TITLE: Finite abelian subgroups of $GL(2,\mathbb C)$ without Pseudoreflection
QUESTION [1 upvotes]: Let $G$ be a finite abelian subgroup of $GL(2,\mathbb C)$ such that $A-I_2$ does not have rank $1$ for all $A\in G.$ Then, is it true that $G$ is cyclic?

REPLY [3 votes]: Yes. Note that this induces a representation $G \to GL(2, \mathbb{C})$. Since $G$ is abelian, this is reducible, and the corresponding $\mathbb{C}G$-module can be written as $V \oplus W$ where $V$ and $W$ are one-dimensional.
Since you assumed $A - I_2$ does not have rank 1 for all $A \in G$, we must have that the maps $G \to \mathbb{C}$ are faithful, which is only possible if $G$ is cyclic.


TITLE: Find the values of the parameters for which the matrices commute
QUESTION [3 upvotes]: I have two matrices:
  $A = \left( \begin{array}{cc}
1 & a \\
b & 6
\end{array} \right)
%
\ \ \ \ B = \left( \begin{array}{cc}
4 & c \\
d & 2
\end{array} \right)$.
  How can I find the values of the parameters $a, b, c, d$ for which the matrices will commute?

I tried calculate $AB$ and $BA$ and compare the matrices elements, but that is not enough to get the result.
I think this have simple solution, but I have difficulties with it.

REPLY [2 votes]: We have
$$AB=\begin{pmatrix}
1&a\\
b&6\\
\end{pmatrix}\begin{pmatrix}
4&c\\
d&2\\
\end{pmatrix}=\begin{pmatrix}
4+ad&c+2a\\
4b+6d&bc+12
\end{pmatrix}$$
and
$$BA=\begin{pmatrix}
4&c\\
d&2\\
\end{pmatrix}\begin{pmatrix}
1&a\\
b&6\\
\end{pmatrix}=\begin{pmatrix}
4+bc&4a+6c\\
d+2b&ad+12
\end{pmatrix}.$$
Hence
\begin{eqnarray}
ad&=&bc\\
0&=&2a+5c\\
2b+5d&=&0\\
\end{eqnarray}
Thus $a=\frac{-5}{2}c$ and $b=\frac{-5}{2}d$ and thus $\frac{-5}{2}cd=\frac{-5}{2}dc$ is automatic. In other words for all $c,d$ and $a=\frac{-5}{2}c$ and $b=\frac{-5}{2}d$ the matrices commute.
Although the equations you get are not linear, a solution to the linear second and third equation automatically gives a solution to the non-linear equation! That makes this easy.


TITLE: dimension of a symmetric matrix with trace zero
QUESTION [2 upvotes]: what will be the dimension of symmetric matrix of order $n\times n(n\geq2)$ with real entries and trace is equal to zero?
The answer is given as :$\frac{n^{2}+n}{2}-1$
can anyone explain how will get this answer?
according to me is should be:$\frac{n^{2}-n}{2}-1$

REPLY [0 votes]: If we define $T:$ your favorite subspace of square matrices$\to\mathbb R$ by $T(M)=$trace $(M)$,  then
$\dim\ker T+\dim im T= \dim$(your favorite subspace of square matrices).
So for zero trace subspace we have $\dim(YFSS)=\dim(YFSS)-1$.


text:
\begin{document}

\prelimpages

\Title{Eigenvalue fluctuations for random regular graphs}
\Author{Tobias Lee Johnson}
\Year{2014}
\Program{Department of Mathematics}

\Chair{Ioana Dumitriu}{Associate Professor}{Department of Mathematics}
\Chair{Soumik Pal}{Associate Professor}{Department of Mathematics}
\Signature{Sara Billey}

\copyrightpage

{\Degreetext{A dissertation
  submitted in partial fulfillment of the\\ requirements for the degree of}

 \titlepage
 }
\setcounter{footnote}{0}

\setcounter{page}{-1}
\abstract{
One of the major themes of random matrix theory is that many asymptotic properties
of traditionally studied distributions of random matrices are \emph{universal}.
We probe the edges of universality by studying the spectral properties of random regular graphs.
Specifically, we prove limit theorems for the fluctuations
of linear spectral statistics of random regular graphs. We find both universal and non-universal behavior.
Our most important tool is Stein's method for Poisson approximation, which we develop for use
on random regular graphs.
}

\tableofcontents

\acknowledgments{

  I owe a debt to many people for their help and friendship.
  I thank Ioana for introducing me to research and to random regular graphs.
  I'm grateful to her for supporting me and for not taking my curmudgeonliness
  too seriously. I'm grateful to Soumik for teaching me probability and shaping my view
  of it. His suggestion that I learn how to use Stein's method and try to apply it
  to cycle counts of random regular graphs has shaped my mathematical career more than I
  would ever have  imagined. I also thank Chris Hoffman, who has been
  a third advisor to me this year, introducing me to new areas of math and collaborating with me.
  I can't thank Elliot Paquette and Matt Junge enough. It's been wonderful to have them as friends
  and collaborators. Most of all,
  I thank my friends and my wife Lindsay.

}

\textpages

\chapter {Introduction}
\input{Tintro.tex}

\chapter{Poisson approximation for cycle counts in random regular graphs}
\label{chap:poisson}
\input{Tpoisson.tex}

\chapter{Fluctuations of linear eigenvalue statistics}
\label{chap:fluctuations}

\section{Fluctuations for random regular graphs: main results}
\label{sec:flucresults}
\input{Tflucresults.tex}
\section{Proof of eigenvalue fluctuation results}
\label{sec:eigenfluc}
\input{Teigenfluc.tex}

\chapter{Minor processes and the Gaussian free field}
\label{chap:GFF}
\input{TPFFintro.tex}
\input{TPFF.tex}
\subsection{Diagonal convergence}
\input{Tdiagonal.tex}
\section{Convergence to the Gaussian free field}
\label{sec:GFFconvergence}
\input{TGFF.tex}

\bibliographystyle{alpha}
\bibliography{thesis}

\appendix
\raggedbottom\sloppy

\end{document}

config: arxiv


TITLE: Matrices whose Linear Combinations are All Singular
QUESTION [10 upvotes]: I'd like to know if the following problem of elementary linear algebra is already solved / solvable.
For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$?

REPLY [2 votes]: This is not a complete answer, just some sufficient and some necessary conditions.
Let's look at the opposite condition. Fix $u,v\in\mathrm{End}_k(V)$ two singular (non proportional) endomorphisms of the finite dimensional $k$-vector space $V$: what conditions are necessary or sufficient for there to be a scalar $\lambda$ such that $u+\lambda v\in\mathrm{GL}(V)$?
The following conditions are necessary:

$\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)$;
$\mathrm{Im}(u)+\mathrm{Im}(u)=V$;
$v\big(\mathrm{Ker}(u)\big)\oplus u\big(\mathrm{Ker}(v)\big)$.

The first is @chaochuang's answer, and the second is @Marc van Leeuwen' comment. To see the necessity of the third condition, write down matrices in a basis adapted to $V=\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)\oplus S$.
The following conditions are sufficient (at least when $k$ is big enough, for instance if $\mathbb Q\subset k$):

The induced morphism $\tilde{v}:\mathrm{Ker}(u)\rightarrow V\rightarrow V/\mathrm{Im}(u)$ is an isomorphism i.e. $v(\mathrm{Ker}(u))\oplus\mathrm{Im}(u)$;
The induced morphism $\tilde{u}:\mathrm{Ker}(v)\rightarrow V\rightarrow V/\mathrm{Im}(v)$ is an isomorphism i.e. $u(\mathrm{Ker}(v))\oplus\mathrm{Im}(v)$.

To see the sufficiency of the first point, write down the matrix for $u+\lambda v$ where the initial base of $V$ is adapted to $S\oplus \mathrm{Ker}(u)=V$, and the final base is the images under $u$ of the basis vectors chosen in $S$ and the images  under $v$ of the basis vectors in $\mathrm{Ker}(u)$, and let $\lambda\rightarrow 0$: the determinant of this matrix is a polynomial in $\lambda$ of the form $\lambda^d(1+\cdots)$ where $d=\dim~\mathrm{Ker}(u)$.
Going back to the initial problem, we get $3$ sufficient conditions for the span of $u$ and $v$ to be completely singular, and $2$ necessary conditions.


text:
\section{Discussion and Outlook} \label{sec:outlook}

We presented a particle method for simulating solutions of FPEs governing the temporal evolution of the probability density for
stochastic dynamical systems of the diffusion type.
By reframing the FPE in a Liouville form,
we obtained an effective dynamics in terms of independent deterministic particle trajectories. Unfortunately, this formulation requires the knowledge of the gradient of the logarithm of the instantaneous probability density of the system state, which is the quantity we try to compute. We circumvented this complication by introducing statistical estimators for the gradient--log--density based on a variational formulation. To combine high flexibility of estimators with computational
efficiency, we employed kernel based estimation together with an additional sparse approximation. For the case of equilibrium systems, we related our framework to
Stein Variational Gradient Descent, a particle based dynamics to approximate
the stationary density, and to a geometric formulation of Fokker--Planck dynamics.
We further discussed extensions of our method to settings with multiplicative noise and to second order Langevin dynamics.

 To demonstrate the performance of our framework, we provided detailed tests
 and comparisons with stochastic simulations and analytic solutions (when possible). We demonstrated the accuracy of our method on conservative and non-conservative model systems
 with different dimensionalities.  In particular, we found, that our framework outperforms stochastic simulations both in linear and nonlinear settings, by delivering more accurate densities for small particle number when the dimensionality is small enough. For increasing particle number, the accuracy of both approaches converges. Yet, our deterministic framework delivered \emph{consistently} results with smaller variance among individual repetitions.
 Furthermore, we showed that our method, even for small particle numbers,
 exhibits low order cumulant trajectories with significantly less temporal fluctuations  when compared against to stochastic simulations of the same particle number.

 We envisage several ways to improve and extend our method.
There is room for improvement by optimising hyper parameters of our algorithm such as
inducing point position and kernel length scale.
Current grid based and uniform random selection of inducing point position may contribute to
the deterioration of solution accuracy in higher dimensions. Other methods,
such as subsampling or clustering of particle positions may lead to further improvements.
On the other hand, a hyper parameter update may not be at all necessary
at each time step in certain settings, such that a further speedup of our algorithm could be achieved.

The implementation of our method depends on the function class chosen to represent the estimator. In this paper we have focused on linear representations, leading to simple
closed form expressions. It would be interesting to see if
other, nonlinear parametric models, such as neural networks,
(see e.g.~\cite{deepnn}) could be employed to represent estimators.
While, in this setting, there would be no closed form solutions, the small changes
in estimates between successive time steps, suggest that
only a few updates of numerical optimisation may be necessary
at each step. Moreover, the ability of neural networks
to automatically learn relevant features from data might help to improve
performance for higher dimensional problems when
particle motion is typically restricted on lower dimensional submanifolds.

From a theoretical point of view, rigorous results on the
accuracy of the particle approximation would be important. These would
depend on the speed of convergence of estimators towards exact gradients
of log--densities. However, to obtain such results may not be easy.
While rates of convergence for kernel based estimators
have been studied in the literature, the methods for proofs usually
rely on the independence of samples and would not necessarily apply to the
case of interacting particles.

We have so far addressed only the forward simulation of FPEs. However, preliminary results indicate that related techniques may be applied to particle based simulations for smoothing (forward--backward) and related control problems for diffusion processes~\cite{reich2015probabilistic}. Such problems involve computations of an effective, controlled drift function in terms of gradient--log--densities. We defer further details and discussions on subsequent publications on the topic.

Taken together, the main advantage of our framework is its minimal requirement in simulated particle trajectories for attaining reliable Fokker--Planck solutions with smoothly evolving transient statistics. Moreover, our proposed method is nearly effortless to set up when compared to classical grid based FPE solvers, while it delivers more reliable results than direct stochastic simulations.


TITLE: Quaternions and spatial translations
QUESTION [5 upvotes]: From my understanding, in spatial applications (3D rendering, games and similar applications) quaternions can only be used to describe rotations/orientations and not translations (like a transformation matrix does).
This seems to be backed up by the fact that most 3D frameworks that use quaternions (OGRE3D for example), use quaternions together with vectors to describe an orientation and translation (which in other frameworks would only need a single transformation matrix).

Did I understand it correctly?
Why can't they describe translations together with rotations?

An argument that I am having trouble countering is: you can describe a translation as a series of rotations around multiple axes and combine those rotations in a single quaternion.
Further research showed me that double quaternions can be used to describe rotation and translation, but I am interested in "single" quaternions.

REPLY [1 votes]: Translations are rotations around axes in the plane at infinity. They can be described by means of unit quaternions once you define such a rotation as $[1, \epsilon \vec{t} ]$ where $\epsilon$ has the property $\epsilon^2=0$.
$\epsilon$ is an infinitesimal quantity that in combination with an axis at infinity gives a finite translation along a straight line.
You can also translate through infinity by means of the unit quaternion $[\epsilon t, \hat{t} ]$, where $t$ is the length of $\vec{t}$ and $\hat{t} = \vec{t} / t$.


TITLE: An integral relating to $\delta$-function
QUESTION [1 upvotes]: How can we prove that
$$
\int_0^{2\pi}\delta(a+b\cos x)dx=2\frac{\theta\big(b^2-a^2\big)}{\sqrt{b^2-a^2}}
$$
where $\delta$ and $\theta$ are Dirac $\delta$ function and Heaviside $\theta$ function respectively.

REPLY [3 votes]: The Heaviside $\theta$ can be better understood if we think of the function inside the $\delta$-function.

Integrating a $\delta$-function only sees the points where the argument of the $\delta$-function is $0$. The above plot is a picture of the argument of the $\delta$-function in the case of your integral. It is easy to see that this splits into three cases.
Case 1: $(a+b)(a-b) <0$. In this case, we have that either $(a+b)$ or $(a-b)$ is a negative quantity with the other one positive. It is clear that we are in the case depicted above; the cosine curve crosses the axis. In this case, there are two places where the function crosses $0$, so the integral only sees those two crossing points. Now, let $c$ and $d$ be the two zeros and let $f(x) = a + b\cos(x)$. We have that
$$
   \int_0^{2\pi} \delta(a+b\cos x) \,dx = \int_{c-\epsilon}^{c+\epsilon} \delta(f(x))\,dx + \int_{d-\epsilon}^{d+\epsilon} \delta(f(x))\,dx.
$$
Here, we truncate the integral to be around just the zeros, since these points are the only ones that matter for this $\delta$ function. Now, by zooming in really close (i.e. making $\epsilon$ small), we can replace $f(x)$ by its linear approximation. Since $f(c) = 0$, this means we can replace $f(x)$ with $f'(c)(x-c)$ in the first integral:
$$
   \int_{c-\epsilon}^{c+\epsilon} \delta(f(x)) \,dx = \int_{c-\epsilon}^{c+\epsilon} \delta(f'(c) (x-c))\,dx = \frac{1}{|f'(c)|} \int_{-\epsilon}^\epsilon \delta(u) \,du = \frac{1}{|f'(c)|}.
$$
Here, we did the substitution $u = |f'(c)| (x-c)$.
Now,
$$
  f'(c) = -b\sin(c) = -b \sqrt{1-\cos^2(c)} = -b\sqrt{1 - \frac{a^2}{b^2}} = - \sqrt{b^2-a^2},
$$
since $\cos(c) = -b/a$ as $c$ was a zero of $f$. So, the first integral gives a factor of $\displaystyle \frac{1}{\sqrt{b^2-a^2}}$. The integral over the second zero gives the exact same factor by symmetry. So, we get
$$ \int_0^{2\pi} \delta(a+b\cos x) \,dx = 2 \frac{1}{\sqrt{b^2-a^2}}.$$
When $(a-b)(a+b) >0$, there are no zeros, so the integral is 0.
The first case is $b^2-a^2 > 0$, and the second case is $b^2-a^2 <0$. Thus, by definition of the heaviside $\theta$, we get the desired result when $b^2-a^2 \neq 0$:
$$
   \int_0^{2\pi} \delta(a+\cos(x))\,dx = 2\frac{\theta(b^2-a^2)}{\sqrt{b^2-a^2}}.
$$
Lastly, we must consider the case when $b^2-a^2=0$. By examining the formula, we see that if $b^2-a^2 =0$, we will get division by $0$ in the expression on the right. So, we expect to get that the integral diverges. Indeed, when $b^2-a^2=0$, we get either $a=b$ or $a=-b$. If $a=b$, the same argument above gives
$$
   \int_0^2\pi \delta(a+b\cos(x))\,dx = \int_0^{2\pi} \delta(f'(\pi) (x-\pi))\,dx = \int_0^{2\pi} \delta(0) \,dx = \infty
$$
as $f'(\pi) = 0$ and so we are integrating $\infty$ over an interval of finite length. Similarly, if $a=-b$, then $f(x) = a - a\cos(x)$, which will have two zeros at $x=0$ and $x=2\pi$. In this case, we can extend the interval of integration to be $(-\epsilon, 2\pi + \epsilon)$ and use the same trick as before; both $f'(0)$ and $f'(\pi)$ are zero, so both integrals around the two zeros will be infinite.
(Note: this is a little bit unrigorous, as there is nothing saying we have to push the interval out; we could just as well push the interval in and get 0 for the interval. I don't have a good reason for this, but maybe someone else can shed some light on this.)


TITLE: Continuous bijection between $[0, 1)$ and $\mathbb{R}$
QUESTION [2 upvotes]: I have to prove the non-existence of a continuous bijection between $[0,1)$ and $\mathbb{R}$.
My attempt:
Since $\mathbb{R}$ is homeomorphic to $(0, 1)$, I tried to prove the non-existence of a continuous bijection $g: [0, 1) \rightarrow (0, 1)$. So, assume by contradiction that such a map exists. I wanted to find a contradiction using the Intermediate Value Theorem, but without success.
Any suggestions? Thanks in advance!

REPLY [6 votes]: Suppose $f:[0,1) \to \Bbb R$ is a continuous bijection.
Then $f[(0,1)]=\Bbb R \setminus \{f(0)\}$ which is a contradiction, as $(0,1)$ is connected and so its continuous image is too, while $\Bbb R\setminus \{p\}$ is disconnected for any $p \in \Bbb R$.

REPLY [3 votes]: If a map $f:[0,1)\to (0,1)$ satifies the intermediate value property, then $f\left[(0,1)\right]=I$ is a subinterval of $(0,1)$. If $f$ is bijective, then $I=(0,1)\setminus\{f(1)\}$. However, $(0,1)\setminus \{c\}$ is never an interval forany $c\in (0,1)$.


TITLE: Single variable justification for the multivariate chain rule.
QUESTION [2 upvotes]: I $\def\d{\mathrm d}\def\p{\partial}$am going to ask everyone to switch their paradigms to that of the real line. I am looking for a "lowbrow" explanation of the following phenomena.
I am talking about the popular method of differentiating functions like $x^x$. First, use the ordinary power rule, then use the exponential rule and then add them together.
$$(x^x)' = x\cdot x^{x-1} + x^x\log(x) = x^x(1+\log(x))$$
Although poorly justified, the result and the method is absolutely correct. From the multivariable chain rule, it can be said that if $y$ depends on functions $f$ and $g$ then
$$\frac{\d y}{\d x} = \frac{\p y}{\p f}\! \frac{\d f}{\d x} + \frac{\p y}{\p g}\! \frac{\d g}{\d x}$$
In our instance, we had $y = f(x)^{g(x)}$ where $f(x) = x$ and $g(x) = x$.
This phenomena can provide simple justification for other rules of differentiation such as the product rule. Assuming this rule, even the generalized product rule is immediately obvious.
I am asking for a way to justify the following statement without explicitly using multivariate calculus beyond partial derivatives.

Theorem: If $f(x)$ can be written explicitly with $n$ instances of $x$ then one may label each of the $x$'s from $x_1\!$ to $x_n$ and then compute $f'(x)$ as follows.
  $$f'(x) = \sum_{k=1}^{n} \frac{\p f}{\p x_k}$$
  And then setting each of the $x_j = x$.

The proof of this need not be formal but I am not looking for anything that treats differentials as fractions. Does anyone have any ideas?

REPLY [2 votes]: Denote $f$ with "$n$ instances labeled" as $F(x_1, \ldots, x_n)$. Hence we have $F(x,\ldots, x) = f(x)$. Then
\begin{align*}
  f(x+h) - f(x) &= F(x+h, \ldots, x+h) - F(x,\ldots, x)\\
                &= \sum_{k=1}^n
                   F(\underbrace{x+h, \ldots, x+h}_{k}, x,\ldots, x) - F(\underbrace{x+h, \ldots, x+h}_{k-1}, x,\ldots, x)
\end{align*}
Hence
$$ \frac{f(x+h) - f(x)}h = \sum_{k=1}^n \frac{ F(\overbrace{x+h, \ldots, x+h}^{k}, x,\ldots, x) - F(\overbrace{x+h, \ldots, x+h}^{k-1}, x,\ldots, x)}h $$
If now $F$ is continuously differentiable, the result follows for $h \to 0$.


TITLE: Bisection Method for intersection of two functions
QUESTION [1 upvotes]: I know how to use the bisection method when finding roots, however I don't know how to use it for when two lines intersect, any help with this would be much appreciated

REPLY [5 votes]: To determine where any two curves $y=f(x)$ and $y=g(x)$ intersect (and lines are considered 'curves' for this purpose), simply set $f(x)=g(x)$.  The reason this works, is that you are looking for pairs $(x,y)$ that satisfy both equations simultaneously, so to ensure the $y$-coordinates are the same, implies that $f(x)=y=g(x)$.  Now rewrite $f(x)=g(x)$ as $f(x)-g(x)=0$, and your question boils down to finding the root(s) of the function $f-g$.


text:
\begin{document}
\title[On the order of Borel subgroups of group amalgams]{On the order of Borel subgroups of group
amalgams and an application to locally-transitive graphs}

\author[L. Morgan, P. Spiga, G. Verret]{Luke Morgan, Pablo Spiga, Gabriel Verret}

\address{Luke Morgan,  School of Mathematics and Statistics,
\newline\indent University of Western Australia, 35 Stirling Highway, Crawley, WA 6009, Australia.}
\email{luke.morgan@uwa.edu.au}

\address{Pablo Spiga,  Dipartimento di Matematica Pura e Applicata,
\newline\indent University of Milano-Bicocca, Milano, 20125 Via Cozzi 55, Italy.}
\email{pablo.spiga@unimib.it}

\address{Gabriel Verret,  School of Mathematics and Statistics,
\newline\indent University of Western Australia, 35 Stirling Highway, Crawley, WA 6009, Australia.
\newline\indent Also affiliated with: FAMNIT, University of Primorska,
\newline\indent Glagolja\v{s}ka 8, SI-6000 Koper, Slovenia.}
\email{gabriel.verret@uwa.edu.au}

\thanks{The research of the first author is supported by the Australian Research Council grant
DP120100446. The last author is supported by UWA as part of the ARC grant DE130101001.}

\subjclass[2010]{Primary 20B25; Secondary 05E18}
\keywords{locally-restrictive, locally-transitive graph, semiprimitive, group amalgams}

\begin{abstract}
A permutation group is called \emph{semiprimitive} if each of its normal subgroups is either
transitive or semiregular. Given nontrivial finite transitive permutation groups $L_1$ and $L_2$
with $L_1$ not semiprimitive, we construct an infinite family of rank two amalgams of permutation
type $[L_1,L_2]$ and Borel subgroups of strictly increasing order. As an application, we show that
there is no bound on the order of edge-stabilisers in locally $[L_1,L_2]$ graphs.

We also consider the corresponding question for amalgams of rank $k\geq 3$. We completely resolve
this by showing that the order of the Borel subgroup is bounded by the permutation type
$[L_1,\dots,L_k]$ only in the trivial case where each of $L_1,\dots,L_k$ is regular.
\end{abstract}

\maketitle

\section{Introduction}
All graphs in this paper  are connected, simple and locally finite. Let $\Gamma$ be a graph, let $v$
be a vertex of $\Gamma$ and let $G$ be a group of automorphisms of $\Gamma$. We denote by
$\Gamma(v)$ the neighbourhood of $v$, by $G_v$ the stabiliser of $v$ in $G$, and by
$G_v^{\Gamma(v)}$ the permutation group induced by $G_v$ on $\Gamma(v)$. We say that $\Gamma$ is
$G$-locally-transitive if $G_v^{\Gamma(v)}$ is transitive for every vertex $v$ of $\Gamma$. (This is
easily seen to imply that $G$ is transitive on the edges of $\Gamma$.)

The starting point for our investigations is a classical result of Goldschmidt~\cite{Goldschmidt}, a
consequence of which states that in a finite $G$-locally-transitive graph of valency three, the
edge-stabilisers have order dividing $128$. Inspired by this result, we introduce the following
terminology.

Let $L_1$ and $L_2$ be finite transitive permutation groups, let $[L_1,L_2]$ denote the multiset
containing $L_1$ and $L_2$  and let $\Gamma$ be a  $G$-locally-transitive graph. We say that
$(\Gamma,G)$ is \emph{locally} $[L_1,L_2]$ if, for some edge $\{u,v\}$ of $\Gamma$, we have
permutation isomorphisms $G_u^{\Gamma(u)}\cong L_1$ and $G_v^{\Gamma(v)}\cong L_2$.

\begin{definition}
The multiset $[L_1,L_2]$ is  \emph{locally-restrictive} if there exists a constant $c \in \mathbb N$
such that, if $\Gamma$ is a finite  $G$-locally-transitive graph with $(\Gamma,G)$ locally
$[L_1,L_2]$ and $\{u,v\}$ is an edge of $\Gamma$, then $|G_{uv}|\leq c$.
\end{definition}

With this terminology, Goldschmidt's result implies that, if $L_1$ and $L_2$ are transitive
permutation groups of degree three then $[L_1,L_2]$ is locally-restrictive. A related conjecture of
Goldschmidt-Sims states that  if $L_1$ and $L_2$ are both primitive permutation groups then
$[L_1,L_2]$ is locally-restrictive. Whilst there has been some progress on the Goldschmidt-Sims
Conjecture (see \cite{fan,fan2,glasner,morgan,rowleytrees,stelledgetrans}), it remains open.
Although the focus of the Goldschmidt-Sims Conjecture is on primitive permutation groups, it is
still possible for $[L_1,L_2]$ to be locally-restrictive even when neither $L_1$ nor $L_2$ is
primitive. For example, it is easy to see that if $L_1$ and $L_2$ are both regular permutation
groups then $[L_1,L_2]$ is locally-restrictive. We therefore pose the following problem.

\begin{problem}\label{problem:main}
Determine which pairs of finite transitive permutation groups are locally-restrictive.
\end{problem}

Our main result, Theorem~\ref{theo:graph}, is a significant step towards solving
Problem~\ref{problem:main}.

\begin{theorem}
\label{theo:graph}
Let $L_1$ and $L_2$ be nontrivial finite transitive permutation groups. If one of $L_1$ or $L_2$ is
not semiprimitive then $[L_1,L_2]$ is not locally-restrictive.
\end{theorem}

(A permutation group is called \emph{semiregular} if the identity is the only element of the group
that fixes a point and  \emph{semiprimitive} if each of its normal subgroups is either transitive or
semiregular.) In view of Theorem~\ref{theo:graph}, we are naturally led to pose the following
question, the answer to which we believe to be positive.

\begin{quest}
\label{question}
Does the converse of Theorem~$\ref{theo:graph}$ hold? In other words, if $L_1$ and $L_2$ are finite
transitive semiprimitive permutation groups, is $[L_1,L_2]$ locally-restrictive?
\end{quest}

Our notion of locally-restrictive is to locally-transitive graphs what the notion of
graph-restrictive (see \cite[Definition 2]{PSVRestrictive}) is to arc-transitive graphs. Many of the
concepts and results we have discussed so far have well-known analogues in the arc-transitive case.
For example, Goldschmidt's Theorem can be seen as the locally-transitive version of Tutte's famous
result on arc-transitive graphs of valency three~\cite{Tutte,Tutte2}. Similarly,  the
Goldschmidt-Sims Conjecture corresponds to the long-standing Weiss Conjecture~\cite{Weiss}
 which asserts that primitive groups are graph-restrictive. The recent Poto\v{c}nik-Spiga-Verret
Conjecture \cite[Conjecture~3]{PSVRestrictive}  asserts that a permutation group is
graph-restrictive if and only if it is semiprimitive. Remarkable evidence towards this conjecture
can be found in~\cite{PabloGab}, where the intransitive case is dealt with. For recent progress on
the transitive case,  see \cite{giudicimorgan,giudicimorgan2,PPSS,PraegerSV,Pablo}.

We remark that an affirmative answer to Question~\ref{question} would show the validity of both the
Weiss and Poto\v{c}nik-Spiga-Verret  Conjectures. In fact, these conjectures can easily be rephrased
using our terminology. Indeed, if $(\Gamma,G)$ is locally $[L,\mathrm C_2]$ (where $C_2$ denotes the
cyclic group of order $2$)
then $ \Gamma $ is simply the barycentric subdivision of a graph $\tilde \Gamma$ on which $G$  acts
faithfully and arc-transitively and such that $(\tilde \Gamma,G)$ is locally $[L,L]$. There is an obvious converse to this procedure,  thus $[L,\mathrm C_2]$ is locally-restrictive if and only if $L$ is graph-restrictive.

The theory of groups acting on trees due to Bass-Serre allows us to interpret
Question~\ref{question}  in terms of  locally-transitive discrete subgroups of the automorphism
group of a bi-regular tree. This equivalence will be proved in Section~\ref{sec:graph}. For now, let
us simply point out that, under this equivalence,  Theorem~\ref{theo:graph} is equivalent to
Theorem~\ref{cor:discsubgroups} below.

\begin{theorem}
\label{cor:discsubgroups}
Let $L_1$ and $L_2$ be nontrivial finite transitive permutation groups and let $\mathfrak{T}$ be the
bi-regular tree with valencies the degrees of $L_1$ and $L_2$. If one of $L_1$ and $L_2$ is not
semiprimitive then, for every integer $c$, there exists a group $G$ of automorphisms of
$\mathfrak{T}$ such that $(\mathfrak{T},G)$ is locally $[L_1,L_2]$  and $c\leq |G_{uv}|<\infty$ for
some edge $\{u,v\}$ of $\mathfrak{T}$.
\end{theorem}

 Theorem ~\ref{cor:discsubgroups} is a significant improvement of  \cite[(7.14)]{basskulkarni} which
shows that the automorphism group of a bi-regular tree with composite valencies contains strictly
ascending chains of locally-transitive discrete subgroups, but with no control over the local
permutation groups.

Question~\ref{question} also has a natural formulation in terms of group amalgams of rank two.
Before presenting our result using this language, we first define amalgams, following~\cite{ivsp}.

\begin{definition}
Let $k\geq 2$. A \emph{rank $k$ amalgam} is a finite set  $\mathcal A$  together with a set of $k$ subsets
$P_1,\ldots,P_k$, where each $P_i$ forms a group, $\bigcup_{i=1}^k P_i=\mathcal A$, $\bigcap_{i=1}^k
P_i\neq\emptyset$ and, for every  $i,j\in\{1,\ldots,k\}$ the group operations defined on $P_i$ and
$P_j$ coincide when restricted to $P_i \cap P_j$.

The \emph{Borel subgroup}  of $\mathcal A$  is $\bigcap_{i=1}^k P_i $ and is denoted
$\mathcal{B}(\mathcal A)$. If there is no nontrivial subgroup of $\mathcal{B}(\mathcal A)$ that is
normalised by each of $P_1,\ldots,P_k$ then we say that $\mathcal A$ is \emph{faithful}. The
\emph{permutation type} of $\mathcal A$  is the multiset $[L_1,\dots,L_k]$ where $L_i$ is the
permutation group induced by $P_i$ in its action on the right cosets of $\mathcal{B}(\mathcal A)$ in
$P_i$.
\end{definition}

In Section~\ref{sec:graph} we will show the equivalence between Theorem~\ref{theo:graph} and
Theorem~\ref{theo:main} below.

\begin{theorem}\label{theo:main}
Let $L_1$ and $L_2$ be nontrivial finite transitive permutation groups. If one of $L_1$ and $L_2$ is
not semiprimitive then, for every integer $c$, there exists a rank two faithful amalgam of
permutation type $[L_1,L_2]$ with Borel subgroup of order at least $c$.
\end{theorem}

Our proof of Theorem~\ref{theo:main} is constructive and can be found in Section~\ref{Sec:ResAreSP}.
The construction used in the proof is a generalisation of the construction that appeared in
\cite[Section~4]{PSVRestrictive} (in fact, Theorem~\ref{theo:graph} generalises \cite[Theorem
4]{PSVRestrictive}) which in turn was inspired by the so-called wreath extension construction
\cite[Chapter IV, 8.1]{malle}. A precursory idea to this construction can also be traced
to~\cite[Section~$4$]{Pablo1}.

Theorem~\ref{theo:main} naturally leads one to wonder about the corresponding statement for amalgams
of rank greater than two. Here is the complete answer in this case.

\begin{theorem}\label{theo:mainmain}
Let $k\geq 3$ and let $L_1,\ldots,L_k$ be nontrivial finite transitive permutation groups. The
following are equivalent:
\begin{enumerate}
\item One of $L_1,\ldots,L_k$ is not regular.
\item For every integer $c$, there exists a rank $k$ faithful amalgam of permutation type
$[L_1,\ldots,L_k]$  with Borel subgroup of order at least $c$.
\end{enumerate}
\end{theorem}

In fact, it is easy to see that if $L_1,\ldots,L_k$ are regular then a faithful amalgam $\mathcal A$
of permutation type $[L_1,\ldots,L_k]$ must have $\mathcal{B}(\mathcal A)=1$ (see
Section~\ref{sec4}). The real meat of Theorem~\ref{theo:mainmain} is therefore the statement that
for rank at least three, these trivial examples are the only ones which admit upper bounds on
$|\mathcal{B}(\mathcal A)|$ depending  upon the permutation type alone. This is in sharp contrast
with the situation in the rank two case; Goldschmidt's result has the highly  nontrivial consequence
that a rank two  faithful amalgam of permutation type $[\Sym(3),\Sym(3)]$ has Borel subgroup of
order at most $128$. In particular, the na\"{i}ve ``$k=2$'' version of Theorem~\ref{theo:mainmain}
is false. We find the relative simplicity of the higher rank case  rather surprising.  Note that
under additional assumptions, the order of the Borel subgroup of a rank three amalgam can be
bounded, see for example~\cite{stelltime}.

\section{Equivalence of Theorems~\ref{theo:graph},~\ref{cor:discsubgroups} and~\ref{theo:main}}
\label{sec:graph}

\begin{lemma}\label{newLemma}
Let $L_1$ and $L_2$ be finite transitive permutation groups and let $B$ be a finite group. The
following are equivalent:
\begin{enumerate}
\item There exists a rank two faithful amalgam of permutation type $[L_1,L_2]$ with Borel subgroup
$B$.
\item There exists a locally $[L_1,L_2]$ pair $(\mathfrak{T},G)$ such that $\mathfrak{T}$ is an
infinite tree and $G_{uv}=B$ for some edge $\{u,v\}$ of $\mathfrak{T}$.
\item There exists a locally $[L_1,L_2]$ pair $(\Gamma,H)$ such that $\Gamma$ is finite and
$H_{uv}=B$ for some edge $\{u,v\}$ of $\Gamma$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)\Longrightarrow  (2):$ Let $\mathcal A$ be a rank two faithful amalgam of permutation type
$[L_1,L_2]$ with Borel subgroup $B$. Let $P_1$ and $P_2$ be the two groups involved in $\mathcal A$
and let $G = P_1 *_B P_2$ (that is, $G$ is the free product of $P_1$ and $P_2$ amalgamated over
$B$). By \cite[I.4.1, Theorem~7]{Serre}, there exists an infinite tree $\mathfrak T$ on which $G$
acts faithfully, edge- but not vertex-transitively, and an edge $\{u,v\}$ of $\mathfrak T$ such that
$G_{uv}=B$, $G_u=P_1$ and $G_v=P_2$. As $\mathfrak T$ is $G$-edge- but not $G$-vertex-transitive, it
must be $G$-locally-transitive. Since $\mathcal A$ is of permutation type $[L_1,L_2]$, it follows
that $(\mathfrak T, G)$ is locally $[L_1,L_2]$.

$(2)\Longrightarrow  (3):$ Let $G'$ be the largest subgroup of $G$ that preserves the bipartition of
$\mathfrak{T}$. Note that $(\mathfrak{T},G')$ is locally $[L_1,L_2]$ and $G'_{uv}=B$. By replacing
$G$ with $G'$, we may thus assume that $G$ is not vertex-transitive. By~\cite[I.4.1,
Theorem~6]{Serre}, it follows that $G$ is  isomorphic to $G_u *_{B} G_v$.

By \cite[I.4.1, Proof of Theorem~7]{Serre}, we may assume that the vertex set of $\mathfrak T$ is
the disjoint union of the right coset spaces $G/G_u$ and  $G/G_v$, with two vertices being adjacent
if they have nonempty intersection, and that the action of $G$ on $\mathfrak T$ is given by right
multiplication. In particular, $G_u$ and $G_v$ are adjacent when viewed as vertices of $\mathfrak
T$. Since $\mathfrak T$ is $G$-locally-transitive it follows that the neighbourhood of $G_u$ is
$\{G_v g\mid g\in G_u\}$.

Let $X=G_uG_vG_u\cup G_vG_uG_v$. Note that, as $L_1$, $L_2$ and $B$ are finite, so are $G_u$, $G_v$
and $X$. By \cite[Theorem~2]{Baumslag}, $G$ is residually finite and hence there exists a  normal
subgroup $R$ of finite index in $G$ with $R \cap X = \{1\}$. Let $H = G / R$ and let $\Gamma$  be
the normal quotient graph $\mathfrak{T}/ R$. (The vertices of $\Gamma$ are the $R$-orbits on the
vertex set of $\mathfrak{T}$, with two such $R$-orbits adjacent in $\Gamma$ if there is an edge
between them in $\mathfrak{T}$.) Note that $\Gamma$ is $H$-locally-transitive and finite.

Since $\mathfrak T$ is locally-transitive, it is bi-regular. We now show that $\Gamma$ is bi-regular
with the same valencies as $\mathfrak{T}$. We argue by contradiction and suppose, without loss of
generality, that the $R$-orbit of $G_u$, viewed as a vertex of $\Gamma$, has valency strictly less
than $G_u$, viewed as a vertex of $\mathfrak{T}$. It follows from the definition of $\Gamma$ that
the vertex $G_u$ of $\mathfrak{T}$ must have two distinct neighbours in the same $R$-orbit. Recall
that  the neighbourhood of $G_u$ in $\mathfrak{T}$ is $\{G_v g\mid g\in G_u\}$. In particular, there
exist $g,h\in G_u$ and $r\in R$ such that $G_vg\neq G_vh$ and $G_vgr=G_vh$. This implies that $r\in
G_uG_vG_u \subseteq X$ and hence $r \in R \cap X = \{1\}$, which is a contradiction.

Let $K/R$ be the kernel of the action of $H=G/R$ on $\Gamma$. By the previous paragraph, $\Gamma$ is
bi-regular with the same valencies as $\mathfrak{T}$ and a standard argument yields that $K$ is
semiregular on $\mathfrak{T}$. In particular, $K=R$ (and $\mathfrak{T}$ is a regular cover of
$\Gamma$) and $H$ acts faithfully on $\Gamma$. It follows that the stabiliser in $H$ of the vertex
$G_uR$ of $\Gamma$ is $G_uR/R\cong G_u$, the stabiliser of the vertex $G_vR$ is  $G_vR/R\cong G_v$
and the stabiliser of the edge $\{G_uR,G_vR\}$ is $BR/R\cong B$. Since $(\mathfrak{T},G)$ is locally
$[L_1,L_2]$, this implies that $(\Gamma,H)$ is locally $[L_1,L_2]$.

$(3)\Longrightarrow  (1):$ Let $\mathcal A$ be the rank two  amalgam of the groups $H_u$ and $H_v$
with Borel subgroup $H_{uv}$. Since $\Gamma$ is $H$-locally-transitive, the group generated by $H_u$
and $H_v$ is transitive on edges of $\Gamma$. In particular, any subgroup of $H_{uv}$ that is
normalised by both $H_u$ and $H_v$ must be trivial. This shows that $\mathcal A$ is faithful.
Clearly, $\mathcal A$ has permutation type $[L_1,L_2]$.
\end{proof}

The following is an immediate corollary to Lemma~\ref{newLemma}.

\begin{corollary}\label{newCor}
Let $L_1$ and $L_2$ be finite transitive permutation groups. The following are equivalent:
\begin{enumerate}
\item For every integer $c$, there exists a rank two faithful amalgam of permutation type
$[L_1,L_2]$ with Borel subgroup of order at least $c$.
\item For every integer $c$, there exists a locally $[L_1,L_2]$ pair $(\mathfrak{T},G)$ such that
$\mathfrak{T}$ is an infinite tree and $c\leq |G_{uv}|<\infty$ for some edge $\{u,v\}$ of
$\mathfrak{T}$.
\item $[L_1,L_2]$ is not locally-restrictive.
\end{enumerate}
\end{corollary}

The equivalence of Theorems~\ref{theo:graph},~\ref{cor:discsubgroups} and~\ref{theo:main} follows
immediately from Corollary~\ref{newCor}.

\section{Proof of Theorem~\ref{theo:main}}\label{Sec:ResAreSP}
All groups mentioned in the next two sections are finite. We adopt the notation and hypothesis of
Theorem~\ref{theo:main} and, without loss of generality, we assume that $L_1$ is not semiprimitive.
To simplify notation, we write $L=L_1$ and $R=L_2$. Let $m_2$ be the degree of $R$, let $\ell$ be a
positive integer, let $m=\ell m_2$ and let $\Omega=\{(y,z)\mid 1\leq y\leq m_2,1\leq z\leq \ell \}$.
Observe that $|\Omega|=\ell m_2=m$ and that the action of $R$ on $\{1,\ldots,m_2\}$ induces an
action of $R$ on $\Omega$: for $r\in R$ and $(y,z)\in \Omega$, we set $$(y,z)^{r}=(y^r,z).$$

We endow the set $\Omega$ with its natural lexicographic order, that is $(y,z)<(y',z')$ if either
$y<y'$, or $y=y'$ and $z<z'$. This total ordering allows us to identify $\Omega$ with
$\{0,\ldots,m-1\}$ in a natural way : $(1,1)$ is identified with $0$, $(m_2,\ell)$ with $m-1$, etc.
We extend the action of $R$ on $\Omega=\{0,\ldots,m-1\}$  to an action of $R$ on $\{0,\ldots,m\}$ by
letting the point $m$ be fixed by every element of $R$.

Since $L$ is not semiprimitive, there exists  a normal subgroup $K$ of $L$ that is neither
transitive nor semiregular. Denote by $\Delta$ the set of orbits of $K$ and let $K'$ be the kernel
of the action of $L$ on $\Delta$. Note that $K'$ is a normal subgroup of $L$ having the same orbits
as $K$ that is neither transitive nor semiregular. We may thus assume that $K=K'$ without loss of
generality. Let $S$ denote the permutation group induced by the action of $L$ on $\Delta$ and let
$\pi: L \rightarrow S$ be the canonical projection with kernel $K$. Fix $\delta\in \Delta$ and
${\lambda}\in \delta$. Since $K$ is transitive on $\delta$, we have $L_{\delta}=KL_{\lambda}$ and
$S_{\delta}\cong L_{\delta} /K=KL_{\lambda} /K\cong L_{\lambda}/(K\cap
L_{\lambda})=L_{\lambda}/K_{\lambda}$.  We sometimes denote by $\pi$ the restriction
$\pi|_{L_{\lambda}}:L_{\lambda}\to S_{\delta}$, slightly abusing notation.

Fix $\mathcal{T}$ a transversal for the set of right cosets of $S_{\delta}$ in $S$ with $1\in
\mathcal{T}$. For every $s\in S$, there exists a unique element of $\mathcal{T}$, which we denote by
 $s^\tau$, such that $S_{\delta} s=S_{\delta} s^\tau$. The correspondence $s\mapsto s^\tau$ defines
a map $\tau:S\to \mathcal{T}$ with $1^\tau=1$.

\begin{lemma}\label{silly1}
If $x,s\in S$, then $(xs^{-1})^\tau s (x^\tau)^{-1}\in S_{\delta}$.
\end{lemma}
\begin{proof}
We have $S_{\delta} xs^{-1}=S_{\delta}(xs^{-1})^\tau$ and hence $S_{\delta} x=S_{\delta}
(xs^{-1})^\tau s$. Furthermore, as $S_{\delta} x=S_{\delta} x^{\tau}$, we obtain $S_{\delta}=
S_{\delta} (xs^{-1})^\tau s(x^{\tau})^{-1}$.
\end{proof}

Let $V$ be the set of all functions from $\Delta$ to $L_{\lambda}$.  Under point-wise
multiplication, $V$ is a group isomorphic to $L_{\lambda}^{\Delta}$. Given $f\in V$ and $g\in L$,
let $f^g$ be the element of $V$ defined by
\begin{equation}\label{eq:def}
f^g(\sigma)=f\left(\sigma^{(g^\pi)^{-1}}\right),\ \sigma\in\Delta.
\end{equation}
This defines a group action of $L$ on $V$ and the semidirect product $V\rtimes L$ is isomorphic to
the standard wreath product $L_{\lambda}\wr_\Delta L$. Moreover, by extending this action of $L$ on
$V$ to the component-wise action of $L$ on $V^m$, we obtain a semidirect product $L\ltimes V^m$
where the multiplication is given by
\begin{eqnarray}\label{eq:binary}
(g,f_1,\ldots,f_m)(g',f_1',\ldots,f_m')&=&(gg',f_1^{g'}f_1',\ldots, f_m^{g'}f_m').
\end{eqnarray}

We now isolate some subgroups of $L\ltimes V^m$ that  provide the backbone for our construction.

\begin{definition}\label{A}
We define the following subsets of $L\ltimes V^m$:
\begin{align*}
&A= && \left \{ (g,f_1,\ldots,f_m)\in L\ltimes V^m\begin{array}{ c | c }   &
(f_i(\delta^x))^\pi=(x(g^\pi)^{-1})^\tau g^\pi(x^\tau)^{-1} \\
& \textrm{ for every } i\in \{1,\ldots,m\} \textrm{ and for every }x\in S    \end{array} \right
\},\\\nonumber
& C  =& & \{(g,f_1,\ldots,f_m)\in A\mid g\in L_{\lambda}\}, \\\nonumber
&M = &&\{(1,f_1,\ldots,f_m)\in L\ltimes V^m    \mid f_i(\Delta) \subseteq K_{\lambda} \textrm{ for
every }i\in\{1,\ldots,m\}\}.
\end{align*}
Let $\varphi:A\to L$ be the map defined by $\varphi : (g,f_1,\ldots,f_m) \mapsto g$.
\end{definition}

Note that, by Lemma~\ref{silly1}, the element $(x(g^\pi)^{-1})^\tau g^\pi(x^\tau)^{-1}$ in the
definition of $A$ lies in $S_{\delta}$.

\begin{lemma}\label{silly2}
The set $A$ is a subgroup of $L\ltimes V^m$.
\end{lemma}
\begin{proof}
Let $(g,f_1,\ldots,f_m),(g',f_1',\ldots,f_m')\in A$ and let $x\in S$. For every $i\in
\{1,\ldots,m\}$, we have
\begin{eqnarray*}
((f_i^{g'}f_i')(\delta^x))^\pi&=&(f_i^{g'}(\delta^x))^\pi\cdot(f_i'(\delta^x))^\pi\\
&\overset{(\ref{eq:def})}{=}&(f_i({\delta^x}^{(g'^\pi)^{-1}}))^\pi\cdot(f_i'(\delta^x))^\pi\\
&\overset{\rm{Def.}~\ref{A}}{=}&(((x(g'^\pi)^{-1})(g^\pi)^{-1})^\tau g^\pi
((x(g'^\pi)^{-1})^\tau)^{-1})\cdot((x(g'^\pi)^{-1})^\tau g'^\pi(x^\tau)^{-1})\\
&=&(x((gg')^\pi)^{-1})^\tau (gg')^\pi(x^{\tau})^{-1}.
\end{eqnarray*}
Using~$(\ref{eq:binary})$ and Definition~\ref{A}, this shows that
$(g,f_1,\ldots,f_m)(g',f_1',\ldots,f_m')\in A$. Clearly, the identity of $L\ltimes V^m$ is contained
in $A$. Since $L\ltimes V^m$ is a finite group, this concludes the proof.
\end{proof}

\begin{lemma}\label{silly3}
The map $\varphi$ is a surjective homomorphism.
\end{lemma}
\begin{proof}
By~$(\ref{eq:binary})$, $\varphi$ is a homomorphism.   For each $s\in S_{\delta}$, choose an element
$s^\varepsilon$ of $L_{\lambda}$ with $(s^\varepsilon)^\pi=s$. Since $\pi:L_{\lambda}\to S_{\delta}$
is surjective, $\varepsilon:S_{\delta}\to L_{\lambda}$ is well-defined. Let $g\in L$ and define
$f_{g}:\Delta\to L_{\lambda}$ with
$$f_g(\delta^x)=((x(g^\pi)^{-1})^{\tau}g^\pi(x^{\tau})^{-1})^\varepsilon,$$
for $x\in S$. First, note that, by Lemma~\ref{silly1},
$(x(g^\pi)^{-1})^{\tau}g^\pi(x^{\tau})^{-1}\in S_{\delta}$ and thus $f_g(\delta^x)\in L_{\lambda}$.
To see that $f_g$ is well-defined, note that for every $y\in S_{\delta}$, we have
$(yx)^\tau=x^{\tau}$ and $(yx(g^\pi)^{-1})^\tau=(x(g^\pi)^{-1})^\tau$, and hence
$f_g(\delta^x)=f_g(\delta^{yx})$. Now
$$(f_g(\delta^x))^\pi=(((x(g^\pi)^{-1})^{\tau}g^\pi(x^{\tau})^{-1})^\varepsilon)^\pi=(x(g^\pi)^{-1})
^{\tau}g^\pi(x^{\tau})^{-1},$$
and hence $(g,f_g,\ldots,f_g)\in A$ and $(g,f_g,\ldots,f_g)^\varphi=g$, which concludes the proof.
\end{proof}

\begin{lemma}\label{lemma:yes}
The kernel of $\varphi$ is $M$ and $M\cong K_{\lambda}^{|\Delta|m}$.
\end{lemma}
\begin{proof}
It is clear that $M\cong K_{\lambda}^{|\Delta|m}$. Suppose first that $(g,f_1,\ldots,f_m)$ is in the
kernel of $\varphi$ then $g=(g,f_1,\ldots,f_m)^\varphi=1$. For every $i\in \{1,\ldots,m\}$ and every
$x\in S$, it follows by Definition~\ref{A} that
$(f_i(\delta^x))^\pi=(x(g^\pi)^{-1})^{\tau}g^\pi(x^\tau)^{-1}=1$ and thus $f_i(\delta^x)\in
K_{\lambda}$. It follows that $(g,f_1,\ldots,f_m)\in M$.

Conversely, if $(g,f_1,\ldots,f_m)\in M$ then $g=1$ and $f_i(\delta^x)\in K_{\lambda}$ for every
$i\in \{1,\ldots,m\}$ and every $x\in S$ and thus
$(f_i(\delta^x))^\pi=1=(x(g^\pi)^{-1})^{\tau}g^\pi(x^\tau)^{-1}$. In particular,
$(g,f_1,\ldots,f_m)\in A$ and hence $(g,f_1,\ldots,f_m)$ is in the kernel of $\varphi$.
\end{proof}

\begin{lemma}\label{silly4}
The set $C$ is a subgroup of $A$, $M$ is the core of $C$ in $A$ and the permutation group induced by
the action of $A$ on the right cosets of $C$ is permutation isomorphic to $L$.
\end{lemma}
\begin{proof}
By Lemmas~\ref{silly3} and~\ref{lemma:yes}, $\varphi$ is a surjective homomorphism with kernel $M$.
In particular, $M\unlhd A$ and $A/M\cong L$. Note that $C$ is the pre-image of $L_{\lambda}$ under
$\varphi$ and thus $M\leq C\leq A$. As $C^{\varphi}=L_{\lambda}$ and $L_{\lambda}$ is core-free in
$L$, it follows that $M$ is the core of $C$ in $A$. Finally, the action of $A$ on the right cosets
of $C$ is permutation isomorphic to the action of $A^{\varphi}=L$ on the right cosets of
$C^{\varphi}=L_\lambda$, that is, to $L$.
\end{proof}

We now introduce an alternative notation for the elements of $A$ that will simplify some later
computations. Let $a=(g,f_1,\ldots,f_m)\in A$. For $i\in\{1,\ldots,m\}$, we write $g_i=f_i(\delta)$
and $h_{i-1}={f_i}|_{ \Delta\setminus\{\delta\}}$. (Note that $f_i$ is completely determined by
$(g_i,h_{i-1})$.) We also write $g_0=g$ and then denote $a$ by
\begin{equation*}
((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1})).
\end{equation*}
Note that, with this notation, the multiplication is not component-wise (in contrast to
$(\ref{eq:binary})$): indeed, if $a'=((g_0',\ldots,g_m'),(h_0',\ldots,h_{m-1}'))$ is another element
of $A$ then
\begin{equation}\label{eq:gogo3}
aa'=((g_0g_0',g_1^{g_0'}g_1'\ldots,g_m^{g_0'}g_m'),(h_0^{g_0'}h_0',\ldots,h_{m-1}^{g_0'}h_{m-1}')).
\end{equation}
 Using the above notation, for each $r\in R$ and $c=((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1}))\in C$,
let
\begin{equation} c^r = ((g_{0r^{-1}},g_{1 r^{-1}}, \dots, g_{m r^{-1}}),
(h_{0r^{-1}},\dots,h_{(m-1)r^{-1}})),  \label{LA}\end{equation}
where, for $i\in \{0,\dots,m\}$, we denote the image of $i$ under $r$ by $ir$.

\begin{lemma}\label{lemma:action}
Equation~\eqref{LA} defines a group action of $R$ on the group $C$.
\end{lemma}
\begin{proof}
In this proof, it is convenient to use both notations for elements of $C$.  Let $c =
(g,f_1,\dots,f_m)=((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1})) \in C$. Since $c \in C$, we have $g_0=g
\in L_\lambda$ and hence $g^\pi \in S_\delta$. Since $c \in A$, for every $x\in S_\delta$ and $i\in
\{1,\dots,m\}$, we have $x^\tau=1=(x (g^\pi)^{-1})^\tau$ and thus
\begin{equation}  g_i^\pi =(f_i(\delta))^\pi = (f_i(\delta^x))^\pi \overset{\rm{Def.}~\ref{A}}{=} (x
(g^\pi)^{-1})^\tau g^\pi (x^\tau)^{-1} =  g^\pi.\label{STAR} \end{equation}

Let $r \in R$ and write $v=r^{-1}$ and $c^r = (g_{0v},f_1',\dots,f_m')$. We first show that $c^r\in
C$. For every $x\in S_\delta$ and $i\in \{1,\dots,m\}$, we have
$$(f_i'(\delta^x))^\pi = (f_i'(\delta))^\pi \overset{(\ref{LA})}{=} (g_{iv})^\pi
\overset{(\ref{STAR})}{=} g_{0v}^\pi = (x (g_{0v}^\pi)^{-1})^\tau g_{0v}^\pi (x^\tau)^{-1} $$
where in the last equality  $x,g_{0v}^\pi\in S_\delta$ is used. Similarly, for every $x\in S
\setminus S_\delta$ and $i\in \{1,\dots,m\}$, we have
$$(f_i'(\delta^x))^\pi \overset{(\ref{LA})}{=} (h_{(i-1)v}(\delta^x))^\pi
\overset{\rm{Def.}~\ref{A}}{=}  (x (g^\pi)^{-1})^\tau g^\pi (x^\tau)^{-1} \overset{(\ref{STAR})}{=}
(x (g_{0v}^\pi)^{-1})^\tau g_{0v}^\pi (x^\tau)^{-1}.$$
This shows that $c^r \in A$. Since $g_i \in L_\lambda$ for all $i \in \{0,\dots,m\}$, we have that
$g_{0v}\in L_\lambda$ and thus $c^r \in C$.

Let $d=((y_0,y_1,\dots,y_m),(z_0,\dots,z_{m-1})) \in C$. Recall that $y_0\in L_\lambda$. Hence, for
$j \in \{1,\dots,m\}$, we have
\begin{equation}
g_j^{y_0} =f_j^{y_0}(\delta)\overset{(\ref{eq:def})}{=} f_j(\delta^{(y_0^\pi)^{-1}}) = f_j(\delta)=
g_j. \label{STARSTAR}
\end{equation}

Now,
\begin{eqnarray*}
cd &\overset{(\ref{eq:gogo3})}{=} &((g_0y_0, g_1^{y_0}y_1,\ldots,g_m^{y_0}y_m),(h_0^{y_0}
z_0,\ldots,h_{m-1}^{y_0} z_{m-1}))\\
&\overset{(\ref{STARSTAR})}{=}& ((g_0y_0,g_1y_1,\ldots,g_m y_m),(h_0^{y_0} z_0,\ldots,h_{m-1}^{y_0}
z_{m-1}))
\end{eqnarray*}
and thus
$$(cd)^r = ((g_{0v} y_{0v},g_{1v}y_{1v},\ldots,g_{mv} y_{mv}),(h_{0 v}^{y_0}
z_{0v},\ldots,h_{(m-1)v}^{y_0} z_{(m-1)v})).$$

Recall that $c^r= ((g_{0v},\ldots, g_{m v}), (h_{0v},\ldots, h_{(m-1)v}))= (g_{0v},f_1',\dots,f_m')$
and thus $f_i'(\delta) = g_{iv}$ and $f_i'(\sigma) = h_{(i-1)v}(\sigma)$ for $i \in \{1,\dots,m\}$
and $\sigma \in \Delta \setminus \{\delta\}$. Similarly, recall that $d^r = ((y_{0v},\ldots, y_{m
v}), (z_{0v},\ldots, z_{(m-1)v}))$ and write $d^r=(y_{0v},e_1',\dots,e_m')$.
By~$(\ref{eq:binary})$, we have $c^rd^r = (g_{0v}y_{0v},f_1'^{y_{0v}} e_1',\dots,f_m'^{y_{0v}}
e_m')$. Since $y_{0v} \in L_\lambda$, we have $y_{0v}^\pi \in S_\delta$ and hence, for $i\in
\{1,\dots,m\}$, we have
$$ (f_i'^{y_{0v}} e_i') (\delta) = f_i'^{y_{0v}}(\delta)e_i'(\delta) \overset{(\ref{eq:def})}{=}
f_i'(\delta^{(y_{0v}^\pi)^{-1}}) y_{iv} = f_i'(\delta) y_{iv} = g_{iv} y_{iv}.$$
Similarly, for $\sigma \in \Delta \setminus \{ \delta \}$ and $i\in \{1,\dots,m\}$, we have
$$f_{i}'^{y_{0v}}(\sigma) \overset{(\ref{eq:def})}{=} f_{i}'(\sigma^{ (y_{0v}^\pi)^{-1}})= h_{(i-1)
v}(\sigma^{ (y_{0v}^\pi)^{-1}}) \overset{(\ref{STAR})}{=} h_{(i-1) v}(\sigma^{ (y_0^\pi)
^{-1}})\overset{(\ref{eq:def})}{=} h_{(i-1) v}^{y_{0}}(\sigma)$$
and thus $(f_{i}'^{y_{0v}}e_i')(\sigma)=(h_{(i-1) v}^{y_{0}}z_{(i-1)v})(\sigma)$. This shows that
$(cd)^r = c^r d^r$.

It is clear from (\ref{LA}) that $(c^r)^{r'}=c^{rr'}$ for every $r'\in R$ and $c^r=1$ if and only if
$c=1$, which concludes the proof.
\end{proof}

By Lemma~\ref{lemma:action}, we can define the semidirect product $C\rtimes R$. Let
\begin{align*}
&P_2=C\rtimes R,\\
&B=C\rtimes R_1,
\end{align*}
with $B$  viewed as a subgroup of $P_2$. From our definitions we have:

\begin{lemma}\label{retard}
The core of $B$ in $P_2$ is $C$. Moreover, the permutation group induced by the action of $P_2$ on
the right cosets of $B$ is permutation isomorphic to $R$.
\end{lemma}

From $(\ref{LA})$, $R_1$ inherits an action on $C$ from $R$. We extend this to an action of $R_1$ on
$A$ in the following way:
given $a=((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1}))\in A$ and $r \in R_1$, let
$$a^r =((g_{0r^{-1}},g_{1 r^{-1}}, \dots, g_{m r^{-1}}), (h_{0r^{-1}},\dots,h_{(m-1)r^{-1}})).$$

With minor changes, the proof of Lemma~\ref{lemma:action} can be adapted to show that this induces a
group action of $R_1$ on $A$. (It is helpful to notice that for all $r \in R_1$ we have $0r=0$.) Let
$$P_1=A\rtimes R_1.$$
We view $B$ as a subgroup of $P_1$ in the obvious way. (Note that the action of $R$ on $C$ cannot be
extended to an action of $R$ on $A$ in any meaningful way.)

\begin{lemma}\label{moreretarded}
The core of $B$ in $P_1$ is $M\rtimes R_1$ and the action of $P_1$ on the right cosets of $B$ is
permutation isomorphic to $L$.
\end{lemma}
\begin{proof}
The proof follows with a computation and from Lemma~\ref{silly4}.
\end{proof}

Let $\mathcal{A}$ be the rank two amalgam of the groups $P_1$ and $P_2$ with $\mathcal{B}(\mathcal
A)=P_1\cap P_2=B$. Lemmas~\ref{retard} and~\ref{moreretarded} show that the permutation type of
$\mathcal{A}$ is  $[L_1,L_2]$.

\begin{proposition}\label{neverends}
The amalgam $\mathcal{A}$ is faithful.
\end{proposition}
\begin{proof}
Let $N$ be a subgroup of $B$ normal in $P_1$ and in $P_2$. We show that $N=1$. By
Lemma~\ref{retard}, the core of $B$ in $P_2$ is $C$ and hence $N\leq C$. By
Lemma~\ref{moreretarded}, the core of $B$ in $P_1$ is $M\rtimes R_1$ and thus $N\leq C\cap (M\rtimes
R_1)=M$.

For $i\in\{0,\ldots,m\}$, let $G(i)$ be the proposition: for every
$((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1}))\in N$, we have $g_i=1$ . Similarly, for $i
\in\{0,\ldots,m-1\}$, let $H(i)$ be the proposition: for every
$((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1}))\in N$, we have $h_i=1$ . We prove the following preliminary
claims.

\smallskip
\noindent\textsc{Claim 1. }Let $i\in \{1,\ldots,m\}$ and let $\sigma\in \Delta$. Suppose that, for
every $(1,f_1,\ldots,f_m)\in N$, we have $f_i(\sigma)=1$. Then $G(i)$ and $H(i-1)$ hold.
\smallskip

\noindent
Let $(1,f_1,\ldots,f_m)\in N$ and let $\mu \in \Delta$. Since $S$ is transitive on $\Delta$ and
$\pi$ is surjective, there exists $g\in L$ such that $\sigma^{ (g^{\pi})^{-1}} = \mu$. By
Lemma~\ref{silly3}, there exists $(f_1',\ldots,f_m')\in V^m$ with $(g,f_1',\ldots,f_m')\in A$. As
$N\unlhd A$,
$$(1,f_1,\ldots,f_m)^{(g,f_1',\ldots,f_m')}=(1,f_1'^{-1}f_1^gf_1',\ldots,f_m'^{-1}f_m^{g}f_m')\in
N.$$
By hypothesis, we have
$1=(f_i'^{-1}f_i^{g}f_i')(\sigma)=f_i'(\sigma)^{-1}f_i(\sigma^{(g^\pi)^{-1}})f_i'(\sigma)=f_i(\mu)$.
Since $\mu$ is an arbitrary element of $\Delta$ we obtain $f_i=1$. Since $(1,f_1,\ldots,f_m)$ was an
arbitrary element of $N$, it follows that $G(i)$ and $H(i-1)$ hold.~$_\blacksquare$
\smallskip

\smallskip
\noindent\textsc{Claim 2. }Let $i\in \{1,\ldots,m\}$. Then $G(i)\Longleftrightarrow   H(i-1)$.
\smallskip

\noindent
Suppose that $G(i)$ holds. Applying Claim~1 with $\sigma=\delta$, we immediately obtain $H(i-1)$.
Conversely, if $H(i-1)$ holds then applying Claim~1 with some $\sigma\in\Delta\setminus\{\delta\}$,
we obtain $G(i)$.~$_\blacksquare$

\smallskip
\noindent\textsc{Claim 3. }Let $i\in \{0,\ldots,m-1\}$ and let $j$ be in the $R$-orbit of $i$. Then
$G(i)\Longrightarrow G(j)$ and $H(i)\Longrightarrow H(j)$.
\smallskip

\noindent
Assume that $G(i)$ holds and let $n=((g_0,\ldots,g_m),(h_0,\ldots,h_{m-1}))\in N$. There exists
$r\in R$ such that $ir^{-1}=j$. Since $R\leq P_2$, $N$ is normalised by $R$ and $n^r\in N$.
By~$(\ref{LA})$, this implies that $((g_{0r^{-1}},g_{1 r^{-1}}, \dots, g_{m r^{-1}}),
(h_{0r^{-1}},\dots,h_{(m-1)r^{-1}}))\in N$. Since $G(i)$ holds, we have that $g_j=g_{ir^{-1}}=1$. As
$n$ was an arbitrary element of $N$, this shows that $G(j)$ holds. The proof that
$H(i)\Longrightarrow H(j)$ is essentially the same and is omitted.~$_\blacksquare$
\smallskip

\smallskip
\noindent\textsc{Claim 4. }$G(i)$ holds for every $i\in\{0,\ldots,m\}$.
\smallskip

\noindent We argue by contradiction and let $z$ be minimal in $\{0,\ldots,m\}$ such that $G(z)$ does
not hold. Since $N\leq M$, we have that $G(0)$ holds and thus $z\geq 1$. By Claim 2, we see that
$H(z-1)$ does not hold.

Let $\mathcal{O}$ be the $R$-orbit on $\{0,\ldots,m\}$ containing $z$. By the minimality of $z$ and
Claim~3, we get that $z$ is the minimum of $\mathcal{O}$. By examining the orbits of $R$ on
$\{0,\ldots,m\}$, we see that this implies that $z-1$ and $z-2$ are in the same $R$-orbit. Since
$H(z-1)$ does not hold, Claim 3 implies that neither does $H(z-2)$. By Claim 2, neither does
$G(z-1)$, contradicting the minimality of $z$. ~$_\blacksquare$
\smallskip

Claim 2 together with Claim 4 implies that $H(i)$ holds for every $i\in\{0,\ldots,m-1\}$ and thus
$N=1$. This concludes the proof.
\end{proof}

Finally, we have $|B|\geq |C| \geq |M|=|K_{\lambda}|^{|\Delta|m}$, where the last equality follows
by Lemma~\ref{lemma:yes}. Recall that $m=\ell m_2$. Since $K$ is not semiregular, we have
$|K_{\lambda}|\geq 2$ and thus $|B|\to\infty$ as  $\ell\to\infty$. This concludes the proof of
Theorem~\ref{theo:main}.

\section{Proof of Theorem~\ref{theo:mainmain}}\label{sec4}
As in the previous section, all groups considered are finite. Suppose first that $L_1,\ldots,L_k$
are regular permutation groups and let $\mathcal{A}=\bigcup_{i=1}^kP_i$ be a rank $k$ faithful
amalgam of permutation type $[L_1,\ldots,L_k]$.  Since $L_i$ is regular, we have
$\mathcal{B}(\mathcal{A})\unlhd P_i$ for every $i\in\{1,\ldots,k\}$. As $\mathcal{A}$ is faithful,
this implies that $\mathcal{B}(\mathcal{A})=1$.  This proves the implication $(2)\Longrightarrow
(1)$ of Theorem~\ref{theo:mainmain}.

We now turn to the proof of the implication $(1)\Longrightarrow (2)$. The following lemma will be
needed.

\begin{lemma}\label{basic}
Let $H$ and $K$ be transitive permutation groups on $\Delta$ and $\Lambda$, respectively. Let
$\delta_0\in \Delta$, $\lambda_0\in \Lambda$ and let $\ell$ be a positive integer. If
$|\Delta|,|\Lambda|\geq 2$ then there exist a set $\Omega$ of cardinality $\ell |\Delta||\Lambda|$,
faithful group actions  $\rho_H:H\to \Sym(\Omega)$ and $\rho_K:K\to \Sym(\Omega)$, and $\omega\in
\Omega$ such that
\begin{enumerate}
\item $\rho_H(H)_\omega=\rho_H(H_{\delta_0})$ and $\rho_K(K)_\omega=\rho_K(K_{\lambda_0})$;
\item $\langle \rho_H(H),\rho_K(K)_\omega\rangle=\rho_H(H)\times\rho_K(K)_\omega$ and $\langle
\rho_H(H)_\omega,\rho_K(K)\rangle=\rho_H(H)_\omega\times\rho_K(K)$; \label{newlabel}
\item $\langle \rho_H(H),\rho_K(K)\rangle$ is transitive on $\Omega$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $\Omega$ be the set $\Delta\times \Lambda\times\mathbb{Z}_\ell$ and let
$\omega=(\delta_0,\lambda_0,0)\in\Omega$. Let $g\in \Sym(\Omega)$ be defined by
$$
(\delta,\lambda,i)^g=\begin{cases}
(\delta,\lambda,i)&\textrm{if }\delta\neq \delta_0 \textrm{ or }\lambda\neq \lambda_0,\\
(\delta_0,\lambda_0,i+1)&\textrm{if }\delta=\delta_0 \textrm{ and }\lambda=\lambda_0.\end{cases}
$$
Define $\rho_{H}:H\to\Sym(\Omega)$ by setting $(\delta,\lambda,i)^{\rho_H(h)}=(\delta^h,\lambda,i)$
for every $h\in H$. Similarly, define $\rho_K,\rho_{K}':K\to \Sym(\Omega)$ by setting
$(\delta,\lambda,i)^{\rho_K'(k)}=(\delta,\lambda^k,i)$ and ${\rho_K(k)}=g^{-1}\rho_K'(k)g$ for every
$k\in K$. It is easy to check that $\rho_H$ and $\rho_K$ define faithful group actions of $H$ and
$K$ on $\Omega$. A simple computation shows that $\rho_H(H)_\omega=\rho_H(H_{\delta_0})$ and
$\rho_K(K)_\omega=\rho_K(K_{\lambda_0})$.

It is easy to check that if $k\in K_{\lambda_0}$ then $\rho_K(k)=\rho'_K(k)$. Since $\rho_H(H)$
centralises  $\rho_K'(K)$, it centralises
$\rho_K'(K_{\lambda_0})=\rho_K(K_{\lambda_0})=\rho_K(K)_\omega$. Similarly, since $H_{\delta_0}$
preserves $\{\delta_0\}$ and $\Delta\setminus\{\delta_0\}$ it follows that $\rho_K(K)$ centralises
$\rho_H(H_{\delta_0})=\rho_H(H)_\omega$. Clearly $\rho_H(H)\cap\rho_K(K)=1$ and hence
$(\ref{newlabel})$ is established.

As $H$ and $K$ are transitive, for every $(\delta,\lambda,i)\in\Omega$, we have
\begin{eqnarray*}
(\delta,\lambda,i)^{\rho_H(H)\rho_K(K)\rho_H(H)}&=&(\Delta\times\{\lambda\}\times\{i\})^{\rho_K(K)\
rho_H(H)}\\
&\supseteq&((\Delta\setminus\{\delta_0\})\times\{\lambda\}\times\{i\})^{\rho_K(K)\rho_H(H)}\\
&=&((\Delta\setminus\{\delta_0\})\times \Lambda\times\{i\})^{\rho_H(H)}\\
&=& \Delta\times \Lambda\times\{i\}.
\end{eqnarray*}
On the other hand, if $k\in K\setminus K_{\lambda_0}$ then
$(\delta_0,\lambda_0,i)^{\rho_K(k)}=(\delta_0,\lambda_0^k,i-1)$. This shows that $\langle
\rho_H(H),\rho_K(K)\rangle$ is transitive on $\Omega$.
\end{proof}

Let $k$ be a positive integer with $k\geq 3$ and let $L_1,\ldots,L_k$ be nontrivial transitive
permutation groups. For $i\in\{1,\ldots,k\}$, let $m_i$ denote the degree of $L_i$ and denote by
$\{0,\ldots,m_i-1\}$ the set acted upon by $L_i$. (Note that $m_i\geq 2$ since $L_i$ is nontrivial.)
Without loss of generality, we may assume that $L_1$ is not regular and thus $V:=(L_1)_0\neq 1$.

Let $\ell$ be a positive integer. By Lemma~\ref{basic}, there exist faithful actions of $L_2$ and
$L_3$ on a set  $\Omega$ of cardinality $\ell m_2m_3$ with $\langle L_2,L_3\rangle$  transitive on
$\Omega$. Moreover, there exists $\omega_0\in \Omega$ such that $(L_2)_{\omega_0}=(L_2)_0$,
$(L_3)_{\omega_0}=(L_3)_0$, $\langle (L_2)_{\omega_0},L_3\rangle=(L_2)_{\omega_0}\times L_3$ and
$\langle L_2,(L_3)_{\omega_0}\rangle=L_2\times (L_3)_{\omega_0}$.

Let $U=\prod_{\omega\in \Omega}V_\omega$ that is, $U$ is the direct product of $|\Omega|$ copies of
$V$, with the copies indexed by $\Omega$. Observe that the action of $\langle L_2, L_3\rangle$ on
$\Omega$ gives rise to a natural group action of $\langle L_2,L_3\rangle$ on $U$ which enables us to
construct the group $U\rtimes \langle L_2,L_3\rangle$. Let $U'=\prod_{\omega\in
\Omega\setminus\{\omega_0\}}V_\omega$, viewed as a subgroup of $U$ in the natural way. Note that, by
the previous paragraph, $(L_2)_{0}\times(L_3)_{0}$ normalises $U'$. Now, consider the following
abstract groups:
\begin{eqnarray*}
P_1&:=& L_1\times \left(U'\rtimes ((L_2)_{0}\times(L_3)_{0})\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times (L_k)_{0},\\
P_2&:=& \left(U\rtimes (L_2\times (L_3)_{0})\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times (L_k)_{0},\\
P_3&:=& \left(U\rtimes ((L_2)_{0}\times L_3)\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times (L_k)_{0},\\
P_4&:=& \left(U\rtimes ((L_2)_{0}\times (L_3)_{0})\right)\times L_4\times\cdots \times
(L_{k-1})_0\times (L_k)_0,\\
&\vdots&\\
P_{k-1}&:=& \left(U\rtimes( (L_2)_{0}\times(L_3)_{0})\right)\times (L_4)_0\times\cdots \times
L_{k-1}\times (L_k)_0,\\
P_k&:=& \left(U\rtimes ((L_2)_{0}\times (L_3)_{0})\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times L_k,\\
B&:=&\left(U\rtimes ((L_2)_{0}\times(L_3)_{0})\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times (L_k)_0.
\end{eqnarray*}
Observe that, for every $i\in\{1,\ldots, k\}$, there is an obvious embedding of $B$ in  $P_i$. (For
$i=1$, this is because $U=V\times U'\leq L_1\times U'$.) Hence, in what follows, we regard $B$ as a
common subgroup of $P_1,\ldots,P_k$. Let $\mathcal{A}=\bigcup_{i=1}^kP_i$. Thus $\mathcal{A}$ is a
rank $k$ amalgam of the groups $P_1,\ldots,P_k$ with $\mathcal{B}(\mathcal A)=B$.

\begin{lemma}\label{new}
The permutation type of $\mathcal{A}$ is  $[L_1,\ldots,L_k]$.
\end{lemma}
\begin{proof}
For every $i\in \{1,\ldots,k\}$, it is immediate from the definitions that the permutation group
induced by the action of $P_i$ on the right cosets of $B$ in $P_i$ is permutation isomorphic to
$L_i$.
\end{proof}

\begin{lemma}\label{newnew}
The amalgam $\mathcal{A}$ is faithful.
\end{lemma}
\begin{proof}
Let $N$ be a subgroup of $B$ with $N\unlhd P_i$ for every $i\in \{1,\ldots,k\}$. Let $K_i$ denote
the core of $B$ in $P_i$. Clearly, we have
\begin{eqnarray*}
K_1&=& \left(U'\rtimes ((L_2)_{0}\times(L_3)_{0})\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times (L_k)_{0},\\
K_2&=& \left(U\rtimes (1\times (L_3)_{0})\right)\times (L_4)_0\times\cdots \times (L_{k-1})_0\times
(L_k)_{0},\\
K_3&=& \left(U\rtimes ((L_2)_{0}\times 1)\right)\times (L_4)_0\times\cdots \times (L_{k-1})_0\times
(L_k)_{0},\\
&\vdots&\\
K_k&=& \left(U\rtimes ((L_2)_{0}\times (L_3)_{0})\right)\times (L_4)_0\times\cdots \times
(L_{k-1})_0\times 1,\\
\end{eqnarray*}
and thus $N\leq \bigcap_{i=1}^kK_i= U'$. Let $n\in N$. As $N\leq U$, we may write
$n=\prod_{\omega\in \Omega} n_\omega$ and, since $N\leq U'$, we have $n_{\omega_0}=1$.  Let
$\omega\in \Omega$. Since $\langle L_2,L_3\rangle$ is transitive on $\Omega$, there exists $x\in
\langle L_2,L_3\rangle$ with $\omega^x=\omega_0$. Recall that $\langle L_2,L_3\rangle\leq \langle
P_2,P_3\rangle$ hence $n^x\in N$ therefore $(n^x)_{\omega_0}=1$. On the other hand
$(n^x)_{\omega_0}=n_{\omega_0^{x^{-1}}}=n_{\omega}$. Since this holds for every $\omega\in\Omega$
and every $n\in N$, we have $N=1$ and thus $\mathcal{A}$ is faithful.
 \end{proof}

We have that $|\mathcal{B}(\mathcal A)|=|B|\geq |U|=|V|^{|\Omega|}=|(L_1)_0|^{\ell m_2m_3}$. Since
$L_1$ is not regular, we have $|(L_1)_0|\geq 2$ and thus $|\mathcal{B}(\mathcal A)|\to\infty$ as
$\ell\to\infty$. This concludes the proof of Theorem~\ref{theo:mainmain}.


TITLE: Why did people originally like Frobenius algebras?
QUESTION [19 upvotes]: These days, lots of people are excited by Frobenius algebras because commutative Frobenius algebras are the same thing as 2D topological quantum field theories.
...but this seems like teaching an old dog new tricks. Can anyone sum up (using only diet representation theory :-P), why Frobenius algebras were invented and what was so good about them?
Also, any nice texts and/or papers along this line would be much appreciated. I'm working through the old Nakayama papers now, but perhaps this material exists in a friendlier and more modernised form somewhere?

REPLY [9 votes]: Frobenius's original turn-of-the-century perspective was the nonvanishing of a determinant. Brauer–Nesbitt–Nakayama studied some equivalent definitions in the late 30s and early 40s.  For instance, an equivalence between the left and right regular representations is a rare and beautiful thing; this gives an equivalence between projectivity and injectivity that is explained in modern language in Lam's Lectures on Modules and Rings.  This also gives a "perfect duality" studied by Dieudonné in the late 50s.  I added the missing early sources to the wikipedia article, including the Brauer–Nesbitt announcement in PNAS which is pretty easy to read.


TITLE: Homeomorphism between pairs of letters
QUESTION [0 upvotes]: How can I check if the pairs of letters T and K , E and W are homeomorphic or not?

REPLY [1 votes]: Look at the cut points.
They'll cut those letters into two or three pieces.
Determine the number of three piece cut points in each letter.