omega },\, g_1\rangle =G_1$. Theorem 4.4 and Remark 4.5 together ensure the existence of an element $\alpha =\sigma \omega \in \operatorname {Alt}(\mathbb {Z})$ so that $\langle g_1,\, s_\alpha \rangle =\langle g_2,\, s_\alpha \rangle =G_1$.

Notation Let $2\mathbb {Z}$ and $2\mathbb {Z}+1$ denote the sets of even and odd integers, respectively.

Proof. First, conjugate $\sigma$ by a suitable power of $t^{2}$ to produce $\sigma _*$, where $\sigma _*$ sends $0$ to a point in $2\mathbb {Z}+1$ and $\sigma _*$ sends no negative even number to $2\mathbb {Z}+1$. Now, given $\omega \in \operatorname {Alt}(\mathbb {Z})$ with $|\operatorname {supp}(\omega )\cap 2\mathbb {Z}|=k>0$, we consider the following steps:

1. i. conjugate $\omega$ by a suitable power of $t^{2}$ to produce $\omega '$ so that $0\in \operatorname {supp}(\omega ')$ and $\operatorname {supp}(\omega ')\cap 2\mathbb {N}=\emptyset$;

2. ii. conjugate $\omega '$ by an appropriate element of $\operatorname {Alt}(2\mathbb {Z}+1)$ to produce $\omega _*$ with $\operatorname {supp}(\sigma _*)\cap \operatorname {supp}(\omega _*)\subset 2\mathbb {Z}$;

3. iii. conjugate $\omega _*$ by $\sigma _*$ to produce $\hat {\omega }$.

Note that $|\operatorname {supp}(\hat {\omega })\cap 2\mathbb {Z}|=|\operatorname {supp}(\omega )\cap 2\mathbb {Z}|-1$. Iterating these steps, therefore, results in an element in $\operatorname {Alt}(2\mathbb {Z}+1)$.

Proof. Again, by relabelling $\mathbb {Z}$, we deal with the equivalent problem of whether there exists $\omega \in \operatorname {Alt}(\mathbb {Z})$ such that $\langle g_1^{\omega },\, t^{2}\rangle =\langle g_2^{\omega },\, t^{2}\rangle =G_2$. If $\operatorname {supp}(g_1)\cap \operatorname {supp}(g_2)=\emptyset$, then Lemma 5.4 implies the result. Otherwise, fix some $c\in \operatorname {supp}(g_1)\cap \operatorname {supp}(g_2)$. We will restrict ourselves to those $\omega \in \operatorname {Alt}(\mathbb {Z})$ such that:

• • $(c)\omega =0$;

• • $(\operatorname {supp}(g_1^{\omega })\cup \operatorname {supp}(g_2^{\omega }))\cap 2\mathbb {Z}=\{0\}$; and

• • the only positive odd integers in $\operatorname {supp}(g_1^{\omega })\cup \operatorname {supp}(g_2^{\omega })$ are $1$ and $3$.

In light of the previous lemma, our aim is to choose $\omega$ so to relabel the points in $\operatorname {supp}(g_1)\cup \operatorname {supp}(g_2)$ so that both $\langle g_1^{\omega },\, t^{2}\rangle$ and $\langle g_1^{\omega },\, t^{2}\rangle$ contain $\operatorname {Alt}(2\mathbb {Z}+1)$.

Note that $|\operatorname {supp}(g_1)|\geqslant 3$. Hence there exist distinct $x,\, y$, both not equal to $c$, with $(x)g_1=y$. We now further restrict our choice of $\omega$ to those such that:

• • $(x)\omega =1$; and

• • $(y)\omega =3$.

To simplify the cases we consider later, we impose that if $y\in \operatorname {supp}(g_2)$, then so is $x$. This is not a problematic choice: if only $y$ were in $\operatorname {supp}(g_2)$, then we could replace $g_1$ with $g_1^{-1}$ and then swap the labels for $x$ and $y$. We first note that any choice of $\omega \in \operatorname {Alt}(\mathbb {Z})$ with the above properties gives $\operatorname {Alt}(2\mathbb {Z}+1)\leqslant \langle g_1^{\omega },\, t^{2}\rangle$. We prove this in two parts. If $|\operatorname {supp}(g_1)|=3$, then we apply Lemma 5.3 to obtain that $\langle g_1^{\omega },\, t^{2}\rangle =G_2$. Otherwise, we apply the following steps, which we have employed numerous times in this paper so far (the first time being in Lemma 5.2).

1. 1. Let $a:=\min (\operatorname {supp}(g_1^{\omega })\setminus \{0,\, 1,\, 3\})$. From our restrictions on $\omega$, $a$ is odd.

2. 2. Conjugate $g_1^{\omega }$ by a power of $t^{2}$ to obtain $\alpha$, an element with $\min (\operatorname {supp}(\alpha ))=3$.

3. 3. Conjugate $g_1^{\omega }$ by $\alpha$ to obtain $g_1'$. Note, for every $z \in \mathbb {Z}\setminus \{1,\, 3,\, (3)g_1^{\omega }\}$, that $(z)g_1^{\omega }=(z)g_1'$.

4. 4. Let $g_1^{*}:=g_1^{\omega } (g_1')^{-1}$. Then $g_1^{*}=(1\;3\;m)$, where $m=(3)\alpha$.

This element $g_1^{*}$ is exactly what we need. If $(3)\alpha \in 2\mathbb {Z}$ then $\langle g_1^{*},\, t^{2}\rangle =G_2$ by Lemma 5.3. If $(3)\alpha \in 2\mathbb {Z}+1$ then, by focusing on the action of $t^{2}$ and $(1\;3\;m)$ on $2\mathbb {Z}+1$, we can apply Lemma 5.1 to see that $\operatorname {Alt}(2\mathbb {Z}+1)\leqslant \langle g_1^{*},\, t^{2}\rangle$.

We must now choose the image of $\omega$ on $\operatorname {supp}(g_2)\setminus \{c,\, x,\, y\}$ so to guarantee that $\operatorname {Alt}(2\mathbb {Z}+1)\leqslant \langle g_2^{\omega },\, t^{2}\rangle$. First, if $1$ and $3$ lie on the same orbit of $g_2^{\omega }$, then there exists $f_2$, a suitable power of $g_2^{\omega }$, such that $(1)f_2=3$. The steps given above then imply that $\langle f_2,\, t^{2}\rangle =G_2$ for any choice of $\omega$ with the properties we have specified above. Similarly, if there exist $a,\, b\in \operatorname {supp}(g_2)\setminus \{c,\, x,\, y\}$ which lie on the same orbit of $g_2$, then we can produce $h_2$, a suitable power of $g_2$ such that $(a)h_2=b$. We can insist that $\omega$ preserves the ordering on $\operatorname {supp}(h_2)\setminus \{c,\, x,\, y,\, a,\, b\}$, and let $q:=\max (\operatorname {supp}(h_2))$. By choosing $\omega$ so that:

• • $a$ is odd;

• • $a$ is minimal in $\operatorname {supp}(g_2^{\omega })$; and

• • $b=a+2$

we see that the steps (1)–(4) above yield the $3$-cycle $((1)h_2^{-1}\;q\;q+2)$. As we saw for $g_1^{*}$, Lemma 5.3 and Lemma 5.1 deal respectively with the cases that $(1)h_2^{-1}$ is even or odd. We now note that if $1$ and $3$ lie on distinct $g_2$ orbits and there are no such $a,\, b\in \operatorname {supp}(g_2)\setminus \{c,\, x,\, y\}$ which lie on the same orbit of $g_2$, then we have one of a finite list of cases. We end by dealing with each such possibility.

If $x,\, y\not \in \operatorname {supp}(g_2)$, then $|\operatorname {supp}(g_2)|\geqslant 3$ and so there exist $a,\, b\in \operatorname {supp}(g_2)\setminus \{c\}$ that lie on the same orbit of $g_2$. If $x\in \operatorname {supp}(g_2)$ but $y\not \in \operatorname {supp}(g_2)$, then either:

• • $g_2^{\omega }$ or its inverse equals $(i\;0\;1)$ for some $i\in \mathbb {Z}\setminus \{0,\, 1\}$, in which case, by choosing the appropriate $\omega$, we can set $i:=-1$ so that $\langle g_2^{\omega },\, t^{2}\rangle =G_2$; or

• • $g_2^{\omega }=(i\;0)(j\;1)$ for some distinct $i,\, j\in \mathbb {Z}\setminus \{0,\, 1\}$. We set $i:=-5$ and $j:=-1$ and note that $\langle g_2^{\omega },\, t^{2}\rangle$ contains $g_2^{\omega }((t^{-2}g_2^{\omega } t^{2})^{-1}g_2^{\omega }(t^{-2}g_2^{\omega } t^{2}))=(-1\;1\;3)$. We can then apply Lemma 5.1 to the orbit of $t^{2}$ containing $2\mathbb {Z}+1$ in order to generate $\operatorname {Alt}(2\mathbb {Z}+1)$, and then apply Lemma 6.4 for our desired conclusion that $\langle g_2^{\omega },\, t^{2}\rangle =G_2$.

We commented above that we could assume that if $y\in \operatorname {supp}(g_2)$, then $x$ must also be in $\operatorname {supp}(g_2)$. So finally, if $x,\, y \in \operatorname {supp}(g_2)$, then $g_2^{\omega }$ or its inverse is either $(i\;1\;0)(j\;3)$, $(i\;0\;3)(j\;1)$, or $(i\;0)(j\;1)(k\;3)$. But then $g_2\not \in \operatorname {Alt}(\mathbb {Z})$, since these are all odd permutations.

Proof. Let $g_1,\, g_2 \in G_1$. We will show that there exists $\alpha \in \operatorname {FSym}(\mathbb {Z})$ such that $\langle s_\alpha ^{2},\, g_1\rangle = \langle s_\alpha ^{2},\, g_2\rangle =G_2$. If $g_1,\, g_2 \in G_2\setminus \operatorname {Alt}(\mathbb {Z})$, then Theorem 4.4 implies the result. If $g_1 \in \operatorname {Alt}(\mathbb {Z})$ and $g_2\in G_2\setminus \operatorname {Alt}(\mathbb {Z})$, then we can apply Lemma 5.4 in the same way to how Lemma 5.2 is used in the proof of Lemma 6.3. The final case was dealt with in the previous proposition.

Proof. We will construct a set $S\subset G_1$, of size 10, such that no $h\in G_1$ generates $G_1$ with every element in $S$. The set will consist of $3$-cycles in $\operatorname {Alt}(\mathbb {Z})$ with support contained within $X=\{1,\, \ldots,\, 5\}$ and chosen such that if $\sigma \in S$ then $\sigma ^{-1}\not \in S$. There are $\binom {5}{3}=10$ such cycles. Let $h\in G_1$. Then $h=\sigma t^{k}$ for some $\sigma \in \operatorname {Alt}(\mathbb {Z})$ and $k\in \mathbb {Z}$. If $k\not \in \{1,\, -1\}$, then $h$ cannot generate $G_1$ with $(1\;2\;3)$.

We start with the observation that, given 5 points in $\mathbb {Z}$, they cannot all lie on the infinite orbit of $h$. To do this, we use Lemma 5.1. Assume that $h$ consists of a single infinite orbit, equal to $\mathbb {Z}$. Without loss of generality, reorder $\mathbb {Z}$ so to replace $h$ with $t$, the translation map. Now, given $\{a,\, b,\, c,\, d,\, e\}\subset \mathbb {Z}$ with $a< b< c< d< e$, at least 3 of these points lie in $2\mathbb {Z}$ or in $2\mathbb {Z}+1$. Let $\sigma$ denote the 3-cycle with support equal to 3 such points. Lemma 5.1 then states that $\langle \sigma,\, t\rangle \ne G_1$.

From the previous paragraph, no $h$ with an infinite orbit equal to $\mathbb {Z}$ can generate $G_1$ with each element from $S$. Hence at least one of $\{1,\, \ldots,\, 5\}$ must lie on a finite orbit of $h$. Let $K$ be the intersection of the infinite orbit of $h$ with $\{1,\, \ldots,\, 5\}$, and let $L:=\{1,\, \ldots,\, 5\}\setminus K$. One of $K$ or $L$ must have size 3, and so contains $m_1< m_2< m_3$ from $\{1,\, \ldots,\, 5\}$. Let $\omega =(m_1\;m_2\;m_3)$. Then $\langle h,\, \omega \rangle \ne G_1$ since the group generated by $h$ and $\omega$ does not act transitively on $\mathbb {Z}$.

Proof. As with the previous proof, let $S$ consist of those $3$-cycles in $\operatorname {Alt}(\mathbb {Z})$ with support contained within $X=\{1,\, \ldots,\, 5\}$ and chosen such that if $\sigma \in S$ then $\sigma ^{-1}\not \in S$. We will show that no $h\in G_2$ generates $G_2$ with every element in $S$. Given $h\in G_2$, we have that $h=\sigma t^{2k}$ where $\sigma \in \operatorname {Alt}(\mathbb {Z})$ and $k \in \mathbb {Z}$. If $k\not \in \{1,\, -1\}$, then $h$ cannot generate $G_2$ with $(1\;2\;3)$. Let $h$ have infinite orbits $\mathcal {O}_1$ and $\mathcal {O}_2$ and define $\phi _h: X\rightarrow \{0,\, 1,\, 2\}$ by

• • $0$ if $x$ lies on a finite orbit of $h$;

• • $1$ if $x\in \mathcal {O}_1$; and

• • $2$ if $x\in \mathcal {O}_2$.

It is clear that the image of $\phi$ must contain $\{1,\, 2\}$, since otherwise $\langle h,\, (a\;b\;c)\rangle$ cannot act transitively on both $\mathcal {O}_1$ and $\mathcal {O}_2$ for any distinct $a,\, b,\, c\in X$. Similarly, we cannot have two points of $X$ sent to 0 by $\phi$, and we cannot have 3 points of $X$ sent to either 1 or to 2. Thus, after relabelling, the only possibility is that $\phi (1)=0$, $\phi (2)=\phi (3)=1$, and $\phi (4)=\phi (5)=2$. But then $\langle h,\, (1\;2\;3)\rangle \ne G_2$.