2.1. The image of a totally symmetric set

The following lemma, due to Kordek and Margalit [Reference Kordek and Margalit19] (Lemma 2.1), is the crux of how totally symmetric sets are used throughout this paper.

In this paper, we often consider whether $\lvert\phi(S)\rvert = \lvert S\rvert$ or not. We say that $\phi$ splits S if $|\phi(S)|=|S|$ .

2.2. Totally symmetric sets with finite order elements

Let $S=\{s_i\}_{i=1}^n$ be a totally symmetric subset of a group G. Since all elements of S are conjugate, every element of S has the same order. Therefore, if one element of S has finite order $k \in \mathbb{N}$ , every element of S has order k.

Since the elements of a totally symmetric set commute, if a totally symmetric set consists of a finite number of elements each of finite order, then $\langle S \rangle$ is a finite group. The following lemma gives a lower bound of the size of this group. A first bound was obtained by Chen, Kordek, and Margalit (a proof of which can be found in [Reference Chudnovsky, Kordek, Li and Partin8]), but we will use an improvement of this bound by Caplinger–Kordek [Reference Caplinger and Kordek7] (Lemma 6).

Combining Lemma 2.4 with work of Chudnovsky–Kordek–Li–Partin, one obtains a lower bound on the size of a group based on the size of a totally symmetric subset consisting of finite order elements [Reference Chudnovsky, Kordek, Li and Partin8].

A restatement of Proposition 2.5 in terms of group homomorphisms is the following: let S be a totally symmetric set of a group G and $\phi \,:\, G \rightarrow H$ a group homomorphism to a finite group H. If $\phi$ splits S, then $|\phi(G)|\geq p^{|S|-1}|S|!$ , where p is the minimal integer such that $s_i^p = s_j^p$ for all $s_i, s_j \in S$ .

3.1. Precursory results

Recall the two totally symmetric sets in $B_n$ of $S_{odd}=\{ \sigma_{2i-1} \}_{i=1}^{\lfloor n/2 \rfloor}$ and $S_{even}=\{ \sigma_{2i} \}_{i=1}^{ \lfloor (n-1)/2 \rfloor }$ . Chudnovsky, Kordek, Li, and Partin used these totally symmetric sets to determine a necessary condition for the existence of a non-cyclic homomorphism of the braid group [Reference Chudnovsky, Kordek, Li and Partin8]. Recently, Caplinger and Kordek obtained a stronger necessary condition than the one found by Chudnovsky–Kordek–Li–Partin [Reference Caplinger and Kordek7].

Lemma 3.1 (Caplinger−Kordek). Let G be a finite group and let $n \geq 5$ . If the homomorphism $B_n \to G$ is non-cyclic, then,

\begin{equation*} \lvert G\rvert \geq 3^{\lfloor n/2 \rfloor -1} \left( \lfloor n/2 \rfloor \right) !. \end{equation*}

In Section 3.2, we strengthen the lower bound found in Lemma 3.1. Before we strengthen this lower bound, we introduce the following well-known facts about the braid group, which can be found in [Reference Artin1]. This lemma provides sufficient conditions for when then image of a homomorphism of the braid group is cyclic.

Proof. Since $n\geq 5$ , both $S_{even}$ and $S_{odd}$ have cardinality at least 2. Suppose that $\phi$ does not split $S_{even}$ . Then $\phi(\sigma_2)=\phi(\sigma_4)$ . Since $\sigma_4$ commutes with $\sigma_1$ , then $\phi(\sigma_2)$ commutes with $\phi(\sigma_1)$ . By Lemma 3.2, $\phi$ must be cyclic, a contradiction.

By a similar computation, if $\phi$ does not split $S_{odd}$ , then $\phi(\sigma_4 )$ commutes with $\phi(\sigma_3 )$ , ultimately forcing $\phi$ to be cyclic.

Notice that both Lemmas 3.2 and 3.3 fail for $n=4$ as there exists the folding homomorphism from $B_4\rightarrow B_3$ which maps $\sigma_1,\sigma_3\mapsto \sigma_1$ and $\sigma_2\mapsto \sigma_2$ .

3.2. Proof of Theorem 1.1

In this section, we prove Theorem 1.1, which provides a strengthened lower bound for the smallest non-cyclic finite quotient of $B_n$ .

We begin by following the proof of Chudnovsky–Kordek–Li–Partin, then further their ideas by applying Lemma 3.2.

Theorem 1. Let $n\geq 5$ , and let $\phi\,:\,B_n\to G$ be a non-cyclic homomorphism to a finite group, G, so that $\phi(B_n)$ is not isomorphic to the symmetric group, $\Sigma_n$ . Then,

\begin{equation*} \lvert \phi(B_n)\rvert \geq \left( \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!.\end{equation*}

Moreover, if p is the smallest integer so that $\phi(\sigma_i)^p=\phi(\sigma_j)^p$ for any i, j, then

\begin{equation*} \lvert \phi(B_n)\rvert \geq \left(\left(\mathrm{lpf}(p)-1\right) \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!,\end{equation*}

where $\mathrm{lpf}(p)$ is the least integer greater than 1 that divides p.

Proof of Theorem 1.1. Denote $\mathcal{O}_\phi=\phi(S_{odd})$ , $\mathcal{E}_\phi=\phi(S_{even})$ , $s_i=\phi(\sigma_i)$ , $k=\left\lfloor n/2 \right\rfloor$ , $\mathcal{B} = \phi(B_n)$ , let d be the order of the $s_i$ ’s and let p be the smallest integer so that $\phi(\sigma_i)^p=\phi(\sigma_j)^p$ for any i, j.

Since we aim for a lower bound, we may assume that $\phi(B_n)$ is the smallest possible quotient not isomorphic to $\Sigma_n$ . In this case, Caplinger and Kordek prove that $p = d$ in Lemma 7 of [Reference Caplinger and Kordek7]. If $d=1$ , then $\phi$ is trivial (hence cyclic). If $d=2$ , $\phi$ factors through the symmetric group $\Sigma_n$ , and the image is either $\Sigma_n$ or cyclic (since the alternating group is the only proper normal subgroup of $\Sigma_n$ for $n\geq 5$ ).

Therefore, we may assume that $d=p \geq 3$ .

As $\phi$ is not cyclic and $n\geq5$ , each $s_i$ is distinct by Lemma 3.3. Thus, $\mathcal{O}_\phi$ is a totally symmetric subset of size k in $\mathcal{B}$ as the injective image of a totally symmetric subset of size k in $B_n$ .

Notice that $\mathcal{B}$ acts by conjugation on the set of totally symmetric subsets of $\mathcal{B}$ of size k, and let $\Gamma= \mathrm{Fix}_{\mathcal{B}}(\mathcal{O}_\phi)$ . This gives us a surjection $\psi \,:\, \Gamma \rightarrow \Sigma_k$ , where $\Sigma_k$ is the symmetric group on k elements.

Under this action by $\mathcal{B}$ , $\mathcal{O}_\phi$ fixes $\mathcal{O}_\phi$ pointwise since the elements of a totally symmetric set pairwise commute. This shows that $\langle \mathcal{O}_\phi \rangle \subseteq \Gamma$ and, in fact, $\langle \mathcal{O}_\phi \rangle \subseteq \mathrm{ker}(\psi)$ . By Lemma 2.4, we have that $\lvert \langle \mathcal{O}_\phi \rangle\rvert \geq p^{k-1}$ , and since $p \geq 3$ , $\lvert \mathrm{ker}(\psi)\rvert \geq 3^{k-1}$ . It follows that

(3.1) \begin{equation} \lvert \mathcal{B}\rvert \geq \lvert \Gamma\rvert = \lvert\Sigma_k\rvert \cdot \lvert\mathrm{ker}(\psi)\rvert\geq k!(\lvert\langle \mathcal{O}_\phi\rangle\rvert)\geq k!3^{k-1}.\end{equation}

We now begin improving the bound on $\lvert \mathcal{B}\rvert $ . Notice that $\mathcal{E}_\phi=\{s_{2i}\}_{i=1}^{\lfloor (n-1)/2 \rfloor}$ is a second totally symmetric set which consists of the images of the remaining generators of $B_n$ . The elements in $\mathcal{E}_\phi$ are currently not accounted for in Equation 3.1. To include these elements in the bound of $\lvert \mathcal{B}\rvert$ , we consider when elements of $\langle \mathcal{E}_\phi \rangle$ are not in $\Gamma$ . The following observations lead us to find elements of $\langle \mathcal{E}_\phi \rangle$ that are not in $\Gamma$ . We then count distinct cosets of $\Gamma$ in $\mathcal{B}$ .

Observation 1. Suppose there exists an $m \in \{ 1, \ldots, p-1 \}$ so that $s_i^m \in \Gamma$ , then $s_i^m\in \ker\psi$ .

By definition of $\Gamma$ , $s_i^m$ acts on $\mathcal{O}_\phi$ by conjugation, fixing $\mathcal{O}_\phi$ setwise. By the relations of the braid group, $s_i^m$ commutes with every element of $\mathcal{O}_\phi$ except for $s_{i\pm1}$ . Since $\mathcal{O}_\phi$ is fixed setwise, then either conjugation by $s_i^m$ swaps the elements $s_{i+1}$ and $s_{i-1}$ , or fixes the elements pointwise. Suppose first that conjugation by $s_i^{m}$ swaps the elements $s_{i+1}$ and $s_{i-1}$ , meaning $s_i^m s_{i-1} s_i^{-m} = s_{i+1}$ . Then,

\begin{equation*}s_{i+2}(s_{i+1})s_{i+2}^{-1} = s_{i+2}\left(s_i^ms_{i-1}s_i^{-m}\right)s_{i+2}^{-1} = s_i^ms_{i-1}s_i^{-m}=s_{i+1},\end{equation*}

which shows that $s_{i+2}$ and $s_{i+1}$ commute. By Lemma 3.2, $\phi$ must be cyclic. If i is large enough so that either $i+1 \gt n-1$ or $i+2 \gt n-1$ , an analogous argument shows that $s_{i-2}$ and $s_{i-1}$ commute and that $\phi$ is cyclic. In both cases, we have contradicted our assumption that $\phi$ is non-cyclic. Thus, conjugation by $s_i^m$ does not swap the elements $s_{i+1}$ and $s_{i-1}$ but rather fixes these elements pointwise. Therefore, for all i, conjugation by $s_i^m$ fixes every element of S pointwise. This implies that if $s_i^m\in \Gamma$ , then $s_i^m\in \ker\psi$ .

Observation 4. $s_i^{m_1}\Gamma\neq s_j^{m_2}\Gamma$ when i and j are even, $m_1$ and $m_2$ are relatively prime to p, and $m_1-m_2$ is relatively prime to p.

From Observation 3, $s_i^{m_1},s_j^{m_2} \notin \Gamma$ for every i, j even and $m_1,m_2$ relatively prime to p. Suppose that $s_i^{m_1}\Gamma=s_j^{m_2}\Gamma$ . This implies that $s_i^{-m_1} s_j^{m_2}\in \Gamma$ .

If $i=j$ , then $s_i^{-m_1} s_j^{m_2}\in \Gamma$ implies that $s_i^{m_2-m_1}\in \Gamma$ . Then, $m_2-m_1$ is not relatively prime to p, a contradiction.

If $i \lt j$ , then $s_j$ commutes with $s_{i-1}$ . Consider the action of $s_i^{-m_1} s_j^{m_2}$ on $s_{i-1}$ by conjugation:

\begin{equation*}s_j^{-m_2}s_i^{m_1}s_{i-1}s_i^{-m_1} s_j^{m_2}=s_i^{m_1}s_{i-1}s_i^{-m_1}.\end{equation*}

Since we supposed that $s_i^{-m_1} s_j^{m_2}\in\Gamma$ , then conjugation by $s_i^{-m_1} s_j^{m_2}$ fixes the set $\mathcal{O}_\phi$ setwise. Therefore, the above equation shows that $s_i^{m_1}s_{i-1}s_i^{-m_1}\in \mathcal{O}_\phi$ . The exponent $m_1$ was chosen so that $s_i^{m_1}\not\in \Gamma$ , which means that $\mathcal{O}_\phi$ is not closed under conjugation by $s_i^{m_1}$ . As described in Observation 1, $s_i^{m_1}s_{i-1}s_i^{-m_1}$ is not an element of $\mathcal{O}_\phi$ when $\phi$ is non-cyclic, a contradiction. Hence, $s_i^{-m_1} s_j^{m_2} \not \in \Gamma$ .

Observation 5. Counting the cosets of $\Gamma$ .

Let $\mathrm{lpf}(p)$ be the least integer greater than 1 that divides p. Notice that the set $\{ 2, \ldots, \mathrm{lpf}(p)-1$ , $\mathrm{lpf}(p)+1 \}$ is a set of $\mathrm{lpf}(p) - 1$ integers which are each relatively prime to p, and pairwise their differences are relatively prime to p. Together with Observation 4, this shows that there are $\mathrm{lpf}(p)-1$ distinct non-intersecting cosets of $\Gamma$ for each $s_i$ with i even. Since there are $|\mathcal{E}_\phi|$ distinct elements $s_i$ with i even, we get $(\mathrm{lpf}(p)-1)|\mathcal{E}_\phi|$ distinct cosets of $\Gamma$ .

By counting the distinct cosets of $\Gamma$ , we obtain a lower bound for the complement of $\Gamma$ in the image of $\phi$ :

\begin{equation*} \lvert \phi(B_n)-\Gamma\rvert \geq (\mathrm{lpf}(p)-1)\lvert\mathcal{E}_\phi\rvert\cdot\lvert \Gamma \rvert\end{equation*}

Combining all of these observations, we arrive at the following lower bound for $\lvert \phi(B_n)\rvert$ :

\begin{align*}\lvert \phi(B_n)\rvert \geq \lvert \Gamma\rvert+((\mathrm{lpf}(p)-1)\lvert \mathcal{E}_\phi\rvert )\lvert \Gamma \rvert & =((\mathrm{lpf}(p)-1)\lvert \mathcal{E}_\phi\rvert +1)\cdot\lvert \Gamma\rvert \\ & \geq((\mathrm{lpf}(p)-1)\lvert \mathcal{E}_\phi\rvert +1)(\lvert\langle \mathcal{O}_\phi\rangle\rvert)k! .\end{align*}

By substituting the values of k, $\lvert \langle \mathcal{O}_\phi \rangle\rvert$ , and $\lvert \mathcal{E}_\phi\rvert$ , we arrive at our final result:

\begin{equation*} \lvert \phi(B_n)\rvert \geq \left(\left(\mathrm{lpf}(p)-1\right) \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!.\end{equation*}

For all even values of p, we have that $\mathrm{lpf}(p)-1 = 1$ , which gives the minimal bound:

\begin{equation*} \lvert \phi(B_n)\rvert \geq \left( \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!.\end{equation*}

4.1. The virtual braid group

Let $vB_n$ denote the virtual braid group on n strands. This group has generators $\sigma_1, \ldots ,\sigma_{n-1}$ and $\tau_1, \ldots, \tau_{n-1}$ . The generators $\sigma_1, \ldots ,\sigma_{n-1}$ satisfy the classical braid group relations, and the generators $\tau_1, \ldots, \tau_{n-1}$ generate the symmetric group. There are also some mixing relations. We list all relations in the virtual braid group, below:

1. (1) $\sigma_i \sigma_j = \sigma_j \sigma_i$ for $\lvert i-j\rvert \gt 1$ (Far Commutativity)

2. (2) $\sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1}$ for $1 \leq i \leq n-2$ (Braid Relation)

3. (3) $\tau_i^2 = 1$ for $1 \leq i \leq n-1$ ( $\tau$ is a Transposition)

4. (4) $\tau_i \tau_j = \tau_j \tau_i$ for $\lvert i-j\rvert \gt 1$ ( $\tau$ Far Commutativity)

5. (5) $\tau_i \tau_{i+1} \tau_i = \tau_{i+1} \tau_i \tau_{i+1}$ for $1 \leq i \leq n-2$ ( $\tau$ Braid Relation)

6. (6) $\sigma_i \tau_j = \tau_j \sigma_i$ for $\lvert i-j\rvert \gt 1$ (Mixed Far Commutativity)

7. (7) $\tau_{i+1} \sigma_i \tau_{i+1} = \tau_i \sigma_{i+1} \tau_i$ for $1 \leq i \leq n-2$ (Mixed Braid Relation)

We note that these relations encode the virtual (or extended) Reidemeister moves. From this point of view, $vB_n$ is the free product of the braid group and the symmetric group modulo relations (6) and (7), $vB_n=B_n*\Sigma_n\ _{(6),(7)}$ . This presentation is nice in the sense that you can “see” the braid group as a subgroup of the virtual braid group. $B_n$ embeds into $vB_n$ , as was first proven by Kamada [Reference Kamada14]; see also Gaudreau [Reference Gaudreau12] and Bellingeri–Paris [Reference Bellingeri and Paris4]. The canonical embeddings of $B_n$ and $\Sigma_n$ in $vB_n$ are $B_n=\langle \sigma_1, \cdots ,\sigma_{n-1}\rangle$ and $\Sigma_n=\langle \tau_1, \cdots ,\tau_{n-1}\rangle$ .

Another presentation of $vB_n$ highlights a key difference between the virtual braids and the classical braids. The pure virtual braid group, $PvB_n$ , is a subgroup of $vB_n$ which is the kernel of the projection $vB_n\to \Sigma_n$ by sending $\sigma_i\mapsto \tau_i$ and $\tau_i\mapsto \tau_i$ . Unlike the classical braid group, $vB_n$ splits as a semidirect product, $vB_n\cong PvB_n\rtimes \Sigma_n$ [Reference Bardakov3].

The subgroup $PvB_n$ is generated by elements denoted $\sigma_{i, j}$ of the form:

\begin{align*}\sigma_{i, j} {}& = \tau_i \tau_{i+1} \ldots \tau_{j-2} \tau_{j-1} \sigma_{j-1} \tau_{j-2} \ldots \tau_{i+1} \tau_i {} \qquad\qquad\qquad \text{when $i<j$, and}\\\sigma_{i, j} {}& = \tau_{i-1} \tau_{i-2} \ldots \tau_{j-2} \tau_{j-1} \sigma_{j} \tau_{j} \tau_{j-1} \ldots \tau_{i-1} \qquad\qquad\qquad {} \text{ when $j<i$.}\end{align*}

These generators are depicted in Figure 2. A presentation for $PvB_n$ is generated by the $\sigma_{i, j}$ , for $i\neq j$ elements and the following two relations [Reference Bardakov3]:

Figure 2.

(a) The element $\sigma_{i, j}$ when $j<i$ . (b) The element $\sigma_{i, j}$ when $i<j$

Commutativity Relation: $\sigma_{i, j}\sigma_{k,l}=\sigma_{k,l}\sigma_{i, j}$ where $\lvert \{ i, j,k,l \}\rvert = 4$

Braid Relation: $\sigma_{i, j}\sigma_{i,k}\sigma_{j,k} = \sigma_{j,k}\sigma_{i,k}\sigma_{i, j}$ where $\lvert \{ i, j,k \}\rvert = 3$

4.2. The welded braid group

The welded braid group, $wB_n$ , is a quotient of $vB_n$ by the Over Crossings Commute relation, or “OC” relation, defined as $\tau_i\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_i\tau_{i+1}$ [Reference Bellingeri and Soulié5]. The welded braid group has also been called the braid-permutation group by Fenn–Rimányi–Rourke [Reference Fenn, Rimányi and Rourke10], a finite-index subgroup of the ring group by Brendle–Hatcher [Reference Brendle and Hatcher6], and the loop braid group by Baez–Crans–Wise [Reference Baez, Wise and Crans2].

Recall from Section 4.1.1, a presentation for $PvB_n$ is generated by the elements denoted $\sigma_{i, j}$ . These elements also generate the pure welded braid group, $PwB_n$ . Analogous to the virtual braid group, $wB_n$ is a semidirect product of the pure welded braid group and the symmetric group, $wB_n=PwB_n\rtimes \Sigma_n$ . Through communication with Dror Bar-Natan, we learned the folklore result that the OC relation implies that $\sigma_{i,k}\sigma_{i, j}=\sigma_{i, j}\sigma_{i,k}$ , and a proof of this fact can be found in [Reference Scherich and Verberne22]. It follows that $PwB_n$ is a quotient of $PvB_n$ by the OC relation, and $PwB_n$ has the following presentation:

Generators: $\{\sigma_{i, j}\}$ for $1\leq i, j\leq n$ with $i\neq j$

Commutativity Relation: $\sigma_{i, j}\sigma_{k,l}=\sigma_{k,l}\sigma_{i, j}$ where $\lvert \{ i, j,k,l \}\rvert = 4$

Braid Relation: $\sigma_{i, j}\sigma_{i,k}\sigma_{j,k} = \sigma_{j,k}\sigma_{i,k}\sigma_{i, j}$ where $\lvert \{ i, j,k \}\rvert = 3$

OC Relation: $\sigma_{i,k}\sigma_{i, j}=\sigma_{i, j}\sigma_{i,k}$

4.2.2. Important lemmas

The classical braid group $B_n$ and the symmetric group $\Sigma_n$ are subgroups of $vB_n$ , and also $wB_n$ , under the canonical embeddings $B_n= \langle \sigma_i \rangle_{i=1}^{n-1}\subseteq vB_n$ and $\Sigma_n= \langle \tau_i \rangle_{i=1}^{n-1}\subseteq vB_n$ . This was proven by Kamada [Reference Kamada15] for the virtual case and Fenn–Rimányi–Rourke [Reference Fenn, Rimányi and Rourke10] for the welded case. The pure subgroup has the canonical presentation $PvB_n= \langle \sigma_{i, j}\rangle _{i\neq j} \subseteq vB_n$ , similarly for $PwB_n$ . From here on, the restriction of a map on $vB_n$ (resp. $wB_n)$ to $B_n$ , $\Sigma_n$ or $PvB_n$ (resp. $PwB_n$ ) refers to the canonical embeddings of these groups. Recall from the introduction that a map $\phi$ is called cyclic (resp. abelian), if its image is cyclic (resp. abelian).

Lemma 4.2. If $\phi \,:\, vB_n \rightarrow G$ is a group homomorphism so that $\phi$ restricted to either $\Sigma_n$ or $B_n$ is abelian, then $\phi$ abelian.

Proof. Suppose that $\phi$ restricted to $\Sigma_n$ is abelian. The $\tau$ braid relation gives that $\phi(\tau_i)=\phi(\tau_{i+1})$ for all i, and so $\phi$ is cyclic on $\Sigma_n$ . Denote $\phi(\tau_i)=g$ . Applying $\phi$ to the mixed braid relation yields

\begin{align*} \phi(\tau_{i+1}\sigma_i\tau_{i+1}) {}& =\phi(\tau_i\sigma_{i+1}\tau_i)\\ g\phi(\sigma_i)g {}& =g\phi(\sigma_{i+t})g\\ \phi(\sigma_i) {}& =\phi(\sigma_{i+1})\end{align*}

This shows that $\phi$ restricted to $B_n$ is also cyclic, and therefore $\phi$ is abelian.

Suppose that $\phi$ restricted to $B_n$ is abelian. A similar argument using the braid relations shows that $\phi$ is cyclic on $B_n$ , and the mixed braid relation shows that $\phi$ is cyclic on $\Sigma_n$ .

Corollary. If $\phi \,:\, wB_n \rightarrow G$ is a group homomorphism so that $\phi$ restricted to either $\Sigma_n$ or $B_n$ is abelian, then $\phi$ is abelian.

Proof. Let $p \,:\, vB_n \rightarrow wB_n$ be the quotient map which sends a virtual braid to its equivalence class modulo the the OC relation. Apply Lemma 4.2 to the map $\phi \circ p$ .

The following lemma is a key step to proving Theorems 1.2 and 1.3. To use totally symmetric sets to count the cardinality of the image of a homomorphism, the homomorphism needs to split a totally symmetric set. This lemma shows that, under the right conditions, when some subset of the totally symmetric sets $\{A_i\}_{i=1}^{n}$ do not split under a map $\phi \,:\, wB_n \rightarrow G$ , the images of the $A_i$ ’s which do not split create a new totally symmetric set in the image. A schematic diagram for Lemma 4.3 is shown in Figure 3.

Lemma 4.3. Let $\{A_{i_1},\cdots ,A_{i_m}\}$ be a subset of $\{A_1,\cdots, A_n\}$ , the totally symmetric sets in $wB_n$ . Let $\phi \,:\, wB_n \rightarrow G$ be a non-abelian group homomorphism. Suppose $\phi$ does not split $A_{i_j}$ for all $i_j$ , and each $\phi(A_{i_j})^2$ is a distinct element in the image. Then the set $\{\phi(A_{i_j})^2\}_{j=1}^m$ is a totally symmetric set in $\phi(wB_n)$ of size m.

Proof. We will prove this lemma for the case where $\{A_{i_1},\cdots A_{i_m}\}=\{A_1,\cdots, A_m\}$ , as all other cases follow from an analogous proof with possible re-indexing.

Figure 3.

Schematic diagram for Lemma 4.3

Let $g_i=\phi(A_i)$ . We will show that the set $\{g_i^2\}_{i=1}^m$ is a totally symmetric set in $\phi(wB_n)$ of size m.

By assumption, the $g_i^2$ ’s are distinct so the set $\{g_i^2\}_{i=1}^m$ has m elements. Notice that every element of $A_i$ is of the form $\sigma_{i, j}$ , where the first index, i, remains fixed. Since $\phi$ does not split any of the totally symmetric sets $A_i$ , $\phi(\sigma_{i, j})$ is determined by its first index i, that is, $\phi(\sigma_{i, j})=g_i$ .

For the commutation condition, applying $\phi$ to the braid relation in $wB_n$ shows

\begin{align*} \sigma_{i, j}\sigma_{i,k}\sigma_{j,k} {}& = \sigma_{j,k}\sigma_{i,k}\sigma_{i, j}\\ \phi(\sigma_{i, j})\phi(\sigma_{i,k})\phi(\sigma_{j,k}) {}& = \phi(\sigma_{j,k})\phi(\sigma_{i,k})\phi(\sigma_{i, j})\\ g_i g_i g_j {}& = g_j g_i g_i, \end{align*}

which shows that for each i and j, $g_i^2$ and $g_j$ commute. In turn, this implies that $g_i^2$ and $g_j^2$ commute.

To show the conjugation condition holds, notice that if $f g_i\, f^{-1}=g_j$ then $f g_i^2\, f^{-1}=g_j^2$ . Therefore, it suffices to show the conjugation condition holds for the set $\{g_i\}$ . The following computations show that conjugation by $\phi(\tau_i)$ swaps $g_i$ and $g_{i+1}$ but fixes all other $g_k$ .

First, we show that conjugation by $\phi(\tau_i)$ swaps $g_i$ and $g_{i+1}$ . There are two cases to consider:

Case 1: Suppose $i\leq n-2$ . A similar computation described in Lemma 4.1 shows that conjugation by $\tau_i$ swaps $\sigma_{i,i+2}$ with $\sigma_{i+1,i+2}$ . Thus,

\begin{align*} \phi(\tau_i\sigma_{i,i+1}\tau_i) {}& = \phi(\sigma_{i+1,i+2})\\ \phi(\tau_i)g_i\phi(\tau_i) {}& = g_{i+1}. \end{align*}

Case 2: Suppose $i>n-2$ , which implies that $i=n-1$ . A similar computation to the one above shows that conjugation by $\tau_{i-1}$ swaps $\sigma_{i,i-1}$ and $\sigma_{i+1,i-1}$ and that conjugation by $\phi (\tau_i)$ swaps $g_i$ and $g_{i+1}$ .

Next, we show that $g_k$ is fixed under conjugation by $\tau_i$ , when $k \neq i, i+1$ . Recall that conjugation by $\tau_i$ on $\sigma_{j,k}$ permutes the indices: $\tau_i\sigma_{j,k}\tau_i=\sigma_{\tau_i(j),\tau_i(k)}$ . So, when $k\neq i, i+1$ , $g_k$ remains fixed under conjugation by $\phi(\tau_i)$ as follows:

\begin{equation*} \phi(\tau_i)g_k\phi(\tau_i)=\phi(\tau_i \sigma_{k,-}\tau_i) = \phi(\sigma_{\tau_i(k),\tau_i(-)}) = \phi(\sigma_{k,-}) = g_k.\end{equation*}

This proves the conjugation condition in the definition of a totally symmetric set holds, and we have proven our claim.

Remark. The Lemma 4.3 is stated for $wB_n$ ; however, it is also true for $vB_n$ . In $vB_n$ , the sets $A_i$ are not totally symmetric, but they do satisfy the conjugation condition, which is the only condition needed in the proof.

5.1. Proof of Theorem 1.2

First, we prove the classification theorem for the welded braid group, $wB_n$ . This is a first step in classifying non-cyclic homomorphisms $wB_n \to G$ , where G is a finite group.

In the statement of Theorem 1.2, the requirement that $n\geq 5$ is only necessary for Part (3) due to the applications of Lemma 3.3 and Theorem 1.1. All other conditions hold for $n\geq 4$ .

The proof of Theorem 1.2 is inspired by Figure 4. We consider cases on whether $\phi$ splits various rows and columns of the diagrams. The rows of the Full diagram, as seen in Figure 4(b), are the totally symmetric sets, $A_i$ , from Lemma 4.1. In the Left diagram, the rows of the outlined triangle are the totally symmetric sets, $L_i$ . The rows above the outlined triangle are the inverses of the columns within the outlined triangle. Both the columns in the outlined triangle and the rows above the outline are not totally symmetric sets but satisfy the conjugation condition. This can be verified by similar computations described in Lemma 4.1. The Right diagram has an analogous form. The rows of the outlined triangle are the totally symmetric sets, $R_i$ . The columns below the outlined triangle are the inverses of the rows within the triangle, and both satisfy the conjugation condition.

Figure 4.

(a) Left diagram. (b) Full diagram. (c) Right diagram

Proof of Theorem 1.2. Let us suppose that $\phi$ is non-abelian and that $\phi$ restricted to $PwB_n$ is non-cyclic. We consider cases on whether or not $\phi$ splits the totally symmetric sets $A_i$ .

Case 1: Suppose that there exists an i so that $\phi$ splits $A_i$ . Since $A_i$ is a totally symmetric set with size $n-1$ , applying Proposition 2.5 yields

\begin{equation*}\lvert \phi(wB_n)\rvert \geq 2^{n-2}(n-1)!.\end{equation*}

Case 2: Suppose $\phi$ does not split any of the $A_i$ ’s. Denote $\phi(A_i)=\{g_i\}$ . Further suppose that there exists $i_0$ and $j_0$ so that $g_{i_0}^2\neq g_{j_0}^2$ . Notice this implies $g_{i_0}\neq g_{j_0}$ . Since $\phi$ is non-abelian, we may assume by Lemma 4.2 that $\phi$ is non-cyclic on $\Sigma_n$ and that $\phi(\tau_i)\neq id$ . We consider cases on $i_0$ and $j_0$ with the goal to apply Lemma 4.3.

Subcase 1: Suppose $i_0,j_0<n$ . We will use the Right diagram in Figure 4 to conclude that $g_1,\cdots, g_{n-1}$ are distinct. By assumption, $g_{i_0}\neq g_{j_0}$ which implies that $\phi\left(\sigma_{{i_0},n}\right)\neq \phi\left(\sigma_{{j_0},n}\right)$ , and therefore $\phi\left(\sigma_{{i_0},n}^{-1}\right)\neq \phi\left(\sigma_{{j_0},n}^{-1}\right)$ . The bottom row of the Right diagram contains both $\sigma_{{i_0},n}^{-1}$ and $\sigma_{{j_0},n}^{-1}$ . Even though the bottom row of the Right diagram is not a totally symmetric set, it does satisfy the conjugation condition. Since $\phi\left(\sigma_{i_0,n}^{-1}\right) \neq \phi\left(\sigma_{i_1,n}^{-1}\right)$ , Remark 2.1 implies that $\phi$ splits the bottom row. Thus, $\phi\left(\sigma_{i,n}^{-1}\right)\neq \phi\left(\sigma_{j,n}^{-1}\right)$ for all $i, j\lt n$ , which shows that $g_i\neq g_j$ , for all $i, j<n$ . Since $g_{i_0}^2\neq g_{j_0}^2$ by assumption, Remark 2.2 shows the each of the $g_i^2$ are unique. Thus, we have shown all of the hypotheses of Lemma 4.3 are satisfied, and $\left\{g_1^2,\cdots, g_{n-1}^2\right\}$ is a totally symmetric set in the image of $\phi$ of size $n-1$ . Proposition 2.5 yields

\begin{equation*}\lvert \phi(wB_n)\rvert \geq 2^{n-2}(n-1)!.\end{equation*}

Subcase 2: Suppose $i_0,j_0>1$ . An analogous argument to Subcase 1 using the Left diagram from Figure 4 concludes that $\left\{g_2^2,\cdots, g_n^2\right\}$ is a totally symmetric set in the image of $\phi$ of size $n-1$ . Proposition 2.5 yields

\begin{equation*}\lvert \phi(wB_n)\rvert \geq 2^{n-2}(n-1)!.\end{equation*}

Subcase 3: Suppose $i_0=1$ and $j_0=n$ , which implies that $g_1 \neq g_n$ , and further that $\phi(\sigma_{1,-}) \neq \phi(\sigma_{n,-})$ . Looking at the Full diagram in Figure 4(b), Subcase 3 analyzes when the top and bottom rows of the Full diagram are mapped to different elements.

We now analyze where $\phi$ can send the second row.

Suppose first that $\phi$ maps $A_2$ , or all the elements of the second row, to $g_1$ . Then in Figure 4(a), the Left diagram, we notice that $L_2$ and $L_{n}$ map to different elements. Therefore, two elements in the top row of Figure 4(a) map to different elements. Since the top row satisfies the conjugation condition, we have that the top row must split. Since $g_{i_0}^2\neq g_{j_0}^2$ by assumption, Remark 2.2 shows that each $g_i^2$ is unique. Therefore, by Lemma 4.3, the set $\left\{ g_2^2, g_3^2, \ldots, g_n^2 \right\}$ is a totally symmetric set of size $n-1$ . Proposition 2.5 yields

\begin{equation*}\lvert \phi(wB_n)\rvert \geq 2^{n-2}(n-1)!.\end{equation*}

A similar argument follows for when $\phi$ maps $A_2$ , or all the elements of the second row, to $g_n$ , but this time we consider Figure 4(c), the Right diagram. Since $R_1$ and $R_2$ map to different elements, the bottom row of the Right diagram must split as it satisfies the conjugation relation. Since $g_{i_0}^2\neq g_{j_0}^2$ by assumption, Remark 2.2 shows that each $g_i^2$ is unique. Therefore, by Lemma 4.3, the set $\left\{ g_1^2, g_2^2, \ldots, g_{n-1}^2 \right\}$ is a totally symmetric set of size $n-1$ . In this case, Proposition 2.5 will again yield

\begin{equation*}\lvert \phi(wB_n)\rvert \geq 2^{n-2}(n-1)!.\end{equation*}

Finally, suppose that $\phi$ sends $A_2$ to an element $g_2$ where $g_2 \neq g_1, g_n$ . Then in Figure 4(a), the Left diagram, we notice that $L_2$ and $L_{n}$ map to different elements. Therefore, two elements in the top row of Figure 4(a) map to different elements. Since the top row satisfies the conjugation condition, we have that the top row must split. Similarly, in Figure 4(c), the Right diagram, we notice that $R_1$ and $R_{2}$ map to different elements. Therefore, two elements in the bottom row of Figure 4(c) map to different elements, and since the bottom row satisfies the conjugation condition the bottom row must split. Notice that we must have that $\phi$ sends each $A_i$ to a unique element. Indeed, suppose that $g_i = g_j$ for some i, j. This implies that either the top row of the Left diagram or the bottom row of the Right diagram cannot split, since these rows have the conjugation relation, which is a contradiction to the above. Since $g_{i_0}^2\neq g_{j_0}^2$ by assumption, Remark 2.2 shows the each $g_i^2$ is unique. By Lemma 4.3, the set $\left\{ g_1^2, g_2^2, \ldots, g_n^2 \right\}$ is a totally symmetric set of size n. In this case, Proposition 2.5 will yield that

\begin{equation*}\lvert \phi(wB_n)\rvert \geq 2^{n-1}(n)!.\end{equation*}

Case 3: Suppose $\phi$ does not split any of the $A_i$ ’s, and $\phi(A_i)^2=\phi(A_j)^2$ for all i and j. Notice that Lemma 4.3 does not apply, and that none of the $A_i$ ’s are split by $\phi$ . In this case, we use the fact that $\phi$ restricted to $B_n$ is non-cyclic. Applying Theorem 1.1 to $\phi$ restricted to $B_n$ , we get

\begin{equation*} \lvert \phi(wB_n)\rvert \geq \lvert \phi(B_n)\rvert \geq \left( \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!.\end{equation*}

5.2. Proof of Theorem 1.3

In this section, we provide a proof for Theorem 1.3 which gives a lower bound on the size of $vB_n$ ’s smallest non-cyclic finite quotient. By considering whether or not a homomorphism factors through $wB_n$ , we may apply our classification of homomorphisms from $wB_n \to G$ , or the necessary condition for the existence of a homomorphism $B_n \to G$ , to determine a classification of the size of finite images of homomorphisms from $vB_n$ .

1. (1) $\phi$ is abelian.

2. (2) $\phi$ factors through $wB_n$ , and either

   1. a. $\phi$ restricted to $PwB_n$ is cyclic.

   2. b. $\lvert \phi(vB_n)\rvert \geq 2^{n-2}(n-1)!$

   3. c. For all i and j, $\phi$ does not split $A_i$ , $\phi(A_i)^2\neq\phi(A_j)^2$ , and

      \begin{equation*} \lvert \phi(vB_n)\rvert \geq \left( \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!.\end{equation*}

3. (3) $\phi$ does not factor through $wB_n$ and

   \begin{equation*} \lvert \phi(vB_n) \rvert \geq \left( \left\lfloor \frac{n-1}{2} \right\rfloor +1\right)\left(3^{\lfloor\frac{n}{2}\rfloor - 1}\right)\left\lfloor \frac{n}{2} \right\rfloor!.\end{equation*}

Proof of Theorem 1.3. Suppose $\phi$ is not abelian. If $\phi$ factors through $wB_n$ , then by Theorem 1.2, one of either (2)(a), (2)(b), or (2)(c) must be true. If $\phi$ does not factor through $wB_n$ and $\phi$ is non-abelian, Lemma 4.2 gives that $\phi$ restricted to $B_n$ is non-abelian, and hence non-cyclic. Applying Theorem 1.1 to $\phi$ restricted to $B_n$ gives that (3) must be true.