Proof. Let $b_1,b_2 \in B$ be representatives for orbits under the action of $H$ whose images in $S/\Gamma$ coincide. This means that there exists $g \in \Gamma$ with $b_1^g = b_2$. In particular, $B^g \cap B \neq \emptyset$ so $g \in H$ since $B$ is a block of $\Gamma$. It follows that $b_1$ and $b_2$ are in the same orbit under the action of $H$ so the required injectivity is established.

Proof. We can find a prime number $p$ satisfying $(n+1)/2 < p \leq n-3$. The stabilizer of $\{1, \ldots, \lfloor n/2 \rfloor \}$ in the action of $S_n$ on the set of all $\lfloor n/2 \rfloor$-element subsets of $\{1, \ldots, n\}$ is the subgroup $S_{\lfloor n/2 \rfloor } \times S_{\lceil n/2 \rceil }$. The $\lfloor n/2 \rfloor$-homogeneity of $H$ is tantamount to the equality $H \cdot (S_{\lfloor n/2 \rfloor } \times S_{\lceil n/2 \rceil }) = S_n$ of subsets of $S_n$. Therefore,

\[ n! = |S_n| = |H \cdot (S_{\lfloor n/2 \rfloor} \times S_{\lceil n/2 \rceil})| = \frac{|H| \cdot |S_{\lfloor n/2 \rfloor} \times S_{\lceil n/2 \rceil}|}{|H \cap (S_{\lfloor n/2 \rfloor} \times S_{\lceil n/2 \rceil})|} = \frac{|H| \cdot \lfloor n/2 \rfloor! \cdot \lceil n/2 \rceil!}{|H \cap (S_{\lfloor n/2 \rfloor} \times S_{\lceil n/2 \rceil})|}, \]

so $n!$ divides $|H| \cdot \lfloor n/2 \rfloor ! \cdot \lceil n/2 \rceil !$, hence $p$ divides this number as well.

Our choice of $p$ guarantees that $p$ divides $|H|$. By Cauchy's theorem, $H$ contains an element of order $p$. An element of order $p$ in $S_n$ is a product of $p$-cycles, but $p> n/2$ so in our case this element is necessarily a $p$-cycle. It is readily checked that an $\lfloor n/2 \rfloor$-homogeneous subgroup of $S_n$ is transitive (equivalently, $1$-homogeneous) and, moreover, primitive (or even $2$-homogeneous). By a theorem of Jordan, the only primitive subgroups of $S_n$ that contain a $p$-cycle for a prime number $p \leq n-3$ are $A_n$ and $S_n$. We conclude that either $H = A_n$ or $H = S_n$ as required.

Proof. We view $S_{n-k}$ as the group of permutations of $\{1, \ldots, n-k\}$ in $S_{n}$ fixing each element of $\{n-k+1, \ldots, n\}$. Denote by $H$ the subgroup of $S_n$ generated by $S_{n-k}$ and a conjugate $\rho$ of $\sigma$. We need to show that $H = S_n$. First we claim that $H$ acts transitively on $\{1, \ldots, n\}$. We take $j \in \{1, \ldots, n\}$ and our task is to show that it lies in the orbit of $1$ under the action of $H$. If $j \in \{1, \ldots, n-k\}$, this is clear since $H$ contains $S_{n-k}$, so we assume that $j \in \{n-k+1, \ldots, n\}$. Since the cycle of $\rho$ in which $j$ lies is of length more than $k$, it contains an element from $\{1, \ldots, n-k\}$. As all the powers of $\rho$ lie in $H$, we conclude that $j$ is indeed in the orbit of $1$ under the action of $H$, as required for transitivity.

Next, we claim that $H$ is primitive. Toward a contradiction, suppose that $\{1, \ldots, n\}$ can be partitioned into disjoint blocks for $H$ with at least two distinct blocks, each block of size at least $2$. As $n-k > n/2$ by assumption, and the size of every block is a proper divisor of $n$, we conclude that $\{1, \ldots, n-k\}$ is not contained in a single block, and some block $B$ meets $\{1, \ldots, n-k\}$ at two distinct elements at least, say $i$ and $j$. Since $\{1, \ldots, n-k\} \nsubseteq B$, we can find $\tau \in S_{n-k}$ with $\tau (i) = i \in B$ and $\tau (j) \notin B$. We see that $\tau \in H$, that $\tau B \cap B \neq \emptyset$, and that $\tau (B) \neq B$. This contradicts our assumption that $B$ is a block for $H$, and concludes the proof of primitivity.

As $n - k > k \geq 1$, there exists a transposition in $S_{n-k}$, so $H$ contains a transposition. By a theorem of Jordan, the only primitive subgroup of $S_n$ that contains a transposition is $S_n$ itself so $H = S_n$ as required.

Proof. Let $\mathcal {G}_1, \ldots, \mathcal {G}_r$ be all the distinct isomorphism types appearing among the groups $G_1, \ldots, G_n$. Then $G \cong \mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_r^{m_r}$ where $m_i = |\{1 \leq j \leq n : G_j \cong \mathcal {G}_i\}|$ for $1 \leq i \leq r$. We claim that $H$ projects onto $\mathcal {G}_i^{m_i}$ for every $1 \leq i \leq r$. Indeed, in case $m_i = 1$ this follows from the second condition that $H$ satisfies by assumption. In case $m_i \geq 2$ this follows from [Reference Bary-Soroker, Gorodetsky, Karidi and SawinBGKS20, Lemma 6.6] applied to the projection of $H$ to $\mathcal {G}_i^{m_i}$, using the first and the third condition that $H$ satisfies. The claim is thus established.

We prove by induction on $1 \leq i \leq r$ that $H$ projects onto $\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_i^{m_i}$. The base case $i=1$ follows from the claim established above. For $i>1$ we apply Goursat's lemma to the projection $H_i$ of $H$ in $(\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}}) \times \mathcal {G}_i^{m_i}$. Since $H_i$ projects onto both $\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}}$ by induction, and onto $\mathcal {G}_i^{m_i}$ by the aforementioned claim, there exists a group $K$ and surjective homomorphisms $\varphi \colon \mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}} \to K$, $\psi \colon \mathcal {G}_i^{m_i} \to K$ such that $H_i = (\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}}) \times _K \mathcal {G}_i^{m_i}$. We claim that $K$ is trivial.

Toward a contradiction, suppose that $K$ admits a simple quotient $S$. Then we have surjections $\overline \varphi \colon \mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}} \to S$ and $\overline \psi \colon \mathcal {G}_i^{m_i} \to S$. Since $S$ is simple, and $\overline \varphi, \overline \psi$ map normal subgroups to normal subgroups, we see that $S$ is a quotient of $\mathcal {G}_t$ for some $1 \leq t \leq i-1$ and a quotient of $\mathcal {G}_i$ because the direct factors are normal subgroups that generate the direct product. As $\mathcal {G}_t \ncong \mathcal {G}_i$, our initial assumption implies that $S$ is abelian. We note that $H_i$ is contained in the proper subgroup $\overline H_i = (\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}}) \times _S \mathcal {G}_i^{m_i}$ of $\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_i^{m_i}$. Since $S$ is abelian, this subgroup $\overline H_i$ contains the commutator subgroup of $\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_i^{m_i}$. It follows that the restriction of the natural homomorphism $\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_i^{m_i} \to (\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_i^{m_i})^{\text {ab}}$ to $\overline H_i$, and therefore also to $H_i$, is not surjective. This contradicts the first condition that $H$ satisfies, and thus proves the claim that $K = \{1\}$.

We conclude that $H_i = (\mathcal {G}_1^{m_1} \times \cdots \times \mathcal {G}_{i-1}^{m_{i-1}}) \times \mathcal {G}_i^{m_i}$ so our induction is complete. Plugging $i=r$, we get that $H = G$, as required.

Proof. Put $K = H \cap (A_{n_1} \times \cdots \times A_{n_r})$. We need to show that $K = A_{n_1} \times \cdots \times A_{n_r}$ and we will do this by invoking Lemma 3.6 whose assumptions we shall verify now. The first assumption is met because the groups $A_n$ for $n \geq 5$ are nonabelian pairwise nonisomorphic simple groups (and the groups $A_n$ for $n < 5$ do not have nonabelian simple quotients). We check next that $K$ satisfies the three conditions in Lemma 3.6.

The first condition is satisfied because the abelianization of $A_{n_1} \times \cdots \times A_{n_r}$ is trivial as $n_1, \ldots, n_r \geq 5$ by assumption. To check the second condition we start by fixing $1 \leq i \leq n$ and noting that in view of our assumptions on $n_i$ and on $H$, we get from Proposition 3.3 that the projection $H_i$ of $H$ to $S_{n_i}$ contains $A_{n_i}$. Since

\[ H/K = H/ (H \cap (A_{n_1} \times \cdots \times A_{n_r})) \cong H \cdot (A_{n_1} \times \cdots \times A_{n_r})/ A_{n_1} \times \cdots \times A_{n_r} \leq (\mathbb{Z}/2 \mathbb{Z})^r, \]

denoting by $K_i$ the projection of $K$ to $S_{n_i}$, we see that $H_{i}/K_i$ is an elementary abelian $2$-group. We conclude that $K_i$ contains $A_{n_i}$ so the second condition is indeed satisfied.

To check the third condition, we take $1 \leq i < j \leq r$ with $n_i = n_j$, and denote by $K_{i,j}$ the projection of $K$ to $A_{n_i} \times A_{n_j}$. As $n_i = n_j$ the function

\[ f \colon \{n_1+ \cdots + n_{i-1}+1, \ldots, n_1 + \cdots + n_i\} \to \{n_1+ \cdots + n_{j-1}+1, \ldots, n_1 + \cdots + n_j\}, \]

given by

\[ f(x) = x + n_{i+1} + \cdots + n_j = x + n_i + \cdots + n_{j-1} \]

is a bijection. This bijection will be silently used in what follows to identify the group $S_{n_i}$ with the group $S_{n_j}$ (and the group $A_{n_i}$ with the group $A_{n_j}$).

Suppose toward a contradiction that $K_{i,j}$ is a proper subgroup of $A_{n_i} \times A_{n_j}$. In view of the second condition verified above, the subgroup $K_{i,j}$ projects onto both $A_{n_i}$ and $A_{n_j}$ so since these two groups are simple, Goursat's lemma tells us that $K_{i,j} = \{(\sigma, \psi (\sigma )) : \sigma \in A_{n_i}\}$ for some automorphism $\psi \colon A_{n_i} \to A_{n_j}$. Our assumption that $n_i \geq 7$ implies that $\operatorname {Aut}(A_{n_i}) = S_{n_i}$, namely there exists $\tau \in S_{n_i}$ for which $K_{i,j} = \{(\sigma, \tau \sigma \tau ^{-1}) : \sigma \in A_{n_i}\}$.

We pick an $\lfloor n_i/2 \rfloor$-element subset $X$ of $\{n_1+ \cdots + n_{i-1}+1, \ldots, n_1 + \cdots + n_i\}$, for instance

\[ X = \{n_1+ \cdots + n_{i-1}+1, \ldots, n_1 + \cdots + n_{i-1} + \lfloor n_i/2 \rfloor\}, \]

and consider the orbit of $(X, \tau (f(X)))$ under the action of $K_{i,j}$. On the one hand, this orbit is $\{(\sigma (X), \tau (\sigma (f(X)))) : \sigma \in A_{n_i}\}$ so its length is at most

\[ |\{\sigma(X) : \sigma \in A_{n_i}\}| \leq {n_i \choose \lfloor n_i/2 \rfloor}. \]

On the other hand, in view of our initial assumption on $H$, the orbit of $(X, \tau (f(X)))$ under the action of the projection $H_{i,j}$ of $H$ to $S_{n_i} \times S_{n_j}$ has length ${n_i \choose \lfloor n_i/2 \rfloor }^2$. Since $K_{i,j}$ is a normal subgroup of $H_{i,j}$ with $[H_{i,j} : K_{i,j}] \leq [H : K] \leq 2^r$, the length of the orbit of $(X, \tau (f(X)))$ under the action of $K_{i,j}$ is at least $2^{-r} {n_i \choose \lfloor n_i/2 \rfloor }^2$. We conclude that

\[ 2^{-r}{n_i \choose \lfloor n_i/2 \rfloor}^2 \leq {n_i \choose \lfloor n_i/2 \rfloor}, \]

so ${n_i \choose \lfloor n_i/2 \rfloor } \leq 2^r$ in contrast to our initial assumption. We have thus shown that $K_{i,j} = A_{n_i} \times A_{n_j}$. We can therefore invoke Lemma 3.6 and conclude that $K = A_{n_1} \times \cdots \times A_{n_r}$ as required.

Proof. Denote by $H$ the subgroup of $S_{n_1} \times \cdots \times S_{n_r}$ generated by $S_{n_1-k} \times \cdots \times S_{n_r-k}$ and a conjugate $\rho$ of $g$. In order to show that $H = S_{n_1} \times \cdots \times S_{n_r}$, we will invoke Lemma 3.6 whose assumptions we verify next. The first assumption is satisfied because for every positive integer $n$, the group $S_n$ does not have a nonabelian simple quotient. We shall now check that $H$ satisfies the three conditions in Lemma 3.6.

For the first condition, we need to check the surjectivity of the restriction of the sign homomorphism $S_{n_1} \times \cdots \times S_{n_r} \to (\mathbb {Z}/ 2\mathbb {Z})^r$ to $H$. This follows at once from the surjectivity of the restriction of this homomorphism to $S_{n_1-k} \times \cdots \times S_{n_r-k}$, a consequence of the fact that $n_i - k > k \geq 1$ for every $1 \leq i \leq r$. The second condition is an immediate consequence of Proposition 3.5.

To check the third condition, we take $1 \leq i < j \leq r$ (with $n_i = n_j$), and denote by $H_{i,j}$ the projection of $H$ to $S_{n_i} \times S_{n_j}$. Suppose toward a contradiction that $H_{i,j}$ is a proper subgroup of $S_{n_i} \times S_{n_j}$. In view of the second condition verified above, the subgroup $H_{i,j}$ projects onto both $S_{n_i}$ and $S_{n_j}$ so Goursat's lemma provides us with a nontrivial group $K$ and surjective homomorphisms $\varphi \colon S_{n_i} \to K$, $\psi \colon S_{n_j} \to K$ such that $H_{i,j} = S_{n_i} \times _{K} S_{n_j}$. Since $K$ is (isomorphic to) a nontrivial quotient of a symmetric group, its abelianization is necessarily nontrivial, so $S_{n_i} \times _{K^\text {ab}} S_{n_j}$ is a proper subgroup of $S_{n_i} \times S_{n_j}$ that contains both $H_{i,j}$ and the commutator subgroup of $S_{n_i} \times S_{n_j}$. It follows that the restriction to $H_{i,j}$ of the natural homomorphism $S_{n_i} \times S_{n_j} \to (S_{n_i} \times S_{n_j})^{\text {ab}}$ is not surjective. This contradicts the first condition verified above, namely the surjectivity of the map $H \to (\mathbb {Z} / 2 \mathbb {Z})^r$. We have thus shown that $H_{i,j} = S_{n_i} \times S_{n_j}$ so the third condition is verified.

We can therefore invoke Lemma 3.6 and conclude that $H = S_{n_1} \times \cdots \times S_{n_r}$ as required.

Proof. Consider first the semigroup $\mathbb {N} \times S$ whose elements we write as $z^{-m}w$ for a nonnegative integer $m$, and $w \in S$. The product is given by

\[ z^{-m}w \cdot z^{-n} v = z^{-(m+n)}wv,\quad m,n \in \mathbb{N},\ v,w \in S_X. \]

There is a natural homomorphism of semigroups $S \to \mathbb {N} \times S$ sending $w \in S_X$ to $z^{-0}w$. We define an equivalence relation $\sim$ on $\mathbb {N} \times S$ by $z^{-m}w \sim z^{-n}v$ if there exists a positive integer $r$ such that

\[ z^{n+r}w = z^{m+r}v \]

in $S$. Since multiplication in $\mathbb {N} \times S$ descends to multiplication on the set of equivalence classes for $\sim$ we can let $S[z^{-1}]$ be the quotient semigroup, and take $\varphi$ to be the composition $S \to \mathbb {N} \times S \to S[z^{-1}]$. We note that $z^{-1}z$ is the identity element of $S[z^{-1}]$ so $S[z^{-1}]$ is indeed a monoid. The inverse of $\varphi (z)$ in $S[z^{-1}]$ is given by $z^{-2} z$ so $\varphi (z)$ is indeed invertible. A routine check establishes the required universal property of $S[z^{-1}]$.

Proof. We first show that the monoid $S_X[z^{-1}]$ is a group. Since $S_X[z^{-1}]$ is generated by $X$ and $z^{-2}z$ it suffices to show that (the image in $S_X[z^{-1}]$ of) every $x \in X$ is invertible in $S_X[z^{-1}]$. Fix an $x \in X$. As $z^{-2}z$ and $y^{m_y}$ lie in the center of $S_X[z^{-1}]$ for every $y \in X$, we have

\[ x \cdot x^{m_x-1} \prod_{y \in X \setminus \{x\}} y^{m_y} \cdot z^{-2}z = 1 = x^{m_x-1} \prod_{y \in X \setminus \{x\}} y^{m_y} \cdot z^{-2}z \cdot x, \]

so $x$ is indeed invertible in $S_X[z^{-1}]$. We have thus shown that $S_X[z^{-1}]$ is a group.

Since $S_X[z^{-1}]$ is a group, the natural map $X \to S_X[z^{-1}]$ is a morphism of braided sets. The universal property of $\Gamma _X$ therefore provides us with a group homomorphism $\Gamma _X \to S_X[z^{-1}]$ which is seen to be the inverse of $S_X[z^{-1}] \to \Gamma _X$ by checking that this is the case on the image of $X$.

Proof. This follows from the fact that $C_1^{n_1} \times \cdots \times C_k^{n_k}$ is a fiber of the map

\[ X(n_1, \ldots, n_k) \to X_{\text{triv}}(n_1, \ldots, n_k), \]

from the fact that this map is a morphism of sets with a right action of $B_n$, and from the fact that the action of $B_n$ on $X_{\text {triv}}(n_1, \ldots, n_k)$ factors through the permutation action of $S_n$ on $X_{\text {triv}}(n_1, \ldots, n_k)$.

Proof. This follows from Propositions 3.1 and 4.20 because the orbit under the action of $B_n$ of every element from $X(n_1, \ldots, n_k)$ contains an element from $C_1^{n_1} \times \cdots \times C_k^{n_k}$.

Proof. Since composition of surjective functions is surjective, it suffices to treat the case $m=1$, namely we assume that $w \in X$, so $w \in C_j$ for some $1 \leq j \leq k$. Take a representative

\[ s \in X^*(n_1, \ldots, n_{j-1}, n_j+1, n_{j+1}, \ldots, n_k) \]

for an orbit under the action of $B_{n+1}$. As $n_j$ is large enough, it follows from the pigeonhole principle that there exists an $x \in C_j$ that appears in $s$ more than $m_x$ times. We can therefore assume, without changing the orbit of $s$ under the action of $B_{n+1}$, that the first $m_x + 1$ entries of $s$ are all equal to $x$. In particular, we can write $s = x^{m_x} s'$ (the equality taking place in $S_X$) for some

\[ s' \in X^*(n_1, \ldots, n_{j-1}, n_j+1-m_x, n_{j+1}, \ldots, n_k)/B_{n+1-m_x}. \]

Since $x$ and $w$ lie in the same connected component of $X$, there exists $g \in \operatorname {Inn}(X)$ such that $x^g = w$. The entries of (a representative of) $s'$ generate $X$, so these entries generate $\operatorname {Inn}(X)$ as a group. In particular, we can write $g$ as a product some of these entries and their inverses (allowing repetition). Therefore, in order to show that $s = w^{m_x}s'$ in $S_X$ and deduce the required surjectivity, it suffices to observe that for every $y \in X$ that appears in (a representative of) $s'$, and every $x_1, \ldots, x_\ell \in X$ for which $x_1 \cdots x_\ell \in Z(S_X)$, we have the equality $x_1 \cdots x_\ell s' = x_1^y \cdots x_\ell ^y s'$ in $S_X$.

Proof. Since composition of injective functions is injective, it suffices to consider the case $w \in C_j$ for some $1 \leq j \leq k$. In this case, it is enough to show that for integers $n_1, \ldots, n_k \geq N$, the function

\[ M_w \colon X^*(n_1, \ldots, n_{j-1},n_j,n_{j+1}, \ldots, n_k)/B_n \to X^*(n_1, \ldots, n_{j-1},n_j+1,n_{j+1}, \ldots, n_k)/B_{n+1} \]

of multiplication by $w$ from the left is injective. It follows from Proposition 4.22 that for large enough $N$, the function $(n_1, \ldots, n_k) \mapsto |X^*(n_1, \ldots, n_k)/B_n|$ is nonincreasing in each coordinate, hence eventually constant because its values are positive integers. Therefore, for large enough $N$, the function $M_w$ maps one finite set onto another finite set of the same cardinality. The required injectivity follows.

Proof. In view of Proposition 4.17, it suffices to show that the natural homomorphism of semigroups $S_X^*(N) \to S_X[z^{-1}]$ is injective. For that, let $u,v \in S_X^*(N)$ for which $z^r u = z^r v$ for some positive integer $r$. Invoking Corollary 4.23 with $w = z^r$, we get that $u = v$ so injectivity follows.

Proof. Since $\Gamma _{\mathcal {T}_2}$ is abelian, we need to show that the natural surjection $\pi \colon \Gamma '_{X \times \mathcal {T}_2} \to \Gamma _X'$ is an isomorphism. We will do so by constructing an inverse. The inclusion of the braided subset $X \times \{0\}$ into $X \times \mathcal {T}_2$ induces a group homomorphism $\psi _0 \colon \Gamma _{X} \stackrel {\sim }{\to } \Gamma _{X \times \{0\}} \to \Gamma _{X \times \mathcal {T}_2}$. We then get a homomorphism $\psi _0' \colon \Gamma _X' \to \Gamma _{X \times \mathcal {T}_2}'$. Arguing in the same way with $X \times \{1\}$ in place of $X \times \{0\}$, we get a homomorphism $\psi _1' \colon \Gamma _X' \to \Gamma _{X \times \mathcal {T}_2}'$. We claim that these homomorphisms coincide, namely $\psi _0' = \psi _1'$.

The commutator subgroup is generated by conjugates of commutators of elements in a generating set, so it suffices to check that $\psi _0'([x,y]^g) = \psi _1'([x,y]^g)$ for every $x,y \in X$ and $g \in \Gamma _X$. Equivalently, it is enough to show that $\psi _0([x,y])^{\psi _0(g)} = \psi _1([x,y])^{\psi _1(g)}$. Since conjugation by $\psi _0(g)$ coincides with conjugation by $\psi _1(g)$, our task is to show that $[\psi _0(x), \psi _0(y)] = [\psi _1(x), \psi _1(y)]$. Indeed, we have

(4.1)\begin{equation} (x,0)(y,0)(x,0)^{-1}(y,0)^{-1} = (x,1)(y,0)(x,1)^{-1}(y,0)^{-1} = (x,1)(y,1)(x,1)^{-1}(y,1)^{-1}. \end{equation}

We have thus proven our claim that $\psi _0' = \psi _1'$. We henceforth denote this map by $\psi '$.

Now we claim that $\psi '$ is an inverse of the natural surjection $\pi$. Since the inclusion of the braided set $X \times \{0\}$ into $X \times \mathcal {T}_2$ is a section of the projection $X \times \mathcal {T}_2 \to X$, it follows that $\pi \circ \psi '$ is the identity on $\Gamma _X'$. It remains to check that the homomorphism $\psi ' \circ \pi$ is the identity on $\Gamma _{X \times \mathcal {T}_2}'$. It suffices to show that $\psi ' \circ \pi$ acts as the identity on some generating set of $\Gamma _{X \times \mathcal {T}_2}'$. Since $X \times \mathcal {T}_2$ is a generating set for $\Gamma _{X \times \mathcal {T}_2}$ which is stable under conjugation, the subgroup $\Gamma _{X \times \mathcal {T}_2}'$ is generated by commutators of elements in $X \times \mathcal {T}_2$. Therefore, it is enough to check that $\psi ' \circ \pi$ acts as the identity on such commutators. This follows at once from (4.1). We have thus shown that $\psi '$ and $\pi$ are inverse to each other.

Proof. Let $a$ be an element in the kernel of our homomorphism. Since $a$ maps to the identity in $\Gamma _{X \times Y}^{\text {ab}}$ we get that $a \in \Gamma _{X \times Y}'$. It follows from the definition of a productive pair that $a = 1$. We have thus shown that the kernel of our homomorphism is trivial, so our homomorphism is indeed injective.

Proof. In view of Corollary 4.27, it suffices to check that the natural homomorphism from $\Gamma _{X \times Y}^{\text {ab}}$, which we identify with $\Gamma _{(X\times Y)_{\text {triv}}}$, to $\Gamma _{X_{\text {triv}} \times Y_{\text {triv}}}$ is injective. This follows at once from our assumption that the pair $(X,Y)$ is synchronized.

Proof. By Theorem 4.24, it suffices to show that $v = w$ in $\Gamma _{X \times Y}$. In view of Corollary 4.29, it is enough to show that the images of $v$ and $w$ in each of the groups $\Gamma _X$, $\Gamma _Y$, $\Gamma _{X_{\text {triv}} \times Y_{\text {triv}}}$ agree. This is guaranteed by the three conditions that $v$ and $w$ satisfy.

Proof. This is an immediate consequence of Proposition 4.20.

Proof. We invoke Corollary 3.7 so given $X_j,Y_j \subseteq D_j$ with $|X_j| = |Y_j| = \lfloor n_j/2 \rfloor$, we need to show that there exists $g \in B_n$ such that $s^g = s$ and $X_j^g = Y_j$ for all $1 \leq j \leq k$. For $1 \leq i \leq n$ put

\[ u_i = \begin{cases} 1 & i \in X_j \text{ for some } 1 \leq j \leq k ,\\ 0 & \text{otherwise,} \end{cases}\quad v_i = \begin{cases} 1 & i \in Y_j \text{ for some } 1 \leq j \leq k ,\\ 0 & \text{otherwise}. \end{cases} \]

Viewing $((x_1, u_1), \ldots, (x_n, u_n))$ and $((x_1, v_1), \ldots, (x_n, v_n))$ as elements of $(X \times \mathcal {T}_2)^n$, our task is to show that $((x_1, u_1), \ldots, (x_n, u_n))^g = ((x_1, v_1), \ldots, (x_n, v_n))$ for some $g \in B_n$.

Since the entries of $s$ generate $X$, they also generate $\operatorname {Inn}(X)$. As $\mathcal {T}_2$ is a trivial braided set, it follows that the entries of $((x_1, u_1), \ldots, (x_n, u_n))$ generate $\operatorname {Inn}(X \times \mathcal {T}_2)$, so these entries also generate $X \times \mathcal {T}_2$ because they map surjectively onto $(X \times \mathcal {T}_2)_{\text {triv}} = X_{\text {triv}} \times \mathcal {T}_2$. Similarly, the entries of $((x_1, v_1), \ldots, (x_n, v_n))$ generate $X \times \mathcal {T}_2$. Therefore, we need to show that $((x_1, u_1), \ldots, (x_n, u_n)) = ((x_1, v_1), \ldots, (x_n, v_n))$ as elements in $S^*_{X \times \mathcal {T}_2}(\lfloor \min \{n_1, \ldots, n_k\}/2 \rfloor )$.