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The tubes with LEDs create exciting lighting effects and look like the path of fireflies flying in the air..
The tubes with LEDs create exciting lighting effects and look like the path of fireflies flying in the air.
The chandelier gives 2 different types of light: decorative mood light and directional light, perfect for lighting up a dining or coffee table.
You can choose to have both decorative and directional light on, or only decorative light, by pressing once or twice on the wall switch.
IKEA. Model T1350 Älvsbyn. This luminaire contains built-in LED lamps. The lamps cannot be changed in the luminaire.


ezs note: These - ahem - "teachers", if these allegations prove true, deserve to fired immediately. They don't deserve to head up a dog training school, never mind a school with young children. If you live in the UK, please write or e-write the Minister of State for School Standards in the Department of Education to let them know of your extreme displeasure regarding this situation. Thanks.


text:
Americans support repercussions for prominent men accused of sexual harassment
Economist/YouGov PollPolitics & current affairs
November 15, 2017, 9:15 p.m.
Many even think the punishments for Harvey Weinstein, Kevin Spacey, and Louis C.K. should have been harsher
Amid the growing allegations of sexual harassment made against major figures in Hollywood, politics and government, the American public is taking a hard line. Not only have they turned on three entertainment figures accused of harassment, but few in the latest Economist/YouGov Poll have any qualms about the punishments that the industry unilaterally has imposed on Harvey Weinstein, Kevin Spacey, and Louis C.K.
In the poll, the public was given a list of actions taken against each of these figures and asked if they felt the punishments were excessive, appropriate, or should have been even tougher. Just over one in ten said the punishments were excessive for Spacey and Louis C.K., both of whom lost jobs and had movies they had completed altered (in the case of Spacey) or kept from public release (in the case of Louis C.K.). Only 7% felt that was the case for Weinstein, who was fired from his company and expelled from the Academy of Motion Picture Arts and Sciences.
Two-thirds of the public say the punishments were either appropriate or not harsh enough. For Weinstein, more than three in four said that.
Those 65 and older are the most likely to believe the punishments were not harsh enough. For example, in Weinstein's case, 51% of men 65 and older (and 42% of women in that age group) said his punishment wasn't harsh enough.
Men are slightly more likely than women to think the punishments were excessive. For example, 16% of men think Louis C.K.'s punishment was excessive, compared with just 10% of women.
Women are also more likely than men to report personal experiences with sexual harassment. Those women who report being harassed are 10 points more likely than women who have not to think that Weinstein deserves even harsher punishments than he has received. They are six points more likely to say that about Spacey; nine points more likely to say that about Louis C.K.
Whatever Americans thought about these men before that accusations, they now have overwhelmingly unfavorable opinions of them.
For more, read complete results here.

redpajama_set_name: RedPajamaCommonCrawl


text:
Call on Mahathir to present Ministerial statement to break five-month government silence on the adverse TI Corruption Perception Index 2001 on Malaysia and how the ACA can return to the national radar and consciousness in a high-powered campaign against corruption
Media Statement
(Petaling Jaya, Wednesday): The five-month-long government silence in Parliament on Transparency International's Corruption Perception Index 2001 that Malaysia is stuck in the lowly 36th world ranking after plunging 13 places in six years from 23rd position in 1995 is most scandalous and outrageous.
I had expected the Minister in the Prime Minister's Department, Datuk Dr. Rais Yatim, to address this issue, declaring a greater government commitment to stamp out corruption and all forms of abuses of power especially as the 2002 Budget has empowered Ministries and statutory agencies with greater delegated spending powers.
On the ground of ensuring speedy implementation of government projects, ministries and statutory agencies were conferred greater delegated authority: tender boards at ministries given the authority to approve tenders of up to RM50 million for works procurement and RM30 million for supplies and services and up to RM5 million restricted tenders without Treasury approval; with the ceiling for tenders of statutory boards raised to RM100 million for all supplies, services and works and up to RM10 million for unrestricted tenders.
It is a matter of grave concern that Parliament, during the budget debate, had not focused on the need and measures to ensure that such greater delegated powers in the 2002 Budget do not lead to more abuses of authority and corrupt practices, and the five-month government silence after the release of Transparency International's Corruption Perception Index 2001 on Malaysia's adverse ranking is symptomatic of the lowly place of the anti-corruption drive in Malaysia.
In fact, in the past six months, the Anti-Corruption Agency (ACA) has disappeared from the national radar and consciousness, after the former ACA director-general Datuk Ahmad Zaki Husin was summarily shunted off to become Chairman of the Advisory Board set up under Article 151(2) of the Malaysian Constitution, being replaced by a policeman to head the ACA for the first time in its 32-year history.
In the past six months, the ACA has so completely disappeared from public view that I do not think there is more than a handful of MPs or the Malaysian public who could offhand remember or know who is now the Director-General of ACA! The question "Who is the director-general of ACA" will draw a blank throughout the country.
I call on the Prime Minister, Datuk Seri Dr. Mahathir Mohamad to give his personal attention to make a Ministerial statement in Parliament to break the government silence on the adverse TI Corruption Perception Index 2001 on Malaysia and how the ACA can return to the national radar and consciousness in a high-powered campaign against corruption - which has become more imperative with the 2002 Budget vesting greater spending powers to ministries and statutory boards without the traditional safeguards and checks.
*Lim Kit Siang - DAP National Chairman


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text:
New lacrosse power Monmouth set for MAAC Tournament
The Monmouth University men's lacrosse team was winless in its first season in 2014, but is now favored to win the MAAC Tournament. The Hawks have the nation's longest winning streak at nine games
@Joshua_Newman
WEST LONG BRANCH -- When Brian Fisher assumed his position as Monmouth University's first men's lacrosse head coach in July 2012, how was he supposed to sell his program?
After all, at that point, the Hawks were still almost two years away from taking the field. That meant no history or perspective to sell, only faith in what Fisher and his coaching staff were attempting to build. In hindsight, the pitch to recruits in those days was simple.
"The pitch was just, 'Do you want to play?' said senior goalie Nick Hreshko, a Warrington, Pa. native. "We had the opportunity to play, and honestly, that's all a lot of guys wanted, the opportunity to play and as a competitor, that's all I wanted. I wanted a chance to play, and on top of that, the academics that Monmouth offers are top-notch. All in all, this is all I ever wanted out of a college, and I'm happy with the choice I made."
Monmouth athlete rushed to Monmouth Medical Center from banquet
"Goal No. 1 was to get enough guys to fill out a team and play our games," Fisher said, sarcastically, on April 21, moments after his team secured the MAAC regular-season title with a win over Canisius on The Great Lawn. "That's a little bit of what it was like in those first 12 months recruiting. One of the benefits of being a college coach is we get to meet a lot of really good people along the way. We have a lot of good people in our program, both the guys and their families. We're proud of that fact."
Four years after Monmouth's winless inaugural season in 2014, things have radically changed in West Long Branch. The Hawks have quickly morphed into a MAAC power, set to open this weekend's MAAC Tournament as the No. 1 seed Friday afternoon against fourth-seeded Detroit Mercy at 4 p.m. at Marist College. The winner there will face the winner of Friday's second semifinal between Canisius and the Red Foxes in Sunday's championship (10 a.m., ESPNU). Sunday's winner gets the MAAC's automatic bid to the NCAA Tournament.
Monmouth (12-3), prohibitive favorites this weekend and owners of the nation's longest winning streak at nine games, are in this position in large part due to early faith in Fisher. That said, at some point, legitimate, college-level talent has to be a factor. Monmouth has shown to have plenty of that, specifically in junior attackman Bryce Wasserman, Monmouth's leader in goal (27), assists (13) and points (40).
Monmouth hoops assistant Ferry off to Pitt
"Coming from Texas (Grapevine Faith Christian School), I knew nothing about the Jersey Shore, nothing about Monmouth," Wasserman said. "When I visited, I loved the place. It's just been so awesome to build this thing up."
Wasserman and his offensive exploits have helped bring Monmouth to uncharted waters, owner of a bull's-eye, two wins from an NCAA Tournament. While Wasserman has emerged as the face of Fisher's program, the defense has been especially stout much of the season.
Hreshko's goals against average of 6.90 is tops in the MAAC, while he ranks in the top 5 in save percentage (.589) and saves per game (9.67). As a team, Monmouth leads the MAAC in caused turnovers per game at 9.20.
As a point of reference, Monmouth outscored the MAAC Tournament field, 29-16, during the regular season, with none of those games being closer than an 11-8 win at the Titans, an associate member of the league on April 15.
"We've been through the ringer, from our first year to some of the close wins this year," Fisher said. "We know it's not going to be easy. There are really good teams in our conference, and there isn't much that separates the top teams in our conference. We're not someone that's going to blow anyone out. We play a good brand of lacrosse that relies on defense and winning close games.
"Winning close games requires having that poise, having that composure, making the plays when it matters, when it's crunch-time."
Staff writer Josh Newman: jnewman@app.com


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text:
If waitresses earned a decent minimum wage, dignity might get a raise
When you are sexually harrassed in a restaurant yet depend on tips, it's harder to voice a protest.
By Annie Quandt
By now, most everyone on social media has seen something about the New Jersey waitress who got an "LOL" as a tip on a $112 check last summer.
While I've never had someone completely stiff me because it took them a while to get their food — the customers' rationale in the New Jersey incident, as they noted on the receipt — I frequently find myself putting up with almost anything from customers in order to get the tips that make up half of my income.
In Maine, 82 percent of all tipped restaurant workers are women, and any woman who has worked for tips will tell you that sexual harassment and rude comments are, sadly, just another part of the job.
When your customers pay your wages instead of your employer, you don't have the luxury of speaking up when you feel uncomfortable or disrespected; if rent is due that week or you have a family to feed, you just have to put up with it.
I've been working at a restaurant in Portland for just about a year now, and I just picked up a second serving job to make ends meet. Recently, two men came in, clearly intoxicated, and sat at their table for an hour and a half trying to look up the waitresses' skirts.
All of the women working that night could feel these men leering and were uncomfortable and anxious the whole shift. When we complained to management, they told us to cut off the men's alcohol consumption, but did nothing else.
These types of incidents are commonplace in the restaurant industry. I have been asked out on dates, with the customer's pen hovering over the tip line as he waited for my answer. I have been asked for my number more times than I can count. I have had customers comment on my outfit or my body while I'm working. I've wanted to say something, but the customer is always right … right?
When female servers can't defend themselves from rude behavior from customers, the entire restaurant culture begins to accept it as the norm. Even management plays a role in harassment in this industry.
If you're not "date ready" when you show up for your shift, in some restaurants, you'll be told to change or unbutton your top or to put on more makeup to make yourself appealing. In my case, the managers have made it clear that the curvier girls are not allowed to wear certain clothing items, while the more slender servers can wear whatever they want to work.
Comments like this about body types and personal style not only make us all feel watched and uncomfortable but also sometimes make it more difficult for us to do our jobs. When I'm sweeping and cleaning and doing outside work in 95-degree heat, the freedom to wear a skirt versus jeans is almost a necessity.
Complaints about sexual harassment from co-workers are rarely taken seriously in restaurants. It is always tough to report unwanted attention or harassment from co-workers or customers, but it is especially difficult if the harassment comes from management.
Where do you turn when the person who holds power over you at your job is the one harassing you? What happens if you do make a formal complaint? The restaurant industry is a tight-knit community, and if any employer thinks you might be a hassle, they won't hire you.
Servers wield so little power in their positions and in their wages, and I am inclined to think that the two are inextricably linked.
According to a Restaurant Opportunities Centers United survey, servers working in states like Maine — where there is a sub-minimum wage for tipped workers — are three times more likely to experience harassment on the job than servers who work in states where everyone makes the same minimum wage.
This is evidence of a systemic problem — combined with the fact that, according to the Equal Employment Opportunity Commission, 7 percent of American women work in restaurants but 37 percent of all EEOC sexual harassment complaints come out of this industry. We're allowing an entire industry full of hardworking women to go to work with the presumption that they will be harassed.
I support the 2016 "wages with dignity" referendum, which would raise the minimum to $12 by 2020 and eliminate the sub-minimum wage for tipped workers by 2024. Earning the same minimum wage as other workers would mean tipped workers wouldn't feel as if they have to ingratiate themselves with their customers regardless of their behavior.
It would mean that management and our co-workers would have to respect us as equals (because when you are paid less, you obviously must be worth less). And it would mean a stable wage for the long winters and tough weekday shifts when servers are more willing to sacrifice dignity at work in order to make ends meet.
I deserve dignity on the job, and one fair minimum wage would help me get it.
Annie Quandt, of Westport Island, is a server working in Portland.
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text:
\section{Introduction}
The starting point for this paper is two simple but beautiful results concerning the probability that two randomly chosen elements of a finite group $G$ commute. If two elements $x,y\in G$ are chosen uniformly at random from $G$ and independently of one another, we define the probability that they commute to be the \emph{degree of commutativity} of $G$, and denote it by $\textup{dc}(G)$.
\begin{theorem}[Gustafson {\cite[\S1]{gustafson}}]\label{gustafson}Let $G$ be a finite group such that $\textup{dc}(G)>\frac{5}{8}$. Then $G$ is abelian.
\end{theorem}
\begin{theorem}[P. M. Neumann {\cite[Theorem 1]{neumann}}]\label{neumann}
Let $G$ be a finite group such that $\textup{dc}(G)\ge\alpha>0$. Then $G$ has a normal subgroup $\Gamma$ of index at most $\alpha^{-1}+1$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{theorem}

There are many natural ways in which one might seek to generalise these results. In this paper we are concerned with generalisations to infinite groups. The first question in this setting is how to define the probability that two group elements commute.

Antol\'in, Martino and Ventura \cite{amv} approach this issue by considering sequences of finitely supported probability measures whose supports converge to the whole of $G$. Given a probability measure $\mu$ on $G$, define the \emph{degree of commutativity} $\textup{dc}_\mu(G)$ of $G$ with respect to $\mu$ via
\[
\textup{dc}_\mu(G)=(\mu\times\mu)(\{(x,y)\in G\times G:xy=yx\}).
\]
Then, given a sequence $M=(\mu_n)_{n=1}^\infty$ of probability measures on $G$, define the \emph{degree of commutativity} $\textup{dc}_M(G)$ of $G$ with respect to $M$ via
\[
\textup{dc}_M(G)=\limsup_{n\to\infty}\textup{dc}_{\mu_n}(G).
\]
Antol\'in, Martino and Ventura \cite[Meta-conjecture 1.3]{amv} suggest the following rather natural analogue of \cref{neumann} for degree of commutativity defined in this way.
\begin{conjecture}[Antol\'in, Martino and Ventura]\label{conj:amv}
For any ``reasonable'' sequence $M=(\mu_n)_{n=1}^\infty$ of probability measures on $G$ we have $\textup{dc}_M(G)>0$ if and only if $G$ is virtually abelian.
\end{conjecture}
They suggest that ``reasonable'' might mean that the measures $\mu_n$ cover $G$ with ``enough homogeneity'' as $n\to\infty$. Of course, the terms ``reasonable'' and ``enough homogeneity'' are somewhat vague, and so they also mention two explicit sequences of measures for which they believe \cref{conj:amv} should hold. The first such sequence is where $S$ is a fixed finite symmetric generating set for $G$, and $\mu_n$ is the uniform measure on the ball of radius $n$ about the identity in $G$ with respect to $S$. In this setting we write $\textup{dc}_S(G)$ instead of $\textup{dc}_M(G)$. The second such sequence is where $\mu$ is some finite probability measure on $G$, and $\mu_n=\mu^{\ast n}$ is defined by letting $\mu^{\ast n}(x)$ be the probability that a random walk of length $n$ on $G$ with respect to $\mu$ ends at $x$. Later, in \cref{def:detect} and \cref{neumann.detect}, we look in more detail at a possible interpretation of what ``enough homogeneity'' might mean in \cref{conj:amv}. First, however, we consider these two explicit sequences of measures.

There has already been some progress towards \cref{conj:amv} in the case of the sequence of uniform measures on balls with respect to a fixed generating set. Antol\'in, Martino and Ventura themselves prove the conjecture completely in that setting for hyperbolic groups \cite[Theorem 1.9]{amv}, whilst Valiunas proves it for right-angled Artin groups \cite[Theorem 1]{valiunas}. Antol\'in, Martino and Ventura also prove the following result, which includes a fairly direct analogue of \cref{gustafson}.
\begin{theorem}[Antol\'in--Martino--Ventura {\cite[Theorem 1.5]{amv}}]\label{amv}
Let $G$ be a finitely generated residually finite group and let $S$ be a finite summetric generating set containing the identity and satisfying
\begin{equation}\label{eq:subexp.growth}
\frac{|S^{n+1}|}{|S^n|}\to1
\end{equation}
as $n\to\infty$. Then $\textup{dc}_S(G)>\frac{5}{8}$ if and only if $G$ is abelian, and $\textup{dc}_S(G)>0$ if and only if $G$ is virtually abelian.
\end{theorem}
In the first main result of the present paper we remove completely from \cref{amv} the hypothesis that $G$ is residually finite, and make the second conclusion more quantitative in the spirit of \cref{neumann}, as follows.
\begin{theorem}\label{main:subexp}
Let $G$ be a finitely generated group, and let $S$ be a finite symmetric generating set containing the identity and satisfying \eqref{eq:subexp.growth}. Then the following hold.
\begin{enumerate}
\item If $\textup{dc}_S(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{dc}_S(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\textup{dc}_S(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{theorem}
We also settle \cref{conj:amv} completely for the case in which the sequence of measures is defined via a symmetric random walk, as follows.
\begin{theorem}\label{main:rw}
Let $G$ be a finitely generated group, and let $\mu$ be a symmetric, finitely supported generating probability measure on $G$ such that $\mu(e)>0$. Let $M=(\mu^{\ast n})_{n=1}^\infty$ be the sequence of measures corresponding to the steps of the random walk on $G$ with respect to $\mu$. Then the following hold.
\begin{enumerate}
\item If $\textup{dc}_M(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{dc}_M(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\textup{dc}_M(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{theorem}
\begin{remark*}
A finitely generated finite-by-abelian group $G$ is virtually abelian, so Theorems \ref{main:subexp} and \ref{main:rw} really do imply the conclusion required by \cref{conj:amv}, and in particular recover \cref{amv} in full. However, the bound on the index of the abelian subgroup of $G$ obtained in this way depends on the rank of $G$ as well as on $\alpha$, whereas the bound given by Theorems \ref{main:subexp} and \ref{main:rw} depends only on $\alpha$, and so the characterisation given by Theorems \ref{main:subexp} and \ref{main:rw} is quantitatively preferable.

This is slightly reminiscent of Gromov's theorem on groups of polynomial growth. Classically, Gromov's theorem says that a group of polynomial growth is virtually nilpotent. However, Breuillard, Green and Tao \cite[Corollary 11.5]{bgt} have given a quantitative refinement of Gromov's theorem showing that a group $G$ of polynomial growth has a finite-index finite-by-nilpotent subgroup. This implies in particular that $G$ is virtually nilpotent, but the bound this gives on the index of a nilpotent subgroup is far worse than the bounds it gives on the index of the finite-by-nilpotent subgroup and the size of the kernel of the projection of that subgroup onto a nilpotent group. As things stand this is a slightly hypothetical remark, since the only known proof of this result renders the bounds in question ineffective, but there are specific classes of group for which effective bounds are known and are much worse for the second characterisation than for the first (see \cite[Corollary 1.8]{resid}, for example). A similar phenomenon occurs in Breuillard and the author's analogous result for finite groups of large diameter \cite[Theorem 4.1]{bt}.
\end{remark*}

Before we move on to consider the meaning of ``enough homogeneity'' in \cref{conj:amv}, let us note that the condition \eqref{eq:subexp.growth} is somewhat suggestive of a particular explicit generalisation, as it implies that the sequence $(S^n)_{n=1}^\infty$ is a \emph{F\o lner sequence} for the group $G$. In general a sequence $(F_n)_{n=1}^\infty$ of finite subsets of $G$ is said to be a \emph{left-F\o lner sequence}, or simply a \emph{F\o lner sequence}, if
\[
\frac{|xF_n\triangle F_n|}{|F_n|}\to0
\]
for every $x\in G$. Even more generally, for a given F\o lner sequence $(F_n)_{n=1}^\infty$, if $\mu_n$ is the uniform probability measure on the set $F_n$ then the sequence $(\mu_n)_{n=1}^\infty$ satisfies the condition
\begin{equation}\label{eq:almost.inv}
\|x\cdot\mu_n-\mu_n\|_1\to0
\end{equation}
for every $x\in G$ (here $x\cdot\mu$ is defined by setting $x\cdot\mu(A)=\mu(x^{-1}A)$). A sequence $(\mu_n)_{n=1}^\infty$ of probability measures satisfying \eqref{eq:almost.inv} is said to be a sequence of \emph{almost-invariant probability measures}. \cref{main:subexp} is therefore a special case of the following result.
\begin{theorem}\label{main:folner}
Let $G$ be a countable amenable group, and let $M=(\mu_n)_{n=1}^\infty$ be a sequence of almost-invariant probability measures on $G$. Then the following hold.
\begin{enumerate}
\item If $\textup{dc}_M(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{dc}_M(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $G$ is finitely generated and $\textup{dc}_M(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{theorem}
It is well known and easy to check that if $G$ is a group of subexponential growth and $S$ is an arbitrary finite symmetric generating set for $G$ containing the identity then there is a sequence $n_1<n_2<\ldots$ such that $(S^{n_i})_{i=1}^\infty$ is a F\o lner sequence for $G$. \cref{main:folner} therefore has the following immediate corollary.
\begin{corollary}
Let $G$ be a finitely generated group of subexponential growth with finite symmetric generating set $S$ containing the identity, and for each $n$ let $\mu_n$ be the uniform probability measure on $S^n$. Then the following hold.
\begin{enumerate}
\item If $\liminf\textup{dc}_{\mu_n}(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\liminf\textup{dc}_{\mu_n}(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\liminf\textup{dc}_{\mu_n}(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{corollary}
Of course, for \cref{conj:amv} to hold in this case we would need to replace the $\liminf$s with $\limsup$s.

Let us emphasise that there are many groups that do not admit generating sets satisfying the condition \eqref{eq:subexp.growth}, but that nonetheless admit sequences of almost-invariant measures. Indeed, it is well known that a countable discrete group admits such a sequence if and only if it belongs to the class of \emph{amenable} groups, which includes all soluble groups, for example, whereas a generating set can satisfy \eqref{eq:subexp.growth} only in a group of subexponential growth.

Unfortunately we do not prove \cref{conj:amv} for the seqence of uniform probability measures on balls with respect to a fixed generating set that does not satisfy \eqref{eq:subexp.growth}. As is noted in \cite{amv}, in order to prove such a result it would be sufficient to prove that if a finite generating set $S$ for a group $G$ did not satisfy \eqref{eq:subexp.growth} then $\textup{dc}_S(G)=0$.

\bigskip
\noindent\textsc{A more precise general statement.} We now suggest a precise version of Antol\'in, Martino and Ventura's idea that \cref{conj:amv} should be satisfied by sequences of measures that cover $G$ with ``enough homogeneity'' in the limit. Specifically, we show that one possible meaning of homogeneity that works in this context is that the measures should not be too concentrated in any coset of a subgroup, in the following sense.
\begin{definition}[uniform detection of index]\label{def:detect}
Let $\pi:(0,1]\to(0,1]$ be a non-decreasing function such that $\pi(\gamma)\to0$ as $\gamma\to0$. We say that a sequence $M=(\mu_n)_{n=1}^\infty$ of probability measures on a group $G$ \emph{detects index uniformly at rate $\pi$} if for every $\varepsilon>0$ there exists $N=N(\varepsilon)\in\mathbb{N}$ such that for every $m\in\mathbb{N}$ if $[G:H]\ge m$ then $\mu_n(H)\le\pi(\frac{1}{m})+\varepsilon$ for every $n\ge N$. We also say simply that $M$ \emph{detects index uniformly} to mean that there exists some $\pi$ such that $M$ detects index uniformly at rate $\pi$.
\end{definition}
The word ``uniform'' in \cref{def:detect} refers to the requirement that the definition be satisfied by the same $N(\varepsilon)$ for all subgroups $H$.
\begin{theorem}[weak Neumann-type theorem for measures that detect index uniformly]\label{neumann.detect}
Let $G$ be a finitely generated group, let $M=(\mu_n)_{n=1}^\infty$ be a sequence of measures on $G$ that detects index uniformly at rate $\pi$, and suppose that $\textup{dc}_M(G)\ge\alpha>0$. Then $G$ has an abelian subgroup of index at most $O_{r,\pi,\alpha}(1)$.
\end{theorem}
\begin{remark*}
In the converse direction, note that if $G$ has an abelian subgroup $H$ such that $\limsup\mu_n(H)\ge\gamma$ then $\textup{dc}_M(G)\ge\gamma^2$.
\end{remark*}
Burillo and Ventura \cite[Proposition 2.4]{bv} show that if $S$ is a finite symmetric generating set for $G$ containing the identity and satisfying \eqref{eq:subexp.growth}, and $M=(\mu_n)_{n=1}^\infty$ is the sequence of probability measures on $G$ defined by setting $\mu_n$ to be the uniform probability measure on $S^n$, then $\mu_n(xH)\to1/[G:H]$ for every subgroup $H$ of $G$ and every $x\in G$ (here, and elsewhere, for notational convenience we take $1/[G:H]$ to be zero when $H$ is an infinite-index subgroup of $G$). This was in turn used in Antol\'in, Martino and Ventura's proof of \cref{amv} \cite{amv}. In the present paper we show that this convergence is uniform over $H$ and $x$, which implies in particular that $M$ detects index uniformly. In order to state this result and others like it more succinctly, we use the following definition.
\begin{definition}[uniform measurement of index]
We say that a sequence $M=(\mu_n)_{n=1}^\infty$ of probability measures on a group $G$ \emph{measures index uniformly} if $\mu_n(xH)\to1/[G:H]$ uniformly over all $x\in G$ and all subgroups $H$ of $G$.
\end{definition}
\begin{remark}\label{measures=>detects}If a sequence of probability measures on a group measures index uniformly then it also detects index uniformly with rate $\iota:(0,1]\to(0,1]$ defined by $\iota(x)=x$.
\end{remark}

\begin{theorem}[almost-invariant measures measure index uniformly]\label{folner.uniform}Let $G$ be a finitely generated group. Then every sequence of almost-invariant measures on $G$ measures index uniformly.
\end{theorem}
\begin{remark*}
By contrast with \cref{folner.uniform}, if $\mu_n$ is the uniform measure on the ball of radius $n$ in a group of exponential growth then the sequence $(\mu_n)_{n=1}^\infty$ may not even \emph{detect} index uniformly. For example, writing $F(x,y)$ for the free group on generators $x,y$, and taking $G=\mathbb{Z}\times F(x,y)$ with generating set $\{(0,e),(\pm1,e),(0,x^{\pm1}),(0,y^{\pm1})\}$, the subgroup $F(x,y)$ has infinite index but $\mu_n(F(x,y))\not\to0$.
\end{remark*}
\begin{theorem}[random walks measure index uniformly]\label{rw.uniform}Let $G$ be a finitely generated group, and let $\mu$ be a symmetric, finitely supported generating probability measure on $G$ such that $\mu(e)>0$. Then the sequence $M_\mu=(\mu^{\ast n})_{n=1}^\infty$ measures index uniformly.
\end{theorem}
\begin{remark*}
The well-known example of $G=\mathbb{Z}$ with $\mu$ the uniform probability measure supported on $\pm1$ and $H=2\mathbb{Z}$ shows that \cref{rw.uniform} is not necessarily true in the absence of the condition $\mu(e)>0$ (or some other aperiodicity assumption).
\end{remark*}

It is this stronger property allows us to give the precise quantitative conclusions of Theorems \ref{main:rw} and \ref{main:folner}, as follows.
\begin{theorem}\label{main:measures}
Let $G$ be a countable group, and let $M=(\mu_n)_{n=1}^\infty$ be a sequence of measures on $G$ that measures index uniformly. Then the following hold.
\begin{enumerate}
\item If $\textup{dc}_M(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{dc}_M(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $G$ is finitely generated and $\textup{dc}_M(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{theorem}

\bigskip
\noindent\textsc{Degree of commutativity with respect to an invariant mean.} We now present a variant of \cref{main:measures} for amenable groups, which is both a natural result in its own right (particularly in the context of \cref{main:folner}) and, as it turns out, an ingredient in the proof of \cref{main:measures} (see \cref{rem:mean} for more details).

By definition, a group is amenable if it admits a \emph{finitely additive left-invariant mean}. A \emph{finitely additive mean} on a group $G$ is a linear functional $\int\diff\mu:\ell^\infty(G)\to\mathbb{R}$ that is positive in the sense that $\int f\diff\mu\ge0$ if $f\ge0$ pointwise, and normalised in the sense that $\int 1_G\diff\mu=1$. Such a mean is said to be \emph{left-invariant} if, defining $g\cdot f$ via $g\cdot f(x)=f(g^{-1}x)$ for $g\in G$, we have $\int g\cdot f\diff\mu=\int f\diff\mu$ for every $f\in\ell^\infty(G)$ and every $g\in G$. In general we abuse notation slightly and denote the mean $\int\diff\mu$ simply by $\mu$. Given a subset $X\subset G$ and a function $f\in\ell^\infty(G)$, we write $1_X$ for the characteristic function of $X$, write
\[
\int_{x\in X}f(x)\diff\mu(x)=\int1_Xf\diff\mu,
\]
and abbreviate
\[
\mu(X)=\int1_X\diff\mu.
\]

It is very natural to define the degree of commutativity on an amenable group $G$ via a finitely additive left-invariant mean $\mu$. Given such a mean on $G$, we define the \emph{product mean} $\mu\times\mu$ on $G\times G$ via
\[
\int_{G\times G}f\diff(\mu\times\mu)=\int_{x\in G}\int_{y\in G}f(x,y)\diff\mu(x)\diff\mu(y).
\]
We then define
\[
\textup{dc}_\mu(G)=(\mu\times\mu)(\{(x,y)\in G\times G:xy=yx\}).
\]

\begin{theorem}\label{main:mean}
Let $G$ be an amenable group with finitely additve left-invariant mean $\mu$. Then the following hold.
\begin{enumerate}
\item If $\textup{dc}_\mu(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{dc}_\mu(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\textup{dc}_\mu(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{theorem}
Let us emphasise that we do not assume in \cref{main:mean} that $G$ is finitely generated.

\begin{prop}[converse to \cref{main:measures} (3) and \cref{main:mean} (3)]\label{neum.converse}
Let $G$ be a countable group with a subgroup $\Gamma$ of index $m$ and a subgroup $H\lhd\Gamma$ of cardinality $d$ such that $\Gamma/H$ is abelian. Then for every finitely additive left-invariant mean $\mu$ on $G$ we have $\textup{dc}_\mu(G)\ge\frac{1}{m^2d}$. Moreover, if $G$ is finitely generated and $M$ is a sequence of probability measures on $G$ that measures index uniformly then $\textup{dc}_M(G)\ge\frac{1}{m^2d}$.
\end{prop}
\begin{remark*}
For non-finitely generated groups it is not the case that finite-by-abelian implies virtually abelian. An example showing this appears on Tao's blog \cite{tao.blog}: if $V$ is an infinite vector space over a finite field $\mathbb{F}$ and $b:V\times V\to\mathbb{F}$ is a non-degenerate anti-symmetric bilinear form then the product set $V\times\mathbb{F}$ with group operation defined by $(v,x)(w,y)=(v+w,x+y+b(v,w))$ for $v,w \in V$ and $x,y\in F$ is finite-by-abelian but not virtually abelian. \cref{neum.converse} therefore shows that the characterisation proposed by \cref{conj:amv} does not hold for groups that are not finitely generated when degree of commutativity is defined with respect to an invariant mean.
\end{remark*}

\bigskip

\noindent\textsc{Independence of degree of commutativity from the method of averaging.} In a forthcoming paper, Antol\'in, Martino and Ventura show that in a virtually abelian group $G$ the value of $\textup{dc}_S(G)$ is independent of the choice of generating set $S$, and that the $\limsup$ appearing in the definition of $\textup{dc}_S(G)$ is actually a limit. This leads immediately to the following corollary of \cref{amv}.
\begin{corollary}[Antol\'in--Martino--Ventura, unpublished]\label{amv.indep}
Let $G$ be a finitely generated residually finite group and let $S,S'$ be two generating sets, each satisfying \eqref{eq:subexp.growth}. Then $\textup{dc}_S(G)=\textup{dc}_{S'}(G)$, and the $\limsup$ appearing in the definition of $\textup{dc}_S(G)$ is actually a limit.
\end{corollary}

Here we prove a similar result, which appears to be a necessary ingredient in our proof of \cref{main:measures} (see \cref{rem:mean}).

\begin{theorem}\label{indep}
Let $G$ be a finitely generated group, and let $M$ be a sequence of probability measures on $G$ that measures index uniformly. Then the $\limsup$ in the definition of $\textup{dc}_M(G)$ is actually a limit, and $\textup{dc}_{M'}(G)=\textup{dc}_M(G)$ for every other sequence $M'$ of probability measures on $G$ that measures index uniformly. Moreover, if $G$ is amenable then $\textup{dc}_\mu(G)=\textup{dc}_M(G)$ for every finitely additive left-invariant mean $\mu$ on $G$.
\end{theorem}

As a by-product of the proofs of our theorems we also obtain the following partial version of \cref{indep} for groups that are not finitely generated.
\begin{corollary}\label{indep.from.mu.and.F}
Let $G$ be a countable group. Then the following hold.
\begin{enumerate}
\item If $\mu$ is a finitely additive left-invariant mean on $G$ such that $\textup{dc}_\mu(G)=0$ then $\textup{dc}_{\mu'}(G)=0$ for every other finitely additive left-invariant mean $\mu'$ on $G$.
\item If $\alpha>\frac{1}{2}$, and if there is a finitely additive left-invariant mean $\mu$ on $G$ such that $\textup{dc}_\mu(G)=\alpha$ or some sequence $M$ of probability measures on $G$ that measures index uniformly such that $\textup{dc}_M(G)=\alpha$, then $\textup{dc}_{\mu'}(G)=\alpha$ for every finitely additive left-invariant mean $\mu'$ on $G$, and $\textup{dc}_{M'}(G)=\alpha$ for every sequence $M'$ of probability measures on $G$ that measures index uniformly. Moreover, the $\limsup$ in the definition of $\textup{dc}_{M'}(G)$ is actually a limit.
\end{enumerate}
\end{corollary}

\begin{remark*}
It follows from \cref{main:mean} and \cref{neum.converse} that in an arbitrary amenable group $G$ the degree of commutativity $\textup{dc}_\mu(G)$ with respect to a given finitely additive left-invariant mean $\mu$ is bounded from below and above in terms of $\textup{dc}_{\mu'}(G)$ for any other finitely additive left-invariant mean $\mu'$. It would be interesting to see whether different invariant means really can give different values of $\textup{dc}$ in situations not governed by \cref{indep} and \cref{indep.from.mu.and.F}, which is to say when they take values in $(0,\frac{1}{2}]$ in a group that is not finitely generated.
\end{remark*}

\bigskip
\noindent\textsc{An application to conjugacy ratios of amenable groups.} Given a finitely generated group $G$ with finite symmetric generating set $S$ containing the identity, and writing $\mathcal{C}(G)$ for the set of conjugacy classes of $G$, Cox \cite{cox} introduces the \emph{conjugacy ratio} $\textup{cr}_S(G)$ of $G$ with respect to $S$ via
\begin{equation}\label{eq:crat}
\textup{cr}_S(G)=\limsup_{n\to\infty}\frac{|\{C\in\mathcal{C}(G):C\cap S^n\ne\varnothing\}|}{|S^n|}.
\end{equation}
For notational convenience, in the present paper we also define a $\liminf$ version
\[
\textup{cr}^\ast_S(G)=\liminf_{n\to\infty}\frac{|\{C\in\mathcal{C}(G):C\cap S^n\ne\varnothing\}|}{|S^n|}.
\]
It is well known that in a finite group the conjugacy ratio coincides with the degree of commutativity; Cox asks whether this remains true for infinite groups \cite[Question 2]{cox}, and also whether $\textup{cr}_S(G)>0$ if and only if $G$ is virtually abelian \cite[Question 1]{cox}. Positive answers to \cref{conj:amv} and the first of these questions would of course imply a postive answer to the second.

Ciobanu, Cox and Martino \cite[Conjecture 1.1]{laura} conjecture explicitly that $\textup{cr}_S(G)>0$ if and only if $G$ is virtually abelian, and prove it for residually finite groups satisfying \eqref{eq:subexp.growth} \cite[Theorem 3.7]{laura}, as well as for hyperbolic groups, lamplighter groups and right-angled Artin groups \cite[\S4]{laura}.

In the present paper we answer both of Cox's questions in the affirmative in the case where $S$ satisfies \eqref{eq:subexp.growth}; in particular, this removes the need for the residually finite hypothesis from the result of Ciobanu, Cox and Martino.

Before we state our results precisely we need some further notation. A sequence $(F_n)_{n=1}^\infty$ of finite subsets of $G$ is said to be a \emph{right-F\o lner sequence} if
\[
\frac{|F_nx\triangle F_n|}{|F_n|}\to0
\]
as $n\to\infty$. We have already noted that if $S$ satisfies \eqref{eq:subexp.growth} then $(S^n)_{n=1}^\infty$ is a left-F\o lner sequence; by symmetry it is also a right-F\o lner sequence.

By analogy with \eqref{eq:crat}, we define the \emph{conjugacy ratio} $\textup{cr}_F(G)$ of $G$ with respect to a F\o lner sequence $F=(F_n)_{n=1}^\infty$ of finite subsets of $G$ via
\[
\textup{cr}_F(G)=\limsup_{n\to\infty}\frac{|\{C\in\mathcal{C}(G):C\cap F_n\ne\varnothing\}|}{|F_n|}.
\]
We also abuse notation slightly and identify $F$ with the sequence $(\mu_n)_{n=1}^\infty$ of uniform probability measures on the sets $F_n$, so that the notation $\textup{dc}_F(G)$ makes sense.

It is an easy exercise to check that in a finite group the conjugacy ratio is always equal to the degree of commutativity. Here we show that this remains true in an infinite group when both are defined with respect to a simultaneous left- and right-F\o lner sequence.
\begin{theorem}\label{cg=dc}
Let $G$ be a finitely generated amenable group, and let $F=(F_n)_{n=1}^\infty$ be both a left- and right-F\o lner sequence for $G$. Then $\textup{cr}_F(G)=\textup{dc}_F(G)$.
\end{theorem}
In particular, combined with \cref{main:folner} this implies the following.
\begin{corollary}\label{cg0.iff.dc0}
Let $G$ be a finitely generated amenable group, and let $F=(F_n)_{n=1}^\infty$ be both a left- and right-F\o lner sequence for $G$. Then the following hold.
\begin{enumerate}
\item If $\textup{cr}_F(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{cr}_F(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\textup{cr}_F(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{corollary}
\begin{remark*}
It is well known that every finitely generated amenable group admits sequences that are simultaneously left- and right-F\o lner, and so \cref{cg=dc} and \cref{cg0.iff.dc0} have content for every finitely generated amenable group.
\end{remark*}
If $S$ is a finite symmetric generating set for $G$ containing the identity and satisfying \eqref{eq:subexp.growth} then $(S^n)_{n=1}^\infty$ is both a left- and right-F\o lner sequence, and so we have the following generalisation of Ciobanu, Cox and Martino's result \cite[Theorem 3.7]{laura}.
\begin{corollary}
Let $G$ be a finitely generated group of subexponential growth, and let $S$ be a finite symmetric generating set containing the identity and satisfying \eqref{eq:subexp.growth}. Then the following hold.
\begin{enumerate}
\item If $\textup{cr}_S(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{cr}_S(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\textup{cr}_S(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{corollary}
Moreover, if $G$ is a finitely generated group of subexponential growth and $S$ is a finite symmetric generating set containing the identity then it is well known and easy to check that some subsequence of $(S^n)_{n=1}^\infty$ is both a left- and right-F\o lner sequence, and so \cref{cg0.iff.dc0} also implies the following.
\begin{corollary}
Let $G$ be a finitely generated group of subexponential growth, and let $S$ be a finite symmetric generating set containing the identity. Then the following hold.
\begin{enumerate}
\item If $\textup{cr}^\ast_S(G)>\frac{5}{8}$ then $G$ is abelian.
\item If $\textup{cr}^\ast_S(G)\ge\frac{1}{2}+\varepsilon$ for some $\varepsilon>0$ then the centre of $G$ has index at most $\frac{1}{\varepsilon}$ in $G$.
\item If $\textup{cr}^\ast_S(G)\ge\alpha>0$ then $G$ has a normal subgroup $\Gamma$ of index at most $\lceil\alpha^{-1}\rceil$ and a normal subgroup $H$ of cardinality at most $\exp(O(\alpha^{-O(1)}))$ such that $H\subset\Gamma$ and $\Gamma/H$ is abelian.
\end{enumerate}
\end{corollary}
\cref{neum.converse} and \cref{cg=dc} combine to give a converse to \cref{cg0.iff.dc0} when $F$ is both a left- and right-F\o lner sequence. However, it turns out to be extremely straightforward to prove this converse directly for just left-F\o lner sequences, as follows.
\begin{prop}[converse to \cref{cg0.iff.dc0}]\label{converse}
Let $G$ be a group with a subgroup $\Gamma$ of index $m$ and a subgroup $H\lhd\Gamma$ of cardinality  $d$ such that $\Gamma/H$ is abelian, and let $F=(F_n)_{n=1}^\infty$ be a left-F\o lner sequence for $G$. Then $\textup{cr}_F(G)\ge\frac{1}{m^2d}$.
\end{prop}
\begin{remark*}
In particular, since a finitely generated finite-by-abelian group $G$ is virtually abelian, if $F$ is both a left- and right-F\o lner sequence then $\textup{cr}_F(G)>0$ if and only if $G$ is virtually abelian.
\end{remark*}

\bigskip
\noindent\textsc{Overall structure of the arguments.} \cref{main:subexp} is a special case of \cref{main:folner}. \cref{main:rw} follows from \cref{rw.uniform} and \cref{main:measures}, whilst \cref{main:folner} follows from \cref{folner.uniform} and \cref{main:measures}.

We prove \cref{neumann.detect} in \cref{sec:weak}, \cref{folner.uniform} in \cref{sec:uniform}, and \cref{rw.uniform} in \cref{sec:uniform.rw}. We prove \cref{indep} in \cref{sec:indep}, and Theorems \ref{main:measures} and \ref{main:mean}, \cref{neum.converse}, and \cref{indep.from.mu.and.F} in \cref{sec:proofs}. Finally, we prove \cref{cg=dc} and \cref{converse} in \cref{sec:cr}.

\bigskip
\noindent\textsc{Acknowledgements}
This project was inspired by an interesting talk on the work \cite{amv} given by Laura Ciobanu at the Universit\'e de Neuch\^atel in May 2017. I am grateful to her both for this talk and for a very useful subsequent discussion. I am also grateful to Itai Benjamini, Emmanuel Breuillard, Ben Green and Alain Valette for helpful conversations, and to Yago Antol\'in and Enric Ventura for helpful comments on earlier versions of this manuscript.

\section{A weak Neumann-type theorem}\label{sec:weak}
In this section we prove \cref{neumann.detect}. We start with the following result, which is based on a simple technique used by P. M. Neumann \cite{neumann} that we use repeatedly throughout this paper.
\begin{prop}\label{large.centraliser.detect}
Let $M=(\mu_n)_{n=1}^\infty$ be a sequence of measures that detects index uniformly at rate $\pi$. Let $\alpha\in(0,1]$, and suppose that $\textup{dc}_M(G)\ge\alpha$. Let $\gamma\in(0,1)$ be such that $\pi(\gamma)<\alpha$, and write
\[
X=\{x\in G:[G:C_G(x)]\le\textstyle\frac{1}{\gamma}\}.
\]
Then $\limsup_{n\to\infty}\mu_n(X)\ge\alpha-\pi(\gamma)$.
\end{prop}
\begin{proof}
By definition of $\textup{dc}_M$ there exists a sequence $n_1<n_2<\ldots$ such that $\textup{dc}_{\mu_{n_i}}(G)\ge\alpha-o(1)$. Writing $\mathbb{E}^{(n)}$ for expectation with respect to $\mu_n$, this means precisely that
\[
\mathbb{E}^{(n_i)}_{x\in G}(\mu_{n_i}(C_G(x)))\ge\alpha-o(1).
\]
Following P. M. Neumann \cite{neumann}, we note that therefore
\begin{align*}
\alpha&\le\mu_{n_i}(X)\mathbb{E}^{(n_i)}_{x\in X}(\mu_{n_i}(C_G(x)))+\mu_{n_i}(G\backslash X)\mathbb{E}^{(n_i)}_{x\in G\backslash X}(\mu_{n_i}(C_G(x)))+o(1)\\
    &\le\mu_{n_i}(X)+\pi(\gamma)+o(1),
\end{align*}
the last inequality being by uniform detection of index, and the result follows.
\end{proof}

\begin{lemma}\label{fin.gen.fin.ind}
Let $m,r\in\mathbb{N}$, and let $G$ be a group generated by $r$ elements. Then $G$ has at most $O_{m,r}(1)$ subgroups of index $m$.
\end{lemma}
\begin{proof}
A subgroup $H$ of index $m$ is the stabiliser of an action of $G$ on the set $G/H$, which has size $m$. There at most $|\text{\textup{Sym}}(m)|^r$ possible actions of $G$ on a set of size $m$, and so there are at most $|\text{\textup{Sym}}(m)|^r$ possibilities for this stabiliser.
\end{proof}

\begin{proof}[Proof of \cref{neumann.detect}]
Let $\gamma=\frac{1}{2}\inf\{\beta\in(0,1]:\pi(\beta)\ge\frac{\alpha}{2}\}$, noting that therefore $\pi(\gamma)<\frac{\alpha}{2}$, and write $X=\{x\in G:[G:C_G(x)]\le\frac{1}{\gamma}\}$, so that \cref{large.centraliser.detect} gives $\limsup_{n\to\infty}\mu_n(X)>\frac{\alpha}{2}$. This implies in particular that the group $\Gamma$ generated by $X$ satisfies $\limsup_{n\to\infty}\mu_n(\Gamma)>\frac{\alpha}{2}$, and hence $[G:\Gamma]\le\frac{1}{\gamma}$ by the definitions of $\gamma$ and uniform detection of index at rate $\pi$.

 \cref{fin.gen.fin.ind} and the definition of $\gamma$ imply that there are at most $O_{r,\pi,\alpha}(1)$ possibilities for $C_G(x)$ with $x\in X$, and so their intersection has index at most $O_{r,\pi,\alpha}(1)$ in $G$. However, this intersection is contained in $C_G(\Gamma)$, and so its intersection with $\Gamma$, which again has index at most $O_{r,\pi,\alpha}(1)$ in $G$, is abelian.
\end{proof}

\section{Uniform measurement of index by almost-invariant measures}\label{sec:uniform}
In this section we prove \cref{folner.uniform}. Throughout the section, in order to avoid having to repeat arguments that are essentially identical for finite- and infinite-index subgroups, we adopt the convention that if $M$ is an infinite quantity (such as the index of a subgroup or the number of vertices in a graph) then $1/M$ takes the value zero.

We start with \cref{folner.uniform}, which shows that sequences of almost-invariant measures measure index uniformly. The only tool we really need other than the almost-invariance of the measures under consideration is the following standard fact.
\begin{lemma}\label{reps.in.bdd.ball}
Let $G$ be a group with finite symmetric generating set $S$ containing the identity, let $H$ be a subgroup of index at least $m\in\mathbb{N}$, and let $x\in G$. Then there exist $y_1,\ldots,y_m\in S^m$ such that the elements $y_ix$ belong to distinct left-cosets of $H$.
\end{lemma}
\begin{proof}
We essentially reproduce the proof of the closely related \cite[Lemma 2.7]{bt}. If $S^{n+1}xH=S^nxH$ for a given $n$ then it follows by induction that $S^rxH=S^nxH$ for every $r\ge n$, and hence that $G=S^nxH$. We may therefore assume that $SxH\subsetneq S^2xH\subsetneq\ldots\subsetneq S^mxH$. However, this implies that the number of left-cosets of $H$ having non-empty intersection with $S^nx$ is strictly increasing for $n=1,\ldots,m$, and so the lemma is proved.
\end{proof}

\begin{proof}[Proof of \cref{folner.uniform}]
It follows from the definition of a sequence of almost-invariant measures that for every $y\in G$ we have
\begin{equation}\label{eq:almost-inv.unif}
|\mu_n(yxH)-\mu_n(xH)|\to0
\end{equation}
uniformly over all $x\in G$ and all subgroups $H$ of $G$. Since this convergence is uniform over all $y\in S^m$, \cref{reps.in.bdd.ball} implies that
\[
\mu_n(xH)\to\frac{1}{[G:H]}
\]
uniformly over all $x\in G$ and all subgroups $H$ of $G$ of index at most $m$. On the other hand, given $\varepsilon>0$, let $m$ be the smallest integer greater than $\frac{1}{\varepsilon}$. If $H$ has index greater than $m$ then \cref{reps.in.bdd.ball} and the fact that the convergence in \eqref{eq:almost-inv.unif} is uniform over $y\in S^{m+1}$ implies that provided $n$ is large enough in terms of $\varepsilon$ we have $\mu_n(xH)<\varepsilon$, and hence in particular
\[
\left|\mu_n(xH)-\frac{1}{[G:H]}\right|\le\varepsilon.
\]
The convergence is therefore uniform over all $x\in G$ and all subgroups $H$ of $G$, as required.
\end{proof}

\section{Uniform measurement of index by random walks}\label{sec:uniform.rw}
In this section we prove \cref{rw.uniform}. Our key tools will be results of Morris and Peres \cite{mp} concerning mixing time and heat-kernel decay of Markov chains. Following the set-up of their paper, let $\{p(x,y)\}$ be the transition probabilities for an irreducible Markov chain on countable state space $V$. A measure $\pi$ on $V$ is said to be \emph{stationary} for $p$ if $\sum_{x\in V}\pi(x)p(x,y)=\pi(y)$ for all $x\in V$; if $V$ is finite we also require $\pi$ to be a probability measure.
\begin{lemma}\label{unif.stationary}
If $p$ is symmetric and $\pi$ is uniform then $\pi$ is stationary for $p$.
\end{lemma}
\begin{proof}
We have $\sum_{x\in V}p(x,y)=\sum_{x\in V}p(y,x)=1$.
\end{proof}
Given a stationary measure $\pi$ on $V$, write $\pi_\ast=\inf_{x\in V}\pi(x)$. For $A\subset V$ write
\[
|\partial A|=\sum_{x\in A,y\notin A}\pi(x)p(x,y),
\]
and define the \emph{conductance} $\Phi_A$ of $A\ne\varnothing$ via
\[
\Phi_A=\frac{|\partial A|}{\pi(A)}.
\]
Define $\Phi:[\pi_\ast,\infty)\to\mathbb{R}$ via
\[
\Phi(r)=\left\{\begin{array}{cl}
                            \inf\{\Phi_A:\pi(A)\le r\} &  \text{if $r\le\frac{1}{2}$}\\
                            \Phi(\frac{1}{2}) & \text{otherwise}
                      \end{array}\right.
\]
if $V$ is finite, or via
\[
\Phi(r)=\inf\{\Phi_A:\pi(A)\le r\}
\]
if $V$ is infinite. Note that in each case irreducibility implies that
\begin{equation}\label{eq:Phi.min}
\Phi(r)\ge\frac{\inf\{\pi(x)p(x,y):p(x,y)>0\}}{r}
\end{equation}
for every $r$.
\begin{theorem}[Morris--Peres \cite{mp}]\label{thm:mp}
Let $\varepsilon>0$, and let $c\in(0,\frac{1}{2}]$ be such that $p(x,x)\ge c$ for all $x\in V$. Then for every
\begin{equation}\label{eq:thm.n.large}
n\ge1+\frac{(1-c)^2}{c^2}\int_{4\pi_\ast}^{4/\varepsilon}\frac{4\diff u}{u\Phi(u)^2}
\end{equation}
we have
\[
\left|\frac{p^n(x,y)-\pi(y)}{\pi(y)}\right|\le\varepsilon
\]
if $V$ is finite \cite[Theorem 5]{mp}, or
\[
\left|\frac{p^n(x,y)}{\pi(y)}\right|\le\varepsilon
\]
if $V$ is infinite \cite[Theorem 2]{mp}.
\end{theorem}
\begin{proof}[Proof of \cref{rw.uniform}]
Write $c=\min_{x\in\text{\textup{supp}}\,\mu}\mu(x)$. Let $\varepsilon>0$ and let $H$ be a subgroup of $G$. We will prove that for every
\begin{equation}\label{eq:n.size}
n\ge1+\frac{32(1-c)^2}{c^4\varepsilon^2}
\end{equation}
we have
\begin{equation}\label{eq:target.finite}
\left|\mu^{\ast n}(xH)-\textstyle\frac{1}{[G:H]}\right|\le\varepsilon
\end{equation}
for every $x\in G$ if $H$ has finite index, or
\begin{equation}\label{eq:target.infinite}
|\mu^{\ast n}(xH)|\le\varepsilon
\end{equation}
for every $x\in G$ if $H$ has infinite index; in particular, this implies the theorem.

The left random walk on $G$ with respect to $\mu$ induces an irreducible Markov chain on the left-cosets $xH$ of $H$. Let $p$ be the transition matrix of this Markov chain. The symmetry of $\mu$ implies that $p$ is symmetric, and hence, by \cref{unif.stationary}, that we may assume that $\pi$ is the uniform probability measure on $G/H$ if $H$ has finite index, or that $\pi$ is everywhere equal to $1$ if $H$ has infinite index. The inequality \eqref{eq:target.finite} is therefore equivalent to the inequality
\[
\left|\frac{p^n(H,xH)-\pi(xH)}{\pi(xH)}\right|\le\varepsilon[G:H],
\]
whilst the inequality \eqref{eq:target.infinite} is equivalent to the inequality
\[
\left|\frac{p^n(H,xH)}{\pi(xH)}\right|\le\varepsilon.
\]
However, since $\inf\{p(xH,yH):p(xH,yH)>0\}\ge c$, it follows from \eqref{eq:Phi.min} that the integrand of \eqref{eq:thm.n.large} is at most $4u[G:H]^2/c^2$ if $H$ has finite index, or at most $4u/c^2$ if $H$ has infinite index, and so \cref{thm:mp} implies that \eqref{eq:target.finite} is satisfied if
\[
n\ge1+\frac{4(1-c)^2[G:H]^2}{c^4}\int_{0}^{4/(\varepsilon[G:H])}u\diff u,
\]
or that \eqref{eq:target.infinite} is satisfied if
\[
n\ge1+\frac{4(1-c)^2}{c^4}\int_{0}^{4/\varepsilon}u\diff u.
\]
In each case this is equivalent to \eqref{eq:n.size}, as reuqired.
\end{proof}

\section{A finite-index subgroup with uniformly bounded conjugacy classes}\label{sec:bfc}
At the heart of P. M. Neumann's original proof of \cref{neumann} lies a result essentially saying that if many elements of $G$ have bounded conjugacy classes then $G$ has a large subgroup in which \emph{every} element has a bounded conjugacy class. In this section we generalise his argument to arbitrary amenable groups as follows.
\begin{prop}\label{indep.cosets}
Let $G$ be an amenable group with finitely additive left-invariant mean $\mu$, let $\alpha\in(0,1]$, and suppose that $\textup{dc}_\mu(G)\ge\alpha$. Given $r\in\mathbb{N}$, write $Y_r=\{x\in G:[G:C_G(x)]\le r\}$, and write $\Gamma_r$ for the group generated by $Y_r$. Then
\begin{enumerate}
\renewcommand{\labelenumi}{(\Alph{enumi})}
\item $[G:\Gamma_r]\le\frac{1}{\alpha-1/r}$, and
\item if $r\ge\frac{2}{\alpha}$ then $[G:C_G(x)]\le r^{6/\alpha+2}$ for every $x\in\Gamma_r$.
\end{enumerate}
\end{prop}
We prove this result shortly. First, let us note that the result that allows Neumann to pass from the uniform bound on the size of conjugacy classes given by \cref{indep.cosets} to the algebraic characterisation given by \cref{neumann} is the following.
\begin{theorem}[P. M. Neumann--Vaughan-Lee {\cite[Theorem 1]{nvl}}]\label{bfc}
Let $G$ be a group in which the size of every conjugacy class is at most $k\in\mathbb{N}$. Then
\[
|[G,G]|\le k^{\frac{1}{2}(3+5\log k)}.
\]
\end{theorem}
\begin{proof}[Remarks on the proof]
The proof in \cite{nvl} assumes that $G$ is finite. However, as is remarked there, one can easily reduce to this case using other known results. First, B. H. Neumann \cite[Theorem 3.1]{b.h.neumann} shows that if an infinite group $G$ has a uniform bound on the size of its conjugacy classes then $|[G,G]|<\infty$. Then, \cite[Lemma 2.1]{macdonald} (which in \cite{macdonald} is attributed to P. Hall) shows that if $|[G,G]|<\infty$ then there exists some finite group $G_0$ whose largest conjugacy class has the same size as the largest conjugacy class of $G$, and such that $|[G_0,G_0]|=|[G,G]|$.
\end{proof}

The first step of P. M. Neumann's original proof of \cref{neumann} is essentially the following observation.
\begin{lemma}\label{large.centraliser}
Let $G$ be an amenable group with finitely additive left-invariant mean $\mu$, let $\alpha\in(0,1]$, and suppose that $\textup{dc}_\mu(G)\ge\alpha$. Let $r>\frac{1}{\alpha}$ and write $X=\{x\in G:[G:C_G(x)]\le r\}$. Then $\mu(X)\ge\alpha-\frac{1}{r}$.
\end{lemma}
\begin{proof}
We have
\begin{align*}
\textup{dc}_\mu(G)&=\int_{x\in X}\mu(C_G(x))\diff\mu(x)+\int_{x\in G\backslash X}\mu(C_G(x))\diff\mu(x)\\
    &\le\mu(X)+\textstyle\frac{1}{r}.
\end{align*}
\end{proof}

\begin{lemma}\label{neum.translates}
Let $G$ be an amenable group with finitely additive left-invariant mean $\mu$, let $X\subset G$ be a symmetric subset containing the identity, and let $m\in\mathbb{N}$. Suppose that
\begin{equation}\label{eq:neum.trans}
\mu(X)\ge\frac{1}{m}.
\end{equation}
Then $\langle X\rangle=X^{3m-1}$.
\end{lemma}
\begin{proof}
We follow Eberhard \cite[Lemma 2.1]{sean}, whose proof for the finite case streamlines part of P. M. Neumann's argument \cite{neumann}. Select for as long as possible a sequence $x_1,x_2,\ldots$ in $G$ such that $x_i\in X^{3i}\backslash X^{3i-1}$. If we have such a sequence $x_1,\ldots,x_j$ then $x_iX\subset X^{3i+1}\backslash X^{3i-2}$, and so the sets $X$ and $x_iX$ are all disjoint. The inequality \eqref{eq:neum.trans} and the left-invariance of $\mu$ therefore imply that $j\le m-1$. It follows that $X^{3m}=X^{3m-1}$, and hence that $\langle X\rangle=X^{3m-1}$, as required.
\end{proof}

\begin{proof}[Proof of \cref{indep.cosets}]
We essentially follow P. M. Neumann \cite[Theorem 1]{neumann}. \cref{large.centraliser} implies that $\mu(Y_r)\ge\alpha-\textstyle\frac{1}{r}$, which implies (A). Moreover, if $r\ge\frac{2}{\alpha}$ then by \cref{neum.translates} it also implies that
\begin{equation}\label{eq:Gamma.in.Y}
\Gamma_r=Y_r^{3\lceil\frac{2}{\alpha}\rceil-1}.
\end{equation}
Writing $k_G(x)$ for the size of the conjugacy class in $G$ of $x\in G$, and noting that $k_G(x)=[G:C_G(x)]$ and $k_G(xy)\le k_G(x)k_G(y)$ for every $x,y\in G$, conclusion (B) then follows from \eqref{eq:Gamma.in.Y} and the definition of $Y_r$.
\end{proof}

\section{Independence of degree of commutativity from the method of averaging}\label{sec:indep}
\begin{proof}[Proof of \cref{indep}]
If $\textup{dc}_M(G)>0$ for some sequence $M$ of probability measures that detects index uniformly then \cref{neumann.detect} and \cref{measures=>detects} imply that $G$ is virtually abelian, and in particular amenable, and it is then trivial that for every finitely additive left-invariant mean $\mu$ we have $\textup{dc}_\mu(G)>0$. We may therefore assume that $G$ is amenable with finitely additive left-invariant mean $\nu$ such that $\textup{dc}_\nu(G)>0$, say $\textup{dc}_\nu(G)=\alpha$. Let $r\in\mathbb{N}$ large enough to apply \cref{indep.cosets} (B) with the mean $\nu$, and define $\Gamma_r$ to be the group generated by $\{x\in G:[G:C_G(x)]\le r\}$ as in that proposition.

Let $\mu$ be an arbitrary finitely additive left-invariant mean on $G$, and let $M=(\mu_n)_{n=1}^\infty$ be an arbitrary sequence of probability measures on $G$ that measures index uniformly, and note that
\begin{equation}\label{eq:pc.split}
\textup{dc}_\mu(G)=\int_{x\in\Gamma_r}\mu(C_G(x))\diff\mu(x)+\int_{x\in G\backslash\Gamma_r}\mu(C_G(x))\diff\mu(x)
\end{equation}
and
\begin{equation}\label{eq:dc.split}
\textup{dc}_{\mu_n}(G)=\sum_{x\in\Gamma_r}\mu_n(C_G(x))\mu_n(x)+\sum_{x\in G\backslash\Gamma_r}\mu_n(C_G(x))\mu_n(x).
\end{equation}
The definition of $\Gamma_r$ implies that the centraliser of any given $x\in G\backslash\Gamma_r$ has index at least $r$, which implies that
\begin{equation}\label{eq:pc.rubbish}
\int_{x\in G\backslash\Gamma_r}\mu(C_G(x))\diff\mu(x)\le\frac{1}{r},
\end{equation}
and also, by definition of uniform measurement of index, that
\begin{equation}\label{eq:dc.rubbish}
\limsup_{n\to\infty}\sum_{x\in G\backslash\Gamma_r}\mu_n(C_G(x))\mu_n(x)\le\frac{1}{r}.
\end{equation}

\cref{indep.cosets} (B) implies that the centraliser of any given $x\in\Gamma_r$ has index at most $r^{6/\alpha+2}$. Since $G$ is finitely generated, \cref{fin.gen.fin.ind} therefore implies that we may enumerate as $C_1,\ldots,C_k$ the finite-index subgroups of $G$ arising as centralisers of elements of $\Gamma_r$. We may then partition $\Gamma_r$ into sets $A_1,\ldots,A_k$ defined by
\[
A_i=\{x\in\Gamma_r:C_G(x)=C_i\},
\]
giving
\begin{equation}\label{eq:int.indep.subgroups}
\int_{x\in\Gamma_r}\mu(C_G(x))\diff\mu(x)=\sum_{i=1}^k\mu(A_i)\mu(C_i)
\end{equation}
and
\begin{equation}\label{eq:Fol.indep.subgroups}
\sum_{x\in\Gamma_r}\mu_n(C_G(x))\mu_n(x)=\sum_{i=1}^k\mu_n(A_i)\mu_n(C_i)
\end{equation}
for every $n$.

However, for each $i$ we have
\[
A_i=C_{\Gamma_r}(C_i)\left\backslash\bigcup_{H\gneqq C_i}C_{\Gamma_r}(H)\right..
\]
Writing $Z$ for the set of subgroups $H\gneqq C_i$ such that $C_{\Gamma_r}(H)$ has infinite index in $G$ and $P$ for the set of subgroups $H\gneqq C_i$ such that $C_{\Gamma_r}(H)$ has finite index in $G$, and noting that each of $Z$ and $P$ is finite by \cref{fin.gen.fin.ind}, we therefore have
\[
A_i=\left(C_{\Gamma_r}(C_i)\left\backslash\bigcup_{H\in P}C_{\Gamma_r}(H)\right.\right)\left\backslash\bigcup_{H\in Z}C_{\Gamma_r}(H)\right.
\]
However, since $P$ is finite the union $\bigcup_{H\in P}C_{\Gamma_r}(H)$ is also the union of finitely many cosets of the finite-index subgroup $\bigcap_{H\in P}C_{\Gamma_r}(H)$, each of which is contained entirely within $C_{\Gamma_r}(C_i)$. This implies that the right-hand side of \eqref{eq:int.indep.subgroups} takes some value $\lambda_r$ that does not depend on the choice of $\mu$, and that the right-hand side of \eqref{eq:Fol.indep.subgroups} converges to $\lambda_r$ as $n\to\infty$ for every $M$ that measures index uniformly. Combined with \eqref{eq:pc.split} and \eqref{eq:pc.rubbish}, this implies that $\lambda_r\to\textup{dc}_\mu(G)$ as $r\to\infty$, whilst combined with \eqref{eq:dc.split} and \eqref{eq:dc.rubbish} it implies that $\textup{dc}_{\mu_n}(G)\to\textup{dc}_M(G)=\textup{dc}_\mu(G)$ as $n\to\infty$.
\end{proof}

\section{Algebraic consequences of a high degree of commutativity}\label{sec:proofs}
In this section we complete the proofs of our main results. We start with \cref{main:measures} (2) and \cref{main:mean} (2), as they are ingredients in the proofs of \cref{main:measures} (1) and \cref{main:mean} (1), respectively.

\begin{proof}[Proof of \cref{main:measures} (2) and \cref{main:mean} (2)]
Given $\delta>0$ define $X_\delta=\{x\in G:[G:C_G(x)]\le2/(1+2\delta)\}$, noting that since $2/(1+2\delta)<2$ the set $X_\delta$ is central. In the setting of \cref{main:mean} (2), \cref{large.centraliser} implies that for $\delta<\varepsilon$ we have $\mu(X_\delta)\ge\varepsilon-\delta$, and so letting $\delta\to0$ we see that $\mu(Z(G))\ge\varepsilon$, and \cref{main:mean} (2) follows. In the setting of \cref{main:measures} (2), \cref{measures=>detects} and \cref{large.centraliser.detect} imply that for $\delta<\varepsilon$ we have $\limsup_{n\to\infty}\mu_n(X_\delta)\ge\varepsilon-\delta$, and so letting $\delta\to0$ we see that $\limsup_{n\to\infty}\mu_n(Z(G))\ge\varepsilon$. \cref{main:measures} (2) then follows from the fact that the $\mu_n$ measure index uniformly.
\end{proof}

The conclusion of \cref{main:measures} (2) and \cref{main:mean} (2) allows us to give an explicit expression for the degree of commutativity, as follows.

\begin{prop}\label{v.central.implies.indep}
Let $G$ be a countable group with a central subgroup $Z$ of finite index, and let $T$ be a left-transversal for $Z$ in $G$. Then if $\mu$ is a finitely additive left-invariant mean on $G$ then
\[
\textup{dc}_\mu(G)=\frac{1}{|G/Z|}\sum_{t\in T}\frac{1}{[G:C_G(t)]}.
\]
Moreover, if $M=(\mu_n)_{n=1}^\infty$ is a sequence of probability measures on $G$ that measure index uniformly then
\[
\textup{dc}_M(G)=\frac{1}{|G/Z|}\sum_{t\in T}\frac{1}{[G:C_G(t)]},
\]
and the $\limsup$ in the definition of $\textup{dc}_M(G)$ is actually a limit.
\end{prop}
\begin{proof}
Elements $x,y\in G$ commute if and only if $xz$ and $y$ commute for every $z\in Z$. This implies that
\begin{align*}
\textup{dc}_\mu(G)&=\sum_{t\in T}\mu(tZ)\mu(C_G(t))\\
     &=\frac{1}{|G/Z|}\sum_{t\in T}\frac{1}{[G:C_G(t)]},
\end{align*}
as required.
It also implies that
\begin{align*}
\textup{dc}_{\mu_n}(G)&=\sum_{t\in T}\mu_n(tZ)\mu_n(C_G(t))\\
     &\to\frac{1}{|G/Z|}\sum_{t\in T}\frac{1}{[G:C_G(t)]} &\text{(by uniform measurement of index)},
\end{align*}
as required.
\end{proof}

\begin{proof}[Proof of \cref{main:measures} (1) and \cref{main:mean} (1)]
In this proof we write $\textup{dc}(G)$ to mean $\textup{dc}_M(G)$ in the context of \cref{main:measures}, or $\textup{dc}_\mu(G)$ in the context of \cref{main:mean}. We may assume that $G$ is not abelian and that $\textup{dc}(G)\ge\frac{5}{8}$, and prove that $\textup{dc}(G)\le\frac{5}{8}$.

Let $T$ be a left-transversal in $G$ of the centre $Z$ of $G$. It follows from \cref{main:measures} (2) or \cref{main:mean} (2), as appropriate, and from \cref{v.central.implies.indep}, that
\[
\textup{dc}(G)=\frac{1}{|G/Z|}\sum_{t\in T}\frac{1}{[G:C_G(t)]}.
\]
Since only one element of $T$ belongs to $Z$, and every other $t\in T$ has $[G:C_G(t)]\ge2$, we conclude that
\begin{align*}
\textup{dc}(G)&\le\frac{1}{|G/Z|}+\frac{|G/Z|-1}{2|G/Z|}\\
     &=\frac{1}{2}+\frac{1}{2|G/Z|}.
\end{align*}
Finally, we observe as Gustafson \cite{gustafson} did that since $G$ is not abelian, the quotient $G/Z$ is neither trivial nor cyclic, and hence $|G/Z|\ge4$. Substituting this into the last inequality then gives $\textup{dc}(G)\le\frac{5}{8}$, as required.
\end{proof}

\begin{proof}[Proof of \cref{main:mean} (3)]
We essentially reproduce P. M. Neumann's proof of \cref{neumann}. Let $r=\frac{1}{\alpha^2}+\frac{1}{\alpha}+1$, let $\Gamma$ be the group $\Gamma_r$ defined in \cref{indep.cosets} and set $H=[\Gamma,\Gamma]$, so that $\Gamma/H$ is abelian. Since the set $Y_r$ generating $\Gamma_r$ in \cref{indep.cosets} is invariant under conjugation by $G$, the subgroup $\Gamma$ is normal in $G$, and since $H$ is characteristic in $\Gamma$ it is normal in $G$.

Conclusion (A) of \cref{indep.cosets} implies that the index of $\Gamma$ in $G$ is less than $\alpha^{-1}+1$, and hence at most $\lceil\alpha^{-1}\rceil$, as required. On the other hand, conclusion (B) implies that the maximum size of a conjugacy class in $\Gamma$ is at most $\exp(O(\alpha^{-O(1)}))$, and so \cref{bfc} implies that the cardinality of $H$ is at most $\exp(O(\alpha^{-O(1)}))$, as required.
\end{proof}

\begin{proof}[Proof of \cref{main:measures} (3)]
Since $G$ is finitely generated, \cref{neumann.detect} and \cref{measures=>detects} imply that $G$ is virtually abelian, and in particular amenable. It follows that $G$ has a finitely additive left-invariant measure $\mu$, and then \cref{indep} implies that $\textup{dc}_\mu(G)=\alpha$. The desired result therefore follows from \cref{main:mean} (3).
\end{proof}

\begin{remark}\label{rem:mean}
In the specific case where $M$ is a sequence of almost-invariant measures, one can prove \cref{main:measures} (3) directly by a similar proof to \cref{main:mean} (3), without using \cref{indep}. However, it does not appear that such an argument is possible when $M$ is assumed more generally to be a sequence of measures that measures index uniformly. This is due to the use of  \cref{indep.cosets} in the proof of \cref{main:mean} (3), and in particular the use of \cref{neum.translates} in the proof of \cref{indep.cosets}. More precisely, \cref{neum.translates} uses the invariance of the invariant mean $\mu$, and whilst the almost-invariance of a sequence of almost-invariant means is enough to prove a version of it, for more general sequences of measures that measure index uniformly there does not appear to be any reason to expect it to hold.

One reason one might be interested in proving \cref{main:measures} (3) directly for almost-invariant measures is that the existence of an invariant mean on a virtually abelian group requires the axiom of choice; one could avoid this in the proof of \cref{main:measures} (3) by replacing \cref{main:mean} (3) with the special case of \cref{main:measures} (3) for almost-invariant measures. We leave it to the interested reader to prove \cref{main:measures} (3) in this way.
\end{remark}

\begin{proof}[Proof of \cref{neum.converse}]
Since $G$ is amenable, \cref{indep} implies that it is sufficient to prove that $\textup{dc}_\mu(G)\ge\frac{1}{m^2d}$ for every finitely additive left-invariant mean $\mu$.

Fix $x\in\Gamma$. The fact that $\Gamma/H$ is abelian implies that $[x,y]\subset H$ for every $y\in\Gamma$. Enumerating the elements of $H$ as $c_1,\ldots,c_d$, we may therefore partition $\Gamma$ into sets $\Gamma_1,\ldots,\Gamma_d$ defined by
\[
\Gamma_i=\{y\in\Gamma:[x,y^{-1}]=c_i\}.
\]
Applying a technique that was used in \cite{nfdl}, for each $i$ with $\Gamma_i\ne\varnothing$ we fix some arbitrary $y_i\in\Gamma_i$ and note that $y_i^{-1}y\in C_G(x)$ for every $y\in\Gamma_i$. In particular,
\begin{equation}\label{eq:partition.central}
y_i^{-1}\Gamma_i\subset C_G(x)
\end{equation}
for every such $i$. Finite additivity and the pigeonhole principle imply that there exists some $i$ with
\begin{align*}
\mu(\Gamma_i)&\ge\frac{1}{d}\mu(\Gamma)\\
    &=\frac{1}{md},
\end{align*}
and then \eqref{eq:partition.central} and left invariance imply that $\mu(C_G(x))\ge\frac{1}{md}$. Since this holds for every $x\in\Gamma$, the proposition is proved.
\end{proof}

\begin{proof}[Proof of \cref{indep.from.mu.and.F}]
Case (1) follows from \cref{main:mean} (3) and \cref{neum.converse}. Case (2) follows from \cref{main:measures} (2), \cref{main:mean} (2) and \cref{v.central.implies.indep}.
\end{proof}

\section{Conjugacy ratios}\label{sec:cr}
In this section we prove \cref{cg=dc} and \cref{converse}. An important ingredient in the proof of \cref{cg=dc} is the following reformulation of the definition of the conjugacy ratio with respect to a left- and right-F\o lner sequence.

\begin{prop}\label{contain.most.classes}
If $F$ is both left- and right-F\o lner then
\[
\frac{|\{C\in\mathcal{C}(G):C\cap F_n\ne\varnothing\text{ and }C\not\subset F_n\}|}{|F_n|}\to 0.
\]
In particular,
\[
\textup{cr}_F(G)=\limsup_{n\to\infty}\frac{|\{C\in\mathcal{C}(G):C\subset F_n\}|}{|F_n|}.
\]
\end{prop}
\begin{proof}
Let $S$ be a finite generating set for $G$. The fact that $F$ is both left- and right-F\o lner implies that
\[
\frac{|S^{-1}F_nS\backslash F_n|}{|F_n|}\to0,
\]
and hence in particular that
\[
\frac{|\{C\in\mathcal{C}(G):C\cap(S^{-1}F_nS\backslash F_n)\ne\varnothing\}|}{|F_n|}\to0.
\]
Note, however, that if $X$ is any subset of a conjugacy class $C$ such that $S^{-1}XS\cap C=X$ then in fact $X=C$, which implies in particular that
\[
\{C\in\mathcal{C}(G):C\cap F_n\ne\varnothing\text{ and }C\not\subset F_n\}\subset\{C\in\mathcal{C}(G):C\cap(S^{-1}F_nS\backslash F_n)\ne\varnothing\},
\]
and so the result follows.
\end{proof}

\begin{proof}[Proof of \cref{cg=dc}]
Writing
\[
\varepsilon_n(x)=\frac{|C_G(x)\cap F_n|}{|F_n|}-\frac{1}{[G:C_G(x)]},
\]
we have
\[
\textup{dc}(F_n)=\frac{1}{|F_n|}\sum_{C\in\mathcal{C}(G):C\cap F_n\ne\varnothing}
                          \left(\sum_{x\in C\cap F_n}\frac{1}{[G:C_G(x)]}\right)+\frac{1}{|F_n|}\sum_{x\in F_n}\varepsilon_n(x),
\]
\cref{folner.uniform} implies that $\varepsilon_n(x)\to0$ uniformly in $x$ as $n\to\infty$, and so this implies that
\begin{equation}\label{eq:dc.with.ccs}
\textup{dc}(F_n)=\frac{1}{|F_n|}\sum_{C\in\mathcal{C}(G):C\cap F_n\ne\varnothing}\left(\sum_{x\in C\cap F_n}\frac{1}{[G:C_G(x)]}\right)+o(1).
\end{equation}
The index of $C_G(x)$ is equal to the cardinality of the conjugacy class of $x$, and so
\[
\sum_{x\in C\cap F_n}\frac{1}{[G:C_G(x)]}\le1
\]
for every $C\in\mathcal{C}(G)$, which with \eqref{eq:dc.with.ccs} implies that $\textup{dc}_F(G)\le\textup{cr}_F(G)$.

On the other hand, \eqref{eq:dc.with.ccs} also implies that
\begin{align*}
\textup{dc}_F(G)&\ge\limsup_{n\to\infty}\frac{1}{|F_n|}\sum_{C\in\mathcal{C}(G):C\subset F_n}\left(\sum_{x\in C}\frac{1}{[G:C_G(x)]}\right)\\
      &=\limsup_{n\to\infty}\frac{1}{|F_n|}\sum_{C\in\mathcal{C}(G):C\subset F_n}\left(\sum_{x\in C}\frac{1}{|C|}\right)\\
      &=\limsup_{n\to\infty}\frac{|\{C\in\mathcal{C}(G):C\subset F_n\}|}{|F_n|}\\
     &=\textup{cr}_F(G),
\end{align*}
the last equality being by \cref{contain.most.classes}.
\end{proof}

\begin{proof}[Proof of \cref{converse}]
Let $x\in\Gamma$. Since $[\Gamma,\Gamma]\subset H$, for every $y\in\Gamma$ we have $y^{-1}xy\in xH$. Fixing a right-transversal $T$ for $\Gamma$ in $G$, for every $t\in T$ and every $y\in\Gamma$ we therefore have $t^{-1}y^{-1}xyt\in t^{-1}xHt$. The conjugacy class of $x$ in $G$ therefore has size at most $|T||H|=md$. Since this is true for every $x\in H$, we conclude that
\[
|\{C\in\mathcal{C}(G):C\cap F_n\ne\varnothing\}|\ge\frac{|\Gamma\cap F_n|}{md}.
\]
Since \cref{folner.uniform} implies that
\[
\frac{|\Gamma\cap F_n|}{|F_n|}\to\frac{1}{m},
\]
the result follows.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Wow what a run...Yesterday against a tough field at the Sam's regional finals in Las Vegaswe managed to win for the third straight contest. This is pretty hard to do and involves some good cooking and luck!
Great job BP! Amazing consistency!
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Attiks is no nonsense company with a highly skilled technical and graphical staff. We handle both on- and offline communication, web development and print design are our middle names. We work internationally and are based in Antwerp, Belgium.
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Japanese vedett site designed by Trainspot and further developed by Attiks.


text:
using System;

public static partial class Extensions
{
    /// <summary>
    ///     A bool extension method that convert this object into a binary representation.
    /// </summary>
    /// <param name="this">The @this to act on.</param>
    /// <returns>A binary represenation of this object.</returns>
    public static byte ToBinary(this bool @this)
    {
        return Convert.ToByte(@this);
    }
}

redpajama_set_name: RedPajamaGithub


text:
Data: US OTT ad spend reached $1.3bn in 2021
As viewership on streaming platforms continues to grow, advertisers are chasing audiences. MediaRadar measures OTT advertising across a sampling of both standalone streaming platforms and TV Anywhere viewing from top linear networks in the US. Among this sample, which includes Hulu, Peacock and other big players in this space, MediaRadar reports that total ad spend in OTT reached nearly $1.3 billion (€1.1bn) in 2021.
On average, OTT represents only 3 per cent of total digital spend per month in the US — however this market is growing considerably.
"Advertisers are increasingly enhancing their ad buys with OTT," said Todd Krizelman, CEO & Co-Founder, MediaRadar. "The rise in viewership coupled with limited advertising per show is a great place for brands to get noticed and not be lost among many ad messages."
Key findings include:
In Q4 2021, Discovery+, HBO Max, Hulu, Paramount+, and Peacock accounted for 65 per cent of total tracked OTT spend (nearly $438 million).
In 2021, 5,530 total advertisers ran ads across these five platforms. Across this set of advertisers, there were almost 500,000 ads being placed in the second half of 2021.
There was an increase of 22 per cent digital spend in Q4 from Q3. Q3 went slightly down to -2 per cent from Q2. Q2 was up 14 per cent from Q1.
Additionally, MediaRadar found that the top 10 categories totaled 81 per cent of the overall quarter spend in Q4 2021.
Key highlights include:
Retail was the top Q4 category buying OTT ads – representing 16 per cent of the quarter's market. This was primarily driven by discount retailers, used car retailers, along with online and brick and mortar department stores. Retail saw a 57 per cent increase from Q3.
Media experienced a 46 per cent increase from Q3. Subscription streaming services, TV dramas, mobile apps and fantasy sports leagues contributed heavily to media holding 14 per cent spend of Q4's total.
Insurance, credit cards and banks contributed to finance also being 14 per cent of the quarter's overall spend, which had a 31 per cent increase from Q3.
Analysing the top industry categories' advertising on OTT channels, the data revealed that all of the five OTT channels are dependent on these same industry categories: Media & Entertainment, Retail, Technology, Financial & Real Estate and Pharma.
Media advertising is the largest ad segment within OTT on the top five providers. Media includes movies, video games, television, and streaming. In total, it accounts for 19 per cent to 33 per cent of total ads running on any given channel.
Retail advertising is 22 per cent of Discovery+, which is nearly 2x more than any other channel.
Financial & Insurance brands represent 19 per cent of HBO Max's advertising volume. This is 2-3x more than most other leading OTT channels.
Pharma advertisers are making a push in Paramount+ and Hulu. As CBS has an older audience in TV, they are well suited to pharma products.
Discovery+ sells 5x more food advertising than other OTT channels followed by Hulu. Food advertising includes CPG products from cereal to soup.
House ads, which refer to the inventory used to promote programming within the same parent company, represent 16 per cent of total ad spots run. House advertising is highest immediately after a new channel launches, but has been declining over time with OTT channels getting better at selling their inventory.
Data: Peacock ad spend up 115%
Data: UK advertisers increase TV spend in pandemic
Data: Lockdown shreds UK ad spend
Data: Marketing ad spend in recovery
Data: Record January ad-spend in US
Categories: Advertising, Articles, Markets, OTT, Research
Tags: advertising, discovery, discovery plus, HBO Max, Hulu, mediaradar, OTT, Paramount Plus, Peacock


Above is a resume sample for a Medial Sales Executive. Use this resume example as a guide when writing yours. Your resume should begin with your personal contact information, your name, address and phone number(s). From here, list a brief summary of your job qualifications – why you are the right person for the job. After this section, list your core skills – items such as experience in medical sales, negotiation skills, strategic planning experience, etc. Note these are listed in bullet point format. Next is the section outlining your professional experience. Jobs are listed in order, starting with your most recent position. Education and computer skills are listed at the bottom.


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text:
Q: WPF datagrid is not empty on instantiation causing error I am having an issue in a window for a WPF app I am creating. I am getting an error  when I tried to bind an ObservableCollection to a DataGrid. The thing that is confusing to me is that the datagrid seems like it is instantiating with a value already in it which causes the following error:

An unhandled exception of type 'System.InvalidOperationException' occurred in PresentationFramework.dll
  Items collection must be empty before using ItemsSource.

XAML code:
<Window x:Class="Store_grade_list_bulk_updater.POModifyWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
    xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
    xmlns:local="clr-namespace:Store_grade_list_bulk_updater"
    mc:Ignorable="d"
    Title="Modify Or Remove Grades & PO Numbers" Height="450" Width="850"
    KeyDown="Escape_KeyDown">
<Grid>
    <Grid.ColumnDefinitions>
        <ColumnDefinition Width="*"/>
        <ColumnDefinition Width="5*"/>
    </Grid.ColumnDefinitions>
    <StackPanel Grid.Column="0"
                VerticalAlignment="Top"
                HorizontalAlignment="Center"
                Margin="0,30,0,0">
        <TextBlock>
            Grade Code:
        </TextBlock>
        <TextBox
            x:Name="GradeBox"
            KeyDown="Grade_KeyDown">

        </TextBox>
        <TextBlock
            Margin="0,10,0,0">
            Purchase Order Code:
        </TextBlock>
        <TextBox
            x:Name="POBOx"
            KeyDown="PO_KeyDown">

        </TextBox>
        <Button
            Margin="0,20,0,0"
            Content="Modify"
            Click="Modify_Click"/>
        <Button
            Margin="0,15,0,0"
            Content="Close Window"
            Click="CloseWindow_Click"/>
    </StackPanel>
    <Grid Grid.Column="1">
        <Grid.ColumnDefinitions>
            <ColumnDefinition Width="*"/>
            <ColumnDefinition Width="2*"/>
        </Grid.ColumnDefinitions>
        <Grid.RowDefinitions>
            <RowDefinition Height="2*"/>
            <RowDefinition Height="3*"/>
        </Grid.RowDefinitions>
        <DataGrid
            Grid.Column="0"
            x:Name="Modify_GradeList"
            Margin="10"
            AutoGenerateColumns="False"
            ItemsSource="{Binding}"
            IsReadOnly="True"
            CanUserAddRows="False"
            SelectionChanged="GradeList_Select">
            <DataGridTextColumn Header="Grade Code" Binding="{Binding Path=Grade}"/>

        </DataGrid>
        <DataGrid
            Grid.Column="1"
            x:Name="Modify_PO_List"
            Margin="10"
            AutoGenerateColumns="False"
            ItemsSource="{Binding}"
            IsReadOnly="True"
            SelectionChanged="PO_List_Select">
            <DataGridTextColumn Header="PO Code" Binding="{Binding PO_Code}" Width="3*"/>
            <DataGridTextColumn Header="Country" Binding="{Binding Country}" Width="*"/>
            <DataGridTextColumn Header="Stores" Binding="{Binding Store_Count}" Width="*"/>

        </DataGrid>
        <DataGrid
            Grid.Column="0"
            Grid.ColumnSpan="2"
            Grid.Row="1"
            x:Name="Store_List"
            Margin="10,0,10,10"
            ItemsSource="{Binding}"
            AutoGenerateColumns="False"
            IsReadOnly="True">
            <DataGridTextColumn Header="Code" Binding="{Binding Store_Code}" Width="*"/>
            <DataGridTextColumn Header="Code" Binding="{Binding Store_Name}" Width="4*"/>
            <DataGridTextColumn Header="Code" Binding="{Binding Grade_Rank}" Width="*"/>

        </DataGrid>
    </Grid>
</Grid>

C# code:
using System;
using System.Collections.ObjectModel;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Input;

namespace Store_grade_list_bulk_updater
{

public partial class POModifyWindow : Window
    {
        public ObservableCollection<GradeTemplate.Template> modify_grades = new ObservableCollection<GradeTemplate.Template>();
        public ObservableCollection<GradeTemplate.PO> modify_po_list = new ObservableCollection<GradeTemplate.PO>();
        public ObservableCollection<GradeTemplate.Grade> modify_grade_details = new ObservableCollection<GradeTemplate.Grade>();

        public POModifyWindow()
        {
            InitializeComponent();
            MainWindow tempMain = (MainWindow)Application.Current.MainWindow;

            // This returns "1"
            MessageBox.Show(Modify_GradeList.Items.Count.ToString());

            // Troubleshooting step
            Modify_GradeList.ItemsSource = modify_grades;
            // End troubleshooting

            modify_grades = tempMain.gradelist;
            Modify_GradeList.ItemsSource = modify_grades;
            Modify_PO_List.ItemsSource = modify_po_list;
            Store_List.ItemsSource = modify_grade_details;
        }

        public void Modify_Click(object sender, EventArgs e)
        {

        }

        public void CloseWindow_Click(object sender, EventArgs e)
        {
            MainWindow tempMain = (MainWindow)Application.Current.MainWindow;
            tempMain.ReceiveTemplates(modify_grades);
            Close();

        }

        public void GradeList_Select (object sender, EventArgs e)
        {
            modify_po_list = Modify_GradeList.SelectedItem as ObservableCollection<GradeTemplate.PO>;
            Modify_PO_List.ItemsSource = modify_po_list;
            GradeTemplate.Template textbox_temp = Modify_GradeList.SelectedItem as GradeTemplate.Template;
            GradeBox.Text = textbox_temp.Grade;
        }

        public void PO_List_Select(object sender, EventArgs e)
        {

        }

        public void Escape_KeyDown(object sender, KeyEventArgs e)
        {
            if (e.Key == Key.Escape)
            {
                MainWindow tempMain = (MainWindow)Application.Current.MainWindow;
                tempMain.ReceiveTemplates(modify_grades);
                Close();
            }
        }

        public void Grade_KeyDown(object sender, KeyEventArgs e)
        {
            if (e.Key == Key.Enter)
            {
                TextBox s = e.Source as TextBox;
                s.MoveFocus(new TraversalRequest(FocusNavigationDirection.Down));
            }

        }

        public void PO_KeyDown(object sender, KeyEventArgs e)
        {
            if (e.Key == Key.Enter)
            {
                GradeBox.Clear();
                POBOx.Clear();
                TextBox s = e.Source as TextBox;
                s.MoveFocus(new TraversalRequest(FocusNavigationDirection.Up));
            }
        }
    }
}

Originally I thought it was due to trying to bind a collection with objects already in it, although this already worked in another window, but when I added in the line of code marked as troubleshooting it still causes the error. From what I can tell it seems to be related to the XAML somehow, as the error only started occurring after I added in the text columns to try and have a consistent format, but I cannot tell where it is occurring or why.

A: If you want to create a column in DataGrid, you need to have <DataGrid.Columns>
and inside you need to define the columns.
by directly, defining the column, it assumes it has data, so it thinks you are providing the itemSource twice and thus the error :-  Collection must be empty before usingItemsSource
        <DataGrid
        Grid.Column="0"
        x:Name="Modify_GradeList"
        Margin="10"
        AutoGenerateColumns="False"
        ItemsSource="{Binding}"
        IsReadOnly="True"
        CanUserAddRows="False"
        SelectionChanged="GradeList_Select">
            <DataGrid.Columns>
            <DataGridTextColumn Header="Grade Code" Binding="{Binding Path=Grade}"/>
            </DataGrid.Columns>
        </DataGrid>

Do the same for other DataGrid.

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