text:
TCA Jump-Ahead: "Curb Your Enthusiasm"
Posted by Will Harris (07/31/2009 @ 12:13 pm)
It occurs to me that, although I'm trying my best to cover the TCA tour in a chronological manner, there are some panels that you'd like to know about more quickly than I might otherwise get to them. As such, I'm instituting a new category called the TCA Jump-Ahead.
First up: "Curb Your Enthusiasm."
It's kind of a running joke in the entertainment industry about how every season of "Curb" should be considered the last season of "Curb" until Larry David says otherwise…and, generally, Larry David is glad to tell you that he's done, he has no more ideas, and he can't be bothered to try and think of any. Thankfully, David announced last year that he would indeed be moving forward with a seventh season of the cringe-worthy comedy (and I mean that in the best possible way), and since then, there have been multiple rumblings about how various members of the cast of "Seinfeld" would be popping up. In the "Curb" panel yesterday, David finally provided some context to exactly how they'd be appearing.
"For years, I've been asked about a 'Seinfeld' reunion, as has Jerry and the other cast members," explained David, "and I would always say, 'No, there's no reunion. There's not going to be a reunion show. We would never do that. It's a lame idea.' And then I thought, 'But it might be very funny to do that on 'Curb.' And I kept thinking about the idea. I started to think of different scenarios and how we could pull this off. I called Jerry, and Jerry was game. And I said, 'Well, I'll call the others,' and I did. And we did it. So we're doing a 'Seinfeld' reunion show on 'Curb.' We're going to see writing. We'll see aspects of the read-through, parts of rehearsals. You'll see the show being filmed. And you'll see it on TV.
"What will you see? You won't see the entire show; you'll see parts of the show. You will get an idea of what happened 11 years later. And within the show, it will be incorporated into regular 'Curb' episodes. So the cast members will be playing themselves on 'Curb' while all this is going on. You're not going to see a 'Seinfeld' show from beginning to end, but you will see parts of the show."
And will there be any reference to Michael Richards' sordid post-"Seinfeld" problems?
"It's possible."
The reunion is scattered through the season, and by David's admittedly questionable recollection, the cast will be on five shows, though they won't all be on the five shows. ("Jerry's on five shows, I think," he said. "The others will be on at least four. Maybe one or two of the others will be on five. I'm not sure.") The season finale will be about the reunion show and will possibly be an hour long, though David admits that he hasn't finished editing it yet and can't say for sure.
There's just one thing, though: anyone who's been watching "Curb" for the previous six seasons has to figure that the odds look good for Larry – the TV Larry, that is – to somehow screw up this reunion.
"He might," said David. "Do you need a staff job for next season? My guy might consider wrecking something like that, yeah. We'll see what happens. My guy could very well wreck it. I'm not saying he did…"
Want a few more tidbits about the upcoming season…?
The obvious: Jeff will be around, Susie will still be giving Larry loads of shit, and Cheryl will be back, of course. But we'll also get to see Larry dating a bit. (We saw a clip of Larry trying to instigate a sexual encounter with a woman in a wheelchair that was non-stop slapstick.)
There will be several other guest stars during the course of the season – count on seeing Richard Lewis, Ted Danson and Mary Steenburgen, Bob Einstein, Vivica A. Fox, J.B. Smoove, Sharon Lawrence, Catherine O'Hara, Meg Ryan, Elisabeth Shue, Christian Slater, and Sherry Stringfield – but the most high-profile during the clips that we saw was Rosie O'Donnell, who gets into a physical altercation with Larry in a restaurant.
"Rosie O'Donnell was hilarious in the show, really funny," said David. "She is in episode 5 and she is a riot. Let me tell you something, she's a force. She is very formidable, physically."
Lastly, I wanted to make mention of the very first scene that we were shown. It showed Larry walking down the street, looking as he always does, and a woman who passes him says, "Smile!" You can imagine how well the advice goes over.
As it turns out, it was more or less inspired by true-life events.
"Yeah, that's happened to me," David admitted. "I would walk down the street scowling because something was bothering me, and this happened a number of times where people would just say, 'Smile!' It's a crazy thing to say to somebody. Mind your own business! What if I yelled out, 'I just found out I have cancer, okay? So shut your fucking mouth!"
Good ol' Larry. Man, are we glad to have you back.
Posted in: Curb Your Enthusiasm, Interviews, News, TCA Blog 2009, TCA Press Tour, TV, TV Comedies
Tags: Cheryl Hines, Curb Your Enthusiasm, Curb Your Enthusiasm blog, HBO, Jason Alexander, Jeff Garlin, Jerry Seinfeld, Julia Louis-Dreyfus, Larry David, Michael Richards, Rosie O'Donnell, Susie Essman, TCA Blog 2009, TCA Press Tour
"Seinfeld" cast to reunite…sort of
Old Show, New Season: "Curb Your Enthusiasm"
The 2010 Primetime Emmy nominations are in!
TCA Tour: "Hung"
TCA Tour, Day 3: "Bored to Death"


text:
Battle Array  (, aṣ-Ṣaff, aka "The Ranks") is the 61st chapter (sūrah) of the Quran, with 14 verses (āyāt). This sura is an Al-Musabbihat sura because it begins with the glorification of Allah.

Summary
1 All things in the universe praise God
2-4 Muslims exhorted to be faithful and to fight for Islam
5 This exhortation enforced by the example of Moses
6 Jesus foretells a Messenger named Ahmad
7-8 Jesus was rejected as a sorcerer notwithstanding his miracles
9 Islam to be exalted above every other religion
10-11 Muslims exhorted to seek wealth by fighting for Islam
12-13 The rewards of those who fight for the faith
14 Muslims exhorted to follow the example of the apostles of Jesus

Hadith
The first and foremost exegesis/tafsir of the Qur'an is found in hadith of Muhammad. Although some scholars, including ibn Taymiyyah, claim that Muhammad has commented on the whole of the Qur'an, others such as Ghazali cite the limited amount of narratives, thus indicating that he has commented only on a portion of the Qur'an. Ḥadīth (حديث) is literally "speech" or "report", that is a recorded saying or tradition of Muhammad validated by isnad; with Sirah Rasul Allah these comprise the sunnah and reveal shariah. According to Aishah, the life of Muhammad was practical implementation of Qur'an. Therefore, mention in hadith elevates the importance of the pertinent surah from a certain perspective.

 Abdullah ibn Salam narrated: "A group of us Companions of the Messenger of Allah (ﷺ) sat talking, and we said: 'If we knew which deed was most beloved to Allah then we would do it.' So Allah, Most High, revealed: Whatsoever is in the heavens and whatsoever is on the earth glorifies Allah. And He is Almighty, the All-Wise. O you who believe! Why do you say that which you do not do? As-Saff (1-2)"...

Notes

References

Saff
Jesus in Islam

fi:Suura#Luettelo

redpajama_set_name: RedPajamaWikipedia


b. in case the winner wishes to receive acknowledgement of winnings, winnings are paid with a payment order issued by a bank or through the new procedure of payment of winnings to the bankaccountsof beneficiariesthat extraordinarily applies from 7/7/2015 (more information on https://corporate.opap.gr/en/oria).
3. For amounts above 1.500,01 EUR, payment will be made exclusively according to what is described in paragraph 2b.


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Home » Features » Food Trends We're Over, And What We're Looking Forward To Eating In 2016
Food Trends We're Over, And What We're Looking Forward To Eating In 2016
by Heather Carney | March 2016 | DIGITAL EDITION
Gold Coast magazine editors talk about food trends from the past year—their favorites and what they're totally over—and what they're most excited to see from the restaurant industry in 2016.
Heather Carney, Managing Editor
Craziest thing you ate this past year: Squid ink pasta at Corsair in Aventura. I also tried frog legs at Pistach in West Palm Beach.
Most memorable dish:The Kobe Beef dinner at NYY Steak at the Seminole Casino Coconut Creek. I've never had steak prepared so many different ways and with so much flavor. Its even marbling was mesmerizing.
Food trend you're most excited for: Simple yet sophisticated food, like the dishes from chef Philip Darmon at Hardy Park Bistro and chef Matthew Byrne at Kitchen in Palm Beach.
Trend you're over: The Southern, farmhouse-style restaurants. I love Southern comfort food, but I'm ready for something new.
Favorite food trend of this past year: I saw burrata and tomato salad on so many menus and loved it. You can't go wrong with a cheese made from mozzarella and cream.
Alyssa Morlacci, Associate Editor
Craziest thing you ate this past year: For a not-so adventurous eater, bone marrow was a courageous texture for me to try. It left the kitchen at Adena Grill still attached to the bone. We scraped it off and spread it on bruschetta toast like jelly.
Most memorable dish: This year I was introduced to the Nauti Dawg Marina Cafe in Lighthouse Point. The low-key spot makes an earth-shattering coconut French toast for brunch.
Food trend you're most excited for: I love the push for health-conscious menus. Less preservatives, additives and hormones is awesome because it means I can eat out more and cook less.
Trend you're over: No more pork belly, please!
Favorite food trend of this past year: It isn't a stroke of genius or anything, but I enjoyed seeing watermelon as a salad topping. Talk about refreshing!
Ileana Llorens, Senior Web Editor
Craziest thing you ate this past year: I'd love to say I branched out and tried something completely out of the ordinary, but for me, this year was all about broadening my horizons with new vegetables and more fish. Hey, baby steps!
Most memorable dish: Hands down the warm lemon madeleines at Cafe Boulud in Palm Beach. They were so light and delicate. I'm not a dessert person, and I couldn't help but sneak an extra two or three when I was left alone at the table. They were life-changing and definitely worth the drive—even if they're the only thing you order.
Food trend you're most excited for: I really love all the restaurants opening up in hotels along the beach. Going out to dinner suddenly becomes a mini escape.
Trend you're over: If we could all agree that kale is overrated and has no place in a smoothie that'd be great. I also don't understand the culinary world's current obsession with cauliflower. A friend recently told me she saw someone on Instagram call a cauliflower steak a cheat-day meal.
Favorite food trend of this past year: I loved the focus on craft cocktails. There's nothing I like more than trying new concoctions. The Kentucky Radler at Hyde Beach Kitchen + Cocktails, for example, mixes whiskey and beer, and the result is delicious. Who knew?!
Lyssa Goldberg, Web Editor
Craziest thing you ate this past year: Grasshoppers. I tried them at Dos Caminos on Fort Lauderdale Beach, where they serve guac topped with chapulines (grasshoppers), which is a popular pairing in Mexico.
Most memorable dish: Pasta carbonara at Taverna Trilussa in Rome during my trip to Italy. More than six months later, I'm still daydreaming about that bowl of pasta. Luckily, Louie Bossi's on Las Olas makes a mean carbonara as well.
Food trend you're most excited for: More clean-eating experiences you don't have to feel guilty about, like DIRT in Miami.
Trend you're over: Prohibition-themed everything. It's a cool aesthetic, but every new restaurant and bar that opens nowadays is doing it, and it's starting to look unoriginal and overdone.
Favorite food trend of this past year: "Middleterranean" cuisine (a blend of Mediterranean and Middle Eastern food options). I fell in love with Mediterranean/Middle Eastern food during a trip to Israel and have been obsessed ever since. I was so happy to see so many Middleterranean restaurants open last year and am excited about others that are still coming.
The 14 Best Restaurants In Fort Lauderdale And Beyond
4 Local Chefs Who Just Might Be The Next Big Stars Of South Florida's Dining Scene
Where To Eat In Fort Lauderdale, According To The Chefs Who Work Here

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: breaking the dataframe and creating new columns for the broken chunks of data in R I have a data frame like this:
Values
1
2
3
.
.
1000

I want to break it into a data frame with 10 columns based on rows. So rows 1-100 goes in one column, rows 101-200 goes in the second column and so on resulting in the final data set as below:
Col1 Col2............... Col10
1    101                  901
2    102                  902
3    103                  903
.     .
.     .
100  200                  1000

A: Here is one method:
dfNew <- data.frame(matrix(unlist(df), 100))
names(dfNew) <- paste0("Col", 1:10)

The first line coerces the data.frame to a vector (unlist), then to a matrix with rows of length 100 (matrix). This matrix is then converted to a data.frame (data.frame).
Here are the first 6 rows:
 head(dfNew)
  Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10
1    1  101  201  301  401  501  601  701  801   901
2    2  102  202  302  402  502  602  702  802   902
3    3  103  203  303  403  503  603  703  803   903
4    4  104  204  304  404  504  604  704  804   904
5    5  105  205  305  405  505  605  705  805   905
6    6  106  206  306  406  506  606  706  806   906

The second line adds the data.frame names.
data
df <- data.frame(Values=1:1000)

A: An alternative is using the do.call which execute and construct a function call coming from rbind which takes a sequence of list consturcted by lapply arguments and combine it by columns
df <- data.frame(Values=1:1000)
mydf <- data.frame(do.call(rbind,lapply(df,matrix, ncol=10)))
names(mydf) <- paste0("Col", 1:10)

If you don't like to use paste0 to set the column names, you can use sprintf
names(mydf) <-sprintf("Col%d", 1:10)

> head(mydf,5)
#  Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10
#1    1  101  201  301  401  501  601  701  801   901
#2    2  102  202  302  402  502  602  702  802   902
#3    3  103  203  303  403  503  603  703  803   903
#4    4  104  204  304  404  504  604  704  804   904
#5    5  105  205  305  405  505  605  705  805   905

redpajama_set_name: RedPajamaStackExchange


text:
Hypopta variegata is een vlinder uit de familie van de houtboorders (Cossidae). De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1924 door Köhler.

Houtboorders


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Besides enjoying my MA New Media and doing research on Web 2.0 developments in China at the UvA I am working 2 days a week as a project assistant at KREM (www.KREM.nl). KREM is a web 2.0 oriented company which specializes in so called corporate social networks. Movies: Koyaanisqatsi, O Brother where art thou?, Amores Perros. When not studying: Golf!, running, traveling, China, Web 2.0. Books: Life of Pi, War and Peace, The World is Flat. Furthermore I am currently doing research on Web 2.0 in China and Chinese Web companies expanding overseas. For this MA thesis research I will travel to China on the 14th of April to interview the most prominent Chinese Web companies that have the ambition to expand internationally.
This post contains parts of a 2 hour interview I did with Tangos Chan from China Web2.0 Review. Among others I asked him about his views on the globalization of Chinese Internet companies, the future of the Chinese Internet market, and the influence of VCs on innovation in China.
Tangos Chan: Actually most of the Chinese companies will not consider a globalization strategy so far because the Chinese market is huge. So if they can serve this market well they could become very big companies. Look at for example Baidu and Tencent. Of course there is also the language barrier. I think an English language service in the US such as Google is automatically a global service because it serves all the English speaking people. In China companies are bound to be a Chinese service. So I think there a really 2 kinds of services that will consider to go abroad. Firstly it is the very big companies such as Baidu, Alibaba, and Tencent. These companies will start to consider to get more users in overseas countries. I think most of these companies will start from some developing countries. Or some eastern Asian countries such as Japan or maybe they are also considering Korea. I think Tencent actually has a dedicated team to learn from new applications in Korea and copy it to Tencent.
Secondly, another kind of service is the new start-ups that have been established by returnees. Or established by someone who has worked for a big international company such as Microsoft or Google. So someone who has international market experience. These kinds of people will launch their service with an English version from the first day.
You should check out a web service called Mojiti. I am not sure whether this start-up is still alive now, it is actually established by one of the ex-Microsoft guys. The service is about video, you can make subtitles or add text on a video. The service was acquired by Hulu, a video-site launched by NBC and News Corp. So the whole Mojiti team is working for Hulu now. I think Hulu just thought that the Mojiti technology was good so they have acquired the team for their technology background. I think these kinds of services will consider a global entry from their first day.
PPW: So do you think that there are enough innovative ideas that stand a chance internationally?
Tangos: I think p2p video streaming technology is quite good in China. So services as PPLive and PPStream. They are leading in the global market. I heard that France Telecom is currently partnering with a Chinese company in order to try to implement the technology in the French market. So I think that there is innovation in the Chinese market but it is still only a small percentage. Most of the Chinese companies just copy ideas from countries overseas, they copy many ideas from the Silicon Valley. But I think that for p2p video-streaming China is quite innovative.
Also IM bots is something that China is good at. So you chat with a robot with your IM service. There is a company called Xiaoi that is the most well known IM bot in China. They are the official partner of Windows Live Messenger in China. So if you want to check the weather you just ask a question to the bot. They received plenty of VC funding, I believe IDT has invested in them. I have blogged about this a while ago. Actually in China there are also some companies that offer IM bots where you can order your lunch. You can just type in what kind of food you want for lunch and maybe half an hour later it is delivered. So you just ad this bot as a contact in your MSN. So then you just type in what you want to order.
Also gaming companies like Shanda and Perfect World are going abroad. I believe Perfect World is besides their operations in the US also starting in Europe. Tencent is also cooperating with AOL to provide casual games. I think for online games, casual games will have a greater chance to succeed overseas when compared to MMO's. This is because Asian and Western companies have very different cultures among MMORPG games. So I think that games that are very popular in China or in Korea might not be as popular in western countries. Also in China over 60% of the netizens play online games and I think that percentage in the States is much lower.
PPW: Do people in lower tier cities play games more often than in metropolitan cities?
Tangos: Because the currently available statistics are only on everybody that plays online games we have no idea about how many of them are playing casual games and how many are playing MMO games. I think in China more and more people are starting to play online casual games. So if we consider the whole landscape including MMO and casual games I don't think that percentage will drop.
According to a recent report the percentage of users who play online games in urban cities is even higher in rural areas. It is about 10% higher. It is a report about Chinese Internet users under the age of 25. I think that the numbers for urban cities are about over 63% and in rural areas it is only about 50%.
PPW: Why is this? I would expect differently because in urban areas people have alternatives for entertainment.
Tangos: According to the report the usage rate of almost all-online services in rural areas is lower than urban areas. Only 2 services share a similar percentage of urban and rural. One is online chatting so QQ. And the other is music download. For the rest of the services the usage rate is higher for urban areas. I think that these numbers are not about how much time users put into online games, but it is about how many of them have played online games. So maybe in urban areas people play short games during their break at work. But also maybe for the people in rural areas, they mainly use the Internet for QQ. Therefore they have similar usage rates compared to urban cities.
PPW: How should companies cope with the rather skewed user base of China; sophisticated users in the eastern metropolitan areas and less experienced netizens in the more rural areas in the west?
Tangos: Actually I think that some companies just focus on one kind of user. For example the SN website called 51.com. I think they are the MySpace of China. They target users in 2nd or 3rd tier cities. I believe they have only a low percentage of users in the big cities. Xiaonei is more targeted on university students so they don't have many members in the more rural big cities. But it is not easy to cover all the users across China. It also depends on your service because if you are a search engine like Baidu it is not really a problem because everyone uses a search engine in more or less the same way. For SN websites it is different. Maybe for video-sites it is not really a big deal either because in both urban and rural areas people will watch dramas.
PPW: And what about MMO companies, is the skewed situation a problem for them?
Tangos: I don't think the big gaming companies will have a problem. Maybe the only problem is whether the Internet cafés in rural areas are able to run the games or not. But for users I don't think it is a big problem. If you look at the figures, for Internet users younger than 18 years old, the percentage of them that has ever played online games is over 76%. So this is the highest group of users that play online games. So when people grow older the percentage decreases.
PPW: Do you think Social Networks have a lifecycle, like games have?
Tangos: I think the challenge is whether Facebook can provide enough applications or new services, which can attract users. Because as a user when you are in a SN for maybe 2 or 3 months you can sometimes feel fatigue. When you log in you just don't know what to do on the network. So I think that is one of the success factors of Facebook; they have launched a developer's platform. So users have many new applications to play with everyday.
I think that will be a trend for Social Networks in China. So far none of the Chinese SN's have opened their developer platform. MySpace China is working on OpenSocial and also some smaller startups are doing this. City!N is working with OpenSocial too.
PPW: You think that there is a Chinese Internet bubble coming up?
Tangos: I don't think that there is a bubble in China. This year investors became more cautious in new investments though. Of course there are some very big investments but generally speaking the investment market environment in Internet start-ups is not so good. There are many small start-ups that have difficulties in raising funding. Especially Chinese companies that do not have a very good business model. I think 2 years ago the market environment was much better. 2 years ago there were plenty of little start-ups without a good business model that have received funding from investors, but after 2 years investors realized that some of their investments are not doing very well now. So now they become more cautious.
PPW: Do you think that another reason for VC's becoming cautious has to do with the fact that there are not so many ideas to copy from other countries anymore; Chinese entrepreneurs have to start coming up with their own stuff?
Tangos: A copy is not a problem for an investor. But also every investor is looking for something unique on a global level. If you have a service, which is truly unique, in China investors will hesitate and they will be more comfortable to invest in copied ideas. I also think there are too many companies that want to invest in China. I think a lot of VC's are managing too much money so they are eager to invest and make some big deals. So for new start-ups it is not easy for them to raise small funds, small rounds. Many VC's just ignore fundraising for under maybe $5 million. There are VC's that used to invest maybe $1 million to $3 million, but currently they will only consider investments from over $3 million. So they have too much money and they need to invest it all within 1 or 2 years and they don't have enough energy and time to invest in many small start-ups. They need to look for other bigger investments.
PPW: It seems that the Chinese government has a pro-active approach in the development of the Chinese economy, they realize that the next step is innovation and setting up Chinese brands in order to sustain the development.
Tangos: Yes. But it is really not easy. If you want to make Chinese more innovative and creative you have to educate them from the bottom up, so start with the children. According to my experience I think the educational system in China, from kindergarten to primary school, does not encourage children to think creative and does not stimulate creativity overall.
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Does anybody out there make leg irons with short chains - say like 6 inches or so? I'm pretty sure I saw these advertised on a web site someplace, but I don't remember where.
: Does anybody out there make leg irons with short chains - say like 6 inches or so? I'm pretty sure I saw these advertised on a web site someplace, but I don't remember where.


text:
Q: groovy/grails - domain object and tooltips Given a grails domain object with properties like
Integer displayOrder
Boolean visible

... what would be the best way to associate meta information about these fields for display on tooltips in edit view?
Thanks a bunch.

A: I would probably just add a message to message.properties with the content:
// messages.properties
myDomain.myField.tooltip = Enter an awesome field value for this field

And then render it appropriately in the markup alongside the field, or wherever:
<g:message code="myDomain.myField.tooltip"/>


text:
Flexepin
Sofort/Klarna
Wazdan Casino Content Now Certified in Spain
Online casino games developer and supplier Wazdan is set to expand into the Spanish iGaming market after its content has received full certification from Spain's gambling regulator, Dirección General de Ordenación del Juego (DGOJ).
Following the certification, the company will now be able to deliver its popular slots and new releases to locally licensed online gambling operators. Wazdan said that securing its Spanish certification before the end of the year was one of its primary goals.
Widely popular across major markets around the globe, Wazdan's ever-growing collection of games includes the likes of 9 Lions, the Power of Gods series – Power of Gods: Egypt and Power of Gods: The Pantheon – as well as Larry the Leprechaun. Aside from its high-quality slots, the provider is also known for its unique range of in-play features.
Spanish online casino players will soon be introduced to Wazdan's portfolio and will be able to take advantage of the supplier's innovative game strategy that enables players to adjust the volatility of some of its most popular slots that are available on the market.
News about Wazdan's entry into the Spanish market emerge less than a month after the company announced that 15 of its top-performing slots were certified in Switzerland.
Entering a Competitive Landscape
Commenting on their Spanish certification, Wazdan Head of Business for Spain and Latin America Albert Bellavista said that they are extremely proud to have received the necessary certification to be able to offer their products in Spain, which ranks among the largest regulated markets in Europe.
Mr. Bellavista went on to say that they are sure their new Spanish audience will be excited by their slots and that their "cutting-edge" features will enable locally licensed iGaming operators to offer their players a brand-new experience. He further noted that Spain is an "extremely competitive space" and that they are ready to be part of it.
The Spanish iGaming space is the latest regulated market Wazdan is set to venture into. Early this year, the company expanded into Hungary through a partnership with locally licensed operator Grand Casino.
It was this year again when the iGaming content maker secured expansion into Latin America and Asia through a tie-up with Salsa Technology, formerly Patagonia. Under that partnership, Wazdan integrated its content into Salsa's Game Aggregation Platform.
Other markets that Wazdan's content is certified and already available in include the UK, Sweden, Denmark, Portugal, Romania, Estonia, Lithuania, and Latvia. The company has partnerships with leading operators in each of these markets to ensure its games can be accessed by an ever-growing gaming audience.
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AC Milan not interested in Modric, says Boban
Ben Spratt
The Croatia international has been linked with a move to Serie A with Inter also thought to be interested, but Milan have distanced themselves
Zvonimir Boban has repeated Paolo Maldini's statement that AC Milan will not be able to sign Luka Modric from Real Madrid.
Modric, the 2018 Ballon d'Or winner, has been linked with Inter and Milan in recent months, although he appears set to stay in La Liga.
Rossoneri technical director Maldini last month said the Croatia midfielder would be "perfect" at San Siro but the deal would not happen.
'Hazard does not deserve to be a starter': How the €100m superstar lost his way at Real Madrid
Messi banned for just two games after first Barcelona red card but appeal still lodged at Camp Nou
Following further speculation, Milan's chief football officer Boban reiterated the point.
"Twenty days ago, Paolo Maldini declared that Luka Modric will not be a Milan player," he told La Gazzetta dello Sport.
"But since we continue to talk about it and write about it without any basis, I would like to repeat that the best player in the world of last year will not be our player.
"As far as I can see, he will still be in Madrid. Obviously, it would be great to have him, but this is the only truth."
Images have previously circulated of a young Modric in Milan colours, and Boban added: "Luka is a person and a player of the highest level.
"We thank him for his good wishes for next season and we are pleased that he has liked our club since he was a child, but it is right to clarify things – out of respect for our fans, too."
While the English transfer window has now closed, the deadline for Italian clubs isn't until August 23, so Milan have plenty of time to wrap up whatever details they are still targeting.
They start the new Serie A season two days later with a trip to Udinese, before their first home fixture of the campaign against Brescia the following weekend.
In Spain, the window remains open until September 2. Madrid will have started their season by then, travelling to Celta Vigo on the opening day on August 17.
Their first home game comes against Real Valladolid the week after.
Messi handed two-game ban for first Barca red card
Transfer LIVE: Man Utd to open Sancho talks in the spring
Copyright © 2021 Goal (United Kingdom) All rights reserved. The information contained in Goal (United Kingdom) may not be published, broadcast, rewritten, or redistributed without the prior written authority of Goal (United Kingdom)


text:
\section{Introduction}

A group $G$ is said to be $\frac{3}{2}$-generated if every non-trivial element of $G$ is part of a generating pair of $G$. In 2000, settling a problem of Steinberg from 1962 \cite{S}, Guralnick and Kantor proved that all finite simple groups are $\frac{3}{2}$-generated \cite{GK}.
In 2008, Breuer, Guralnick and Kantor \cite{BGK} observed that if a group $G$ is $\frac{3}{2}$-generated then every proper quotient of it must be cyclic. They conjectured that for finite groups this is also a sufficient condition.
The conjecture was proved in 2021 by Burness, Guralnick and Harper \cite{BGH}.

Note that for an infinite group $G$ every proper quotient being cyclic is not a sufficient condition for the group  being $\frac{3}{2}$-generated. Indeed, the infinite alternating group $A_{\infty}$ is simple but not finitely generated. Moreover, there are finitely generated infinite simple groups which are not $2$-generated (see \cite{Guba})
and in particular, are not $\frac{3}{2}$-generated. Recently, Cox \cite{Cox} constructed an example of an infinite $2$-generated group $G$ such that every proper quotient of $G$ is cyclic and yet $G$ is not $\frac{3}{2}$-generated.

Obvious examples of infinite $\frac{3}{2}$-generated groups are the Tarski monsters constructed by Olshanskii \cite{O}. Recall that Tarski monsters are infinite finitely generated non-cyclic groups where every proper subgroup is cyclic\footnote{There are two types of Tarski monsters. One where every proper subgroup is infinite cyclic and one where every proper subgroup is cyclic of order $p$ for some fixed prime $p$.}. In particular, if $T$ is a Tarski monster, then $T$ is generated by any pair of non-commuting elements of $T$. Since the center of $T$ is trivial, every non-trivial element of $T$ is part of a generating pair of $T$.

In 2019, Donoven and Harper gave the first examples of infinite non-cyclic $\frac{3}{2}$-generated groups, other than Tarski monsters. Indeed, they proved that Thompson's group $V$ is $\frac{3}{2}$-generated.
More generally, they proved that all  Higman–Thompson groups $V_n$ (see \cite{Higman}) and all Brin–Thompson groups $nV$ (see \cite{Brin}) are $\frac{3}{2}$-generated.
In 2020, Cox constructed two more examples of infinite $\frac{3}{2}$-generated groups with some special properties (see \cite{Cox}). Quite recently (in 2022), Bleak, Harper and Skipper proved that Thompson's group $T$ is also $\frac{3}{2}$-generated \cite{BHS}.

In this paper, we study Thompson's group $F$. Recall that Thompson's group $F$ is the group of all piecewise-linear homeomorphisms  of the interval $[0,1]$ with finitely many breakpoints where all breakpoints are dyadic fractions (i.e., numbers from $\mathbb{Z}[\frac{1}{2}]\cap (0,1)$) and all slopes are integer powers of $2$. Thompson's group $F$ is $2$-generated. The derived subgroup of $F$ is infinite and simple and can be characterized as the subgroup of $F$ of all functions $f$ with slope $1$ both at $0^+$ and at $1^-$ (see \cite{CFP}).
The abelianization $F/[F,F]$ is isomorphic to $\mathbb{Z}^2$. The standard abelianization map $\pi_{ab}\colon F\to \mathbb{Z}^2$ maps every function $f\in F$ to $(\log_2(f'(0^+)),\log_2(f'(1^-)))$ (see \cite{CFP}). Since the abelianization of $F$ is $\mathbb{Z}^2$, Thompson's group $F$ cannot be $\frac{3}{2}$-generated. However, we prove that it is ``almost'' $\frac{3}{2}$-generated in the sense that the following theorem holds.

\begin{theorem}\label{thm:almost3/2}
	Every element of $F$ whose image in the abelianization $\mathbb{Z}^2$ is part of a generating pair of $\mathbb{Z}^2$ is part of a generating pair of $F$.
\end{theorem}

In fact, we have the following more general theorem.

\begin{theorem}\label{thm:main intro}
	Let $(a,b),(c,d)\in\mathbb{Z}^2$ be such that $\{a,c\}\neq\{0\}$ and $\{b,d\}\neq\{0\}$. Let $f\in F$ be a non-trivial element such that $\pi_{ab}(f)=(a,b)$, then there is an element $g\in F$ such that $\pi_{ab}(g)=(c,d)$ and such that $$\langle f,g\rangle =\pi_{ab}^{-1}(\langle (a,b),(c,d)\rangle).$$
\end{theorem}

Recall that a subgroup $H$ of $F$ is normal if and only if it contains the derived subgroup of $F$ \cite{CFP} (in particular, all finite index subgroups of $F$ are normal subgroups of $F$).
It follows that the normal subgroups of $F$ are exactly the subgroups $\pi_{ab}^{-1}(\langle(a,b),(c,d)\rangle)$ for $(a,b),(c,d)\in\mathbb{Z}^2$. Note that if $a=c=0$ or $b=d=0$, then $\pi_{ab}^{-1}(\langle(a,b),(c,d)\rangle)$ is not finitely generated (see Remark \ref{rem:not finitely generated} below). Theorem \ref{thm:main intro} implies that all other normal subgroups of $F$ are $2$-generated. Hence, we have the following.

\begin{corollary}
	Every finitely generated normal subgroup of $F$ is $2$-generated. %
\end{corollary}

In particular, every finite index subgroup of $F$ is $2$-generated.
Recall that in \cite{BW} (see also \cite{BCR}), the finite index subgroups of $F$ that are isomorphic to $F$ were characterized. Let $p,q\in\mathbb{N}$. We denote by $F_{p,q}$ the subgroup $\pi_{ab}^{-1}(p\mathbb{Z}\times q\mathbb{Z})$ of $F$.
Then for every $p,q\in\mathbb{N}$ the subgroup $F_{p,q}$ is isomorphic to $F$ and these are the only finite index subgroups of $F$ that are isomorphic to $F$ \cite{BW}. In particular, the finite index  subgroups $F_{p,q}$ of $F$ are known to be $2$-generated.

   It was conjectured in \cite[Conjecture 12.6]{G16} that all finite index subgroups of $F$ are $2$-generated.

Note  that for every non-trivial $(a,b)\in\mathbb{Z}^2$, there exists $(c,d)\in\mathbb{Z}^2$ such that $\langle(a,b),(c,d)\rangle$ is a finite index subgroup of $\mathbb{Z}^2$ of the form $p\mathbb{Z}\times q\mathbb{Z}$ for some $p,q\in\mathbb{N}$ (see Lemma \ref{lem:linear algebra} below). Hence, Theorem \ref{thm:main intro} implies the following.

\begin{corollary}
	Let $f\in F$ be an element whose image in the abelianization of $F$ is non-trivial. Then there is an element $g\in F$ such that the subgroup $\langle f,g\rangle$ is isomorphic to $F$ and has finite index in $F$.
\end{corollary}

Note that Theorem \ref{thm:almost3/2}  shows that in some sense it is ``easy'' to generate Thompson's group $F$.
Several other results demonstrate (in different ways) the abundance of generating pairs of Thompson's group $F$. Recall that in \cite{G-RG}, we prove that in the two  natural probabilistic models studied in \cite{CERT}, a random pair of elements of $F$ generates $F$ with positive probability.
In \cite{GGJ}, Gelander, Juschenko and the author proved that Thomspon's group $F$ is invariably generated by $3$ elements (i.e., there are $3$ elements $f_1,f_2,f_3\in F$ such that regardless of how each one of them is conjugated, together they generate $F$.) Using results from \cite{G22}, one can show that in fact, there is a  pair of elements in $F$ which invariably generates $F$.

\section{Preliminaries}\label{s:FT}

\subsection{F as a group of homeomorphisms}

Recall that Thompson group $F$ is the group of all piecewise linear homeomorphisms of the interval $[0,1]$ with finitely many breakpoints where all breakpoints are  dyadic fractions and all slopes are integer powers of $2$.
The group $F$ is generated by two functions $x_0$ and $x_1$ defined as follows \cite{CFP}.

\[
x_0(t) =
\begin{cases}
	2t &  \hbox{ if }  0\le t\le \frac{1}{4} \\
	t+\frac14       & \hbox{ if } \frac14\le t\le \frac12 \\
	\frac{t}{2}+\frac12       & \hbox{ if } \frac12\le t\le 1
\end{cases} 	\qquad
x_1(t) =
\begin{cases}
	t &  \hbox{ if } 0\le t\le \frac12 \\
	2t-\frac12       & \hbox{ if } \frac12\le t\le \frac{5}{8} \\
	t+\frac18       & \hbox{ if } \frac{5}{8}\le t\le \frac34 \\
	\frac{t}{2}+\frac12       & \hbox{ if } \frac34\le t\le 1
\end{cases}
\]

The composition in $F$ is from left to right.

Every element of $F$ is completely determined by how it acts on the set $\mathbb{Z}[\frac{1}{2}]$. Every number in $(0,1)$ can be described as $.s$ where $s$ is an infinite word in $\{0,1\}$. For each element $g\in F$ there exists a finite collection of pairs of (finite) words $(u_i,v_i)$ in the alphabet $\{0,1\}$ such that every infinite word in $\{0,1\}$ starts with exactly one of the $u_i$'s. The action of $F$ on a number $.s$ is the following: if $s$ starts with $u_i$, we replace $u_i$ by $v_i$. For example, $x_0$ and $x_1$  are the following functions:

\[
x_0(t) =
\begin{cases}
	.0\alpha &  \hbox{ if }  t=.00\alpha \\
	.10\alpha       & \hbox{ if } t=.01\alpha\\
	.11\alpha       & \hbox{ if } t=.1\alpha\
\end{cases} 	\qquad
x_1(t) =
\begin{cases}
	.0\alpha &  \hbox{ if } t=.0\alpha\\
	.10\alpha  &   \hbox{ if } t=.100\alpha\\
	.110\alpha            &  \hbox{ if } t=.101\alpha\\
	.111\alpha                      & \hbox{ if } t=.11\alpha\
\end{cases}
\]
where $\alpha$ is any infinite binary word.

The group $F$ has the following finite presentation \cite{CFP}.
$$F=\langle x_0,x_1\mid [x_0x_1^{-1},x_1^{x_0}]=1,[x_0x_1^{-1},x_1^{x_0^2}]=1\rangle,$$ where $a^b$ denotes $b^{-1} ab$.

\subsection{Elements of F as pairs of binary trees} \label{sec:tree}

Often, it is more convenient to describe elements of $F$ using pairs of finite binary trees (see \cite{CFP} for a detailed exposition). The considered binary trees are rooted \emph{full} binary trees; that is, each vertex is either a leaf or has two outgoing edges: a left edge and a right edge. A  \emph{branch} in a binary tree is a simple path from the root to a leaf. If every left edge in the tree is labeled ``0'' and every right edge is labeled ``1'', then a branch in $T$ has a natural binary label. We rarely distinguish between a branch and its label.

Let $(T_+,T_-)$ be a pair of finite binary trees with the same number of leaves. The pair $(T_+,T_-)$ is called a \emph{tree-diagram}. Let $u_1,\dots,u_n$ be the (labels of) branches in $T_+$, listed from left to right. Let $v_1,\dots,v_n$ be the (labels of) branches in $T_-$, listed from left to right. We say that the tree-diagram $(T_+,T_-)$ has the \emph{pair of branches} $u_i\rightarrow v_i$ for $i=1,\dots,n$. The tree-diagram $(T_+,T_-)$ \emph{represents} the function $g\in F$ which takes binary fraction $.u_i\alpha$ to $.v_i\alpha$ for every $i$ and every infinite binary word $\alpha$. We also say that the element $g$ takes the branch $u_i$ to the branch $v_i$.
For a finite binary word $u$, we denote by $[u]$  the dyadic interval $[.u,.u1^{\mathbb{N}}]$. If $u\rightarrow v$ is a pair of branches of $(T_+,T_-)$, then $g$ maps the interval $[u]$ linearly onto $[v]$.

A \emph{caret} is a binary tree composed of a root with two children. If $(T_+,T_-)$ is a tree-diagram and one attaches a caret to the $i^{th}$ leaf of $T_+$ and the $i^{th}$ leaf of $T_-$ then the resulting tree diagram is \emph{equivalent} to $(T_+,T_-)$ and represents the same function in $F$. The opposite operation is that of \emph{reducing} common carets. A tree diagram $(T_+,T_-)$ is called \emph{reduced} if it has no common carets; i.e, if there is no $i$ for which the  $i$ and ${i+1}$ leaves of both $T_+$ and $T_-$ have a common father. The reduced tree-diagrams of the generators $x_0$ and $x_1$ of $F$ are depicted in Figure \ref{fig:x0x1}.

\begin{figure}[ht]
	\centering
	\begin{subfigure}{.55\textwidth}
		\centering
		\includegraphics[width=.55\linewidth]{x0.png}
		\caption{}
		\label{fig:x0}
	\end{subfigure}%
	\begin{subfigure}{.55\textwidth}
		\centering
		\includegraphics[width=.55\linewidth]{x1.png}
		\caption{
		\label{fig:x1}
	\end{subfigure}
	\caption{(A) The reduced tree-diagram of $x_0$. (B) The reduced tree-diagram of $x_1$. In both figures, $T_+$ is on the left and $T_-$ is on the right.}
	\label{fig:x0x1}
\end{figure}

 When we say that a function $f\in F$ has a pair of branches $u_i\rightarrow v_i$, the meaning is that some tree-diagram representing $f$ has this pair of branches. In other words, this is equivalent to saying that $f$ maps the dyadic interval $[u_i]$ linearly onto $[v_i]$.
Clearly, if $u\rightarrow v$ is a pair of branches of $f$, then for any finite binary word $w$, $uw\rightarrow vw$ is also a pair of branches of $f$. Similarly, if $f$ has the pair of branches $u\rightarrow v$ and $g$ has the pair of branches $v\rightarrow w$ then $fg$ has the pair of branches $u\rightarrow w$.

Let $\mathcal B$ be the set of all finite binary words and let $u,v\in\mathcal B$. We say that $v$ is a \emph{descendant} of $u$ if $u$ is a strict prefix of $v$. We say that $u$ and $v$ are \emph{incomparable} if $u$ is not a prefix of $v$ and $v$ is not a prefix of $u$. Note that if $u$ and $v$ are incomparable then the interiors of $[u]$ and $[v]$ have empty intersection. In that case, we will write  $[u]<[v]$ if for every $x$ in the interior of $[u]$ and every $y$ in the interior of $[v]$ we have $x<y$. The following lemma will be useful.

\begin{lemma}\label{lem:obvious}
	Let $f$ be a non-trivial element of $F$. Then there exist $u,v,w\in\mathcal B$ such that $[u]<[v]<[w]$ and such that $f$ or $f^{-1}$ has the pairs of branches $u\to v$ and $v\to w$.
\end{lemma}

\begin{proof}
	Let $\alpha$ be the maximal number in $[0,1]$ such that $f$ fixes the interval $[0,\alpha]$ pointwise. Since $f$ is non-trivial, $\alpha<1$.
	Note that $f'(\alpha^+)\neq 1$. If $f'(\alpha^+)>1$, let $g=f$, otherwise, let $g=f^{-1}$, so that $g'(\alpha^+)>1$.  Since $\alpha$ is a breakpoint of $f$, it must be dyadic. Hence, there exists a finite binary word $s$

	such that $\alpha=.s$.
	 Since $g'(\alpha^+)> 1$, there exist $n>m$ in $\mathbb{N}$ such that $g$ has the pair of branches $.s0^n\to .s0^m$ (see \cite[Lemma 2.6]{G16}).
	Now, let $u\equiv s0^{n+(n-m)}1$, $v\equiv s0^n1$ and $w\equiv s0^m1$. Then $[u]<[v]<[w]$ and the function $g$ has the pairs of branches $u\to v$ and $v\to w$, as necessary.
\end{proof}

\subsection{The derived subgroup and the abelianization of $F$}\label{sec:derived}

Recall that the derived subgroup of $F$ is an infinite simple group which can be characterized as the subgroup of $F$ of all functions $f$ with slope $1$ both at $0^+$ and at $1^-$ (see \cite{CFP}).
That is, a function $f\in F$ belongs to $[F,F]$ if and only if the reduced (equiv. any) tree-diagram of $f$ has pairs of branches of the form $0^m\rightarrow 0^m$ and $1^n\rightarrow 1^n$ for some $m,n\in\mathbb{N}$.

As noted above, a subgroup $H$ of $F$ is normal if and only if it contains the derived subgroup of $F$. In particular, every finite index subgroup of $F$ is a normal subgroup of $F$ \cite{CFP}.
Recall that the abelianization of $F$ is isomorphic to $\mathbb{Z}^2$ and that the standard abelianization map $\pi_{ab}\colon F\to\mathbb{Z}^2$ maps an element $f\in F$ to $(\log_2(f'(0^+)),\log_2(f'(1^-)))$.
We make the following observation.

\begin{remark}\label{rem:not finitely generated}
	Let $(a,b),(c,d)\in\mathbb{Z}^2$ be such that $a=c=0$ or $b=d=0$. Then the subgroup $\pi_{ab}^{-1}(\langle (a,b),(c,d)\rangle)\leq F$ is not finitely generated.
\end{remark}

\begin{proof}
	We prove it in the case where $a=c=0$. Note that in that case, $\langle (a,b),(c,d)\rangle=\langle (0,g)\rangle$, where $g=\gcd(b,d)$. Let $H=\pi_{ab}^{-1}(\langle (0,g)\rangle)$. We  claim that $H$ is not finitely generated. Assume by contradiction that $H$ is generated by $f_1,\dots,f_n$, for $n\in\mathbb{N}$. Then for each $i=1,\dots,n$ we have $f_i'(0^+)=1$. Hence, for each $i$, $f_i$ fixes a right neighborhood of $0$. Let $\alpha\in (0,1)$ be a small enough dyadic fraction such that for each $i=1,\dots,n$, $f_i$ fixes the interval $[0,\alpha]$ pointwise. Since $H$ is generated by $f_1,\dots,f_n$, it follows that $H$ fixes the interval $[0,\alpha]$ pointwise. But it is easy to construct an element $f\in F$ such that  $\pi_{ab}(f)=(0,g)$ and such that $f$ does not fix $\alpha$, a contradiction.
\end{proof}

Recall that for every pair of elements $(a,b),(c,d)\in\mathbb{Z}^2$ the index of the subgroup $\langle (a,b),(c,d)\rangle$ of $\mathbb{Z}^2$ is the absolute value of the determinant $ad-bc$ (where if $ad-bc=0$, the index is infinite). The following simple lemma was referred to in the introduction. As we were unable to find a reference for it, we provide a proof.

\begin{lemma}\label{lem:linear algebra}
	Let $(a,b)\in\mathbb{Z}^2$ be non-trivial. Then there exists $(c,d)\in\mathbb{Z}^2$ and $p,q\in\mathbb{N}$ such that
	$\langle (a,b),(c,d)\rangle=p\mathbb{Z}\times q\mathbb{Z}$ and such that $pq=\gcd(a,b)$.
\end{lemma}

\begin{proof}
	If $a=0$ then for $(c,d)=(1,0)$ and $p=1$, $q=|b|$ we have the result. Indeed, $$\langle (a,b),(c,d)\rangle=\langle (0,b),(1,0)\rangle=\mathbb{Z}\times |b|\mathbb{Z}.$$ Similarly, if $b=0$, we are done. Hence, we can assume that $a,b\neq 0$.
	Let $g=\gcd(a,b)$ and let $a',b'\in\mathbb{Z}$ be such that $a=ga'$ and $b=gb'$. Clearly, $\gcd(a',b')=1$. Let $g=p_1^{n_1}\cdots p_m^{n_m}$ be the prime factorization of $g$. Since $\gcd(a',b')=1$, each of the primes $p_i$ for $i=1,\dots,m$ divides at most one of the numbers $a'$ or $b'$. Let $q$ be the product of all $p_i^{n_i}$, $i=1,\dots,m$ such that $p_i$ divides $a'$ (if there are no such $p_i$'s, we let $q=1$). Let $p\in\mathbb{N}$ be such that $g=pq$ (in other words, $p$ is the product of all $p_i^{n_i}$ such that $p_i$ does not divide $a'$). By construction, $q$ and $b'$ are co-prime, $p$ and $a'$ are co-prime and $p$ and $q$ are co-prime. It follows that $qa'$ and $pb'$ are co-prime. Hence, by the Euclidean algorithm there exist $m,n\in\mathbb{Z}$ such that $m(qa')-n(pb')=1$. Let $(c,d)=(np,mq)$.
	Then the index of the subgroup $\langle (a,b),(c,d)\rangle$ in $\mathbb{Z}^2$ is
	$$|ad-bc|=|(ga')(mq)-(gb')(np)|=g|mqa'-npb'|=g=pq.$$
	On the other hand, since $p$ divides both $a=ga'$ and $c=np$, we have that $p|\gcd(a,c)$.  Similarly, $q$ divides $\gcd(b,d)$. Hence,
	$$\langle (a,b),(c,d)\rangle\leq \gcd(a,c)\mathbb{Z}\times \gcd(b,d)\mathbb{Z}\leq p\mathbb{Z}\times q\mathbb{Z}.$$
	Since $p\mathbb{Z}\times q\mathbb{Z}$ is an over-group of $\langle (a,b),(c,d)\rangle$ and they both have index $pq$ in $\mathbb{Z}^2$, they must coincide. Hence,
	$$\langle (a,b),(c,d)\rangle= p\mathbb{Z}\times q\mathbb{Z},$$
	as necessary.
\end{proof}

\subsection{Generating normal subgroups of $F$}

Let $H$ be a subgroup of $F$. A function $f\in F$ is said to be a \emph{piecewise-$H$} function if there is a finite subdivision of the interval $[0,1]$ such that on each interval in the subdivision, $f$ coincides with some function in $H$. Note that since all breakpoints of elements in $F$ are dyadic fractions, a function $f\in F$ is a piecewise-$H$ function if and only if there is a  dyadic subdivision of the interval $[0,1]$ into finitely many pieces such that on each dyadic interval in the subdivision, $f$ coincides with some function in $H$.

Following \cite{GS,G16}, we define the \emph{closure} of a subgroup $H$ of $F$, denoted $\mathrm{Cl}(H)$, to be the subgroup of $F$ of all piecewise-$H$ functions. A subgroup $H$ of $F$ is \emph{closed} if $H=\mathrm{Cl}(H)$. In \cite{G16}, we gave the following characterization of subgroups of $F$ which contain the derived subgroup of $F$ (as well as an algorithm for determining if a finitely generated subgroup of $F$ contains the derived subgroup of $F$).

\begin{theorem}\cite[Theorem 7.10]{G16}\label{gen}
	Let $H$ be a subgroup of $F$. Then $H$ contains the derived subgroup of $F$ (equiv. $H$ is a normal subgroup of $F$) if and only if the following conditions hold.
	\begin{enumerate}
		\item[(1)] $\mathrm{Cl}(H)$ contains the derived subgroup of $F$.
		\item[(2)] There is an element $h\in H$ and a dyadic fraction $\alpha\in (0,1)$ such that $h$ fixes $\alpha$, $h'(\alpha^-)=1$ and
		$h'(\alpha^+)=2$.
	\end{enumerate}
\end{theorem}

Note that if a subgroup $H$ of $F$ contains the derived subgroup of $F$, then the image of $H$ in the abelianization of $F$ completely determines the subgroup $H$.

\section{Subgroups $H$ of $F$ whose closure is a normal subgroup of $F$}

In the next section, we apply Theorem \ref{gen} to prove that a given subset of $F$ generates a normal subgroup $H$ of $F$.
To do so, we will have to prove in particular that $H$ satisfies Condition (1) of Theorem \ref{gen}, i.e., that the closure of $H$ contains the derived subgroup of $F$. To that end, with each subgroup of $F$, we associate an equivalence relation on the set of finite binary words $\mathcal B$.

\begin{definition}
	Let $H$ be a subgroup of $F$.
 The \emph{equivalence relation induced by $H$} on the set of finite binary words $\mathcal B$, denoted $\sim_H$, is defined as follows. For every pair of finite binary words $u,v\in\mathcal B$ we have $u\sim_H v$ if and only if there is an element in $H$ with the pair of branches $u\to v$.
\end{definition}

Note that if $u\sim_Hv$ then for every finite binary word $w$, we have $uw\sim_H vw$.
Moreover, if $H$ is closed, we have the following.

\begin{lemma}\label{coherent}
	Let $H$ be a closed subgroup of $F$. Then for every pair of finite binary words $u,v\in\mathcal B$ we have $u\sim_H v$ if and only if $u0\sim_H v0$ and $u1\sim_H v1$.
\end{lemma}

\begin{proof}
	Assume that $u0\sim_H v0$ and $u1\sim_Hv1$. We claim that there is an element in $H$ with the pair of branches $u\to v$.
	Since $u0\sim_H v0$, there is an element $h_1\in H$ with the pair of branches $u0\to v0$.
	Similarly, since $u1\sim_H v1$, there is an element  $h_2\in H$ with the pair of branches $u1\to v1$. Let $\alpha=.u01^\mathbb{N}=.u1$ and $\beta=.v01^\mathbb{N}=.v1$. Then $h_1(\alpha)=\beta$ and $h_2(\alpha)=\beta$. Let $h_3\in F$ be the following function
	\[h_3(x)=\begin{cases}
		h_1(x) & x\in [0,\alpha]\\
		h_2(x) & x\in [\alpha,1]
	\end{cases}
	\]
	(Since $\alpha$ is dyadic, $h_3$ is indeed in $F$.) By construction, $h_3$ is a piecewise-$H$ function.
	 Hence, since $H$ is closed, the element $h_3\in H$. But the element $h_3$ has the pairs of branches $u0\to v0$ and $u1\to v1$. Hence, $h_3$ maps the interval $[u0]$ linearly onto $[v0]$ and the interval $[u1]$ linearly onto $[v1]$. It follows that $h_3$ maps the interval $[u]$ linearly onto the interval $[v]$. Hence, it has the pair of branches $u\to v$. Therefore, $u\sim_H v$.
\end{proof}

\begin{remark}
	An equivalence relation $\sim$ on the set of finite binary words $\mathcal B$ such that for every $u,v\in\mathcal B$ we have $u\sim v$ if and only if $u0\sim v0$ and $u1\sim v1$ is called a \emph{coherent} equivalence relation in \cite{BBQS}. Coherent equivalence relations were used in \cite{BBQS} to study maximal subgroups of Thompson's group $V$.
\end{remark}

The following corollary follows from Lemma \ref{coherent} by induction on $k$.

\begin{corollary}\label{cor:coherent}
	Let $H$ be a closed subgroup of $F$. Let $u,v\in\mathcal B$ and let $k\in\mathbb{N}$. Assume that for every finite binary word $w$ of length $k$ we have $uw\sim_H vw$. Then $u\sim_H v$.
\end{corollary}

Let $\mathcal B'$ be the set of all finite binary words which contain both digits ``0'' and ``1''. In other words, $\mathcal B'=\mathcal B\setminus\{\emptyset, 0^n,1^n\mid n\in\mathbb{N}\}$.

\begin{lemma}\label{lem:all inner words identified}
	Let $H$ be a closed subgroup of $F$. Assume that for every pair of finite binary words $u,v\in\mathcal B'$ we have $u\sim_H v$. Then $H$ contains the derived subgroup of $F$.
\end{lemma}

\begin{proof}
	 Let $f\in [F,F]$ . Then the reduced tree-diagram of $f$ consists of pairs of branches
	\[
	f :
	\begin{cases}
		0^m & \rightarrow 0^m\\
		u_i  & \rightarrow v_i \mbox{ for } i=1,\dots,k \\
		1^n & \rightarrow 1^n\\
	\end{cases}
	\]
	where $k,m,n\in\mathbb{N}$ and where for each $i=1,\dots,k$, the binary words $u_i$ and $v_i$ contain both digits $``0"$ and $``1"$ and as such belong to $\mathcal B'$. By assumption, for each $i=1,\dots,k$ we have $u_i\sim_H v_i$ and as such there is an element $h_i\in H$ with the pair of branches $u_i\rightarrow v_i$. Then $h_i$ coincides with $f$ on the interval $[u_i]$. We note also that $f$ coincides with the identity function {\bfseries{1}} $\in H$ on $[0^m]$ and on $[1^n]$. Since $[0^m],[u_1],\dots,[u_k],[1^n]$ is a subdivision of the interval $[0,1]$ and on each of these intervals $f$ coincides with a function in $H$, $f$ is a piecewise-$H$ function. Since $H$ is closed, $f\in H$.
\end{proof}

\begin{lemma}\label{lem:identified of length k}
	Let $H$ be a subgroup of $F$. Assume that there is $k\in\mathbb{N}$ such that for every $u,v\in\mathcal B'$ and every $s\in\mathcal B$ such that the length $|s|=k$ we have $us\sim_H vs$. Then $\mathrm{Cl}(H)$ contains the derived subgroup of $F$.
\end{lemma}

\begin{proof}
	Since $H$ is contained in $\mathrm{Cl}(H)$, the equivalence relation $\sim_H$ is contained in $\sim_{\mathrm{Cl}(H)}$. By Lemma \ref{lem:all inner words identified}, to prove that $\mathrm{Cl}(H)$ contains the derived subgroup of $F$ it suffices to prove that for every $u,v\in\mathcal B'$ we have $u\sim_{\mathrm{Cl}(H)} v$. Let $u,v\in \mathcal B'$. By assumption, there exists $k\in\mathbb{N}$ such that for every finite binary word $s$ of length $k$ we have $us\sim_{\mathrm{Cl}(H)} vs$. But then, since $\mathrm{Cl}(H)$ is closed, by Corollary \ref{cor:coherent}, we have $u\sim_{\mathrm{Cl}(H)} v$, as necessary.
\end{proof}

The next lemma is used in the next section to show that for certain subgroups $H$ of $F$, the closure of $H$ contains the derived subgroup of $F$.

\begin{lemma}\label{lem:tree}
	Let $H$ be a subgroup of $F$ and let $T$ be a non-empty finite binary tree with branches $u_1,\dots,u_n$. Let $w\in B$ and assume that the following assertions hold.
	\begin{enumerate}
		\item $w\sim_H w0\sim_H w1$.
		\item For every $i=2,\dots,n-1$ we have $u_i\sim_H w$.
		\item for every $i\geq 0$, we have $u_10^i1\sim_H w$.
		\item for every $i\geq 0$, we have $u_n1^i0\sim_H w$.
	\end{enumerate}
	Then $\mathrm{Cl}(H)$ contains the derived subgroup of $F$.
\end{lemma}

\begin{proof}
	First, note that since $w\sim_H w0\sim_H w1$, it follows by induction that for every finite binary word $q$, we have $wq\sim_H w$. In other words, every descendant $v$ of $w$ is $\sim_H$-equivalent to $w$. It follows, that for every $p\in \mathcal B$ such that $p\sim_H w$, every descendant of $p$ is also $\sim_H$-equivalent to $w$.  Now, let $k$ be the maximum length of a branch of $T$ and let $u,v\in\mathcal B'$.
	By Lemma \ref{lem:identified of length k}, to prove that $\mathrm{Cl}(H)$ contains the derived subgroup of $F$ it suffices to prove that for every $s\in \mathcal B$ of length $k$ we have $us\sim_H vs$. Let $s\in \mathcal B$ of length $k$. We claim that $us\sim_H w$.
	Indeed, since $|us|>k$, there is  $i\in\{1,\dots,n\}$ such that the branch $u_i$ of $T$ is a strict prefix of $us$. If $i\in\{2,\dots,n\}$, then since $u_i\sim_H w$ and $us$ is a descendant of $u_i$, we have that $us\sim_H w$ as necessary. Otherwise, either $u_1$ or $u_n$ is a strict prefix of $us$. We consider the case where $u_1$ is a strict prefix of $us$, the other case being similar. Since $u_1$ is a string of zeros and $u$ contains the digit $1$, the word $u_1$ must be a strict prefix of $u$. In fact, there exists $j\geq 0$ such that $u_10^j1$ is a prefix of $u$. Since $u_10^j1\sim_H w$ by assumption and since  $us$ is a descendant of $u_10^j1$, we get that $us\sim_H w$, as necessary. In a similar way, one can show that $vs\sim_H w$. Hence, $us\sim_H vs$, as required.
\end{proof}

\section{Proof of the main theorems}

To prove Theorem \ref{thm:main intro}, we will need the following proposition.

\begin{proposition}\label{prop:main}
	Let $f$ be a non-trivial element of $F$ and let $c,d\in\mathbb{Z}\setminus\{0\}$. Then the following assertions hold.
	\begin{enumerate}
		\item[(1)] There is an element $g\in F$ such that $\pi_{ab}(g)=(c,d)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$.
		\item[(2)] If $f$ has non-trivial slope at $1^-$ then there is an element $g\in F$ such that $\pi_{ab}(g)=(c,0)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$.
		\item[(3)] If $f$ has non-trivial slope at $0^+$ then there is an element $g\in F$ such that $\pi_{ab}(g)=(0,d)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$.
		\item[(4)] If $f$ has non-trivial slope both at $0^+$ and at $1^-$ then there is an element $g\in F$ such that $\pi_{ab}(g)=(0,0)$ (i.e., such that $g\in [F,F]$) and such that $\langle f,g\rangle$ contains the derived subgroup of $F$.
	\end{enumerate}
\end{proposition}

\begin{proof}
	We prove parts (1) and (2). The proofs of parts (3) and (4) are similar.

	(1) Since $f$ is non-trivial, by Lemma \ref{lem:obvious}, there exist finite binary words $u,v$ and $w$ such that $[u]<[v]<[w]$ and such that $f$ or $f^{-1}$ has the pairs of branches $u\to v$ and $v\to w$. It is easy to see that $u,v$ and $w$ must belong to $\mathcal B'$. Note that if (1) holds for $(c,d)$ it also holds for $(-c,-d)$. Hence, we can assume without loss of generality that $c>0$.
	We will also assume that $d>0$, and we will explain below how to modify the proof for the case where $c>0$ and $d<0$.

	Let $T$ be a finite binary tree such that $u,v0,v1,w0,w10$ and $w11$ are branches of $T$ and such that $T$ has at least three branches after the branch $w11$. Let $(S_+,S_-)$ be the reduced tree-diagram of $x_1$ (see Figure \ref{fig:x0x1}(B)).
	Let $T_1$ be the minimal binary tree with branch $0^{c+1}$ and let $T_2$ be the minimal binary tree with branch $1^c$. Let $C_1$ and $C_2$ be trees which consist of a single caret each. Let $T_3$ be the minimal binary tree with branch $1^{d+1}$ and $T_4$ be the minimal binary tree with branch $0^{d}$.

	Let $u_1,\dots,u_n$ be the branches of $T$.	 We will use two copies of the tree $T$ to construct a new tree-diagram in $F$.
	Let $R_+$ be the tree obtained from the first copy of $T$ by performing the following operations:
	\begin{enumerate}
		\item[(1a)] Attaching the tree $T_1$ to the end of the branch $u_1$;
		\item[(2a)] Attaching the tree $S_1$ to the end of the branch $w10$;
		\item[(3a)] Attaching the tree $T_3$ to the end of the branch $u_n$.
	\end{enumerate}

	Let $R_-$ be the tree obtained from the second copy of $T$ by performing the following operations:
	\begin{enumerate}
		\item[(1b)] Attaching the tree $T_2$ to the end of the branch $u_1$.
		\item[(2b)] Attaching the caret $C_1$ to the end of the branch $w0$.
		\item[(3b)] Attaching the tree $S_2$ to the end of the branch $w10$.
		\item[(4b)] Attaching the caret $C_2$ to the end of the branch $w11$.
		\item[(5b)] Attaching the tree $T_4$ to the end of the branch $u_n$.
	\end{enumerate}

	Note that in the construction of $R_+$, we added $(c+1)+3+(d+1)$ carets to the tree $T$.
	Similarly, in the construction of $R_-$, $c+1+3+1+d$ carets were added to the tree $T$. Hence $R_+$ and $R_-$ have the same number of carets. Let $g$ be the element represented by the tree-diagram $(R_+,R_-)$. We claim that the assertion in Proposition \ref{prop:main}(1) holds for $g$.

	Recall that $w0,w10,w11$ are (necessarily consecutive) branches of $T$. In addition, $w0$ cannot be one of the first $4$ branches of $T$, since $u_1,u,v0$ and $v1$ precede it. By assumption, there are at least three branches in $T$ to the right of the branch $w11$. Hence, there exists $5\leq k\leq n-5$ such that $u_k\equiv w0$, $u_{k+1}\equiv w10$ and $u_{k+2}\equiv w11$. By construction, the tree-diagram $(R_+,R_-)$ has the following sets of pairs of branches (we recommend to the reader verifying it to read separately the list of branches of $R_+$ (the branches on the left hand-side of $\mbox{(A),(B) and (C)}$) and the list of branches of $R_-$ (the branches on the right hand-side of $\mbox{(A),(B) and (C)}$)).
	\begin{align*}
	\mbox{(A)}& \begin{cases}
	u_10^{c+1}\to u_10\\
	u_10^{c+1-i}1\to u_11^i0, & \mbox{ for } 1\leq i\leq c-1\\
	u_101\to u_11^c\\
	u_11\to u_2\\
	{u_i}\to u_{i+1} & \mbox{for } 2\leq i\leq k-2\\
	u_{k-1}\to w00\\
	u_k\equiv w0\to w01\\
\end{cases}\\
\mbox{(B)}& \begin{cases}
	w100 \to    w100& \\
	w10100\to		w1010& \\
	w10101\to		w10110& \\
	w1011\to		w10111& \\
\end{cases}
\end{align*}
\begin{align*}
\mbox{(C)}& \begin{cases}
	w11\to w110& \\
	u_{k+3}	\to w111\\
	u_{i} \to u_{i-1} & \mbox{for } k+4\leq i\leq n-1\\
	u_n0\to u_{n-1}\\
	u_n10\to u_n0^d\\
	u_n1^i0\to u_10^{d+1-i}1   & \mbox{ for } 2\leq i\leq d\\
	u_n1^{d+1}\to u_n 1\\
\end{cases}
\end{align*}

Note that the pairs of branches in $\mbox{(B)}$ are the pairs of branches of $(R_+,R_-)$ which result from the attachment of the trees $S_1$ and $S_2$ to the branch $w10$ of both copies of $T$ (see Figure \ref{fig:x0x1}(B) for the branches of $S_1$ and $S_2$).
 The preceding pairs of branches of $(R_+,R_-)$ are all in $\mbox{(A)}$. In particular, $\mbox{(A)}$ contains all the pairs of branches of $(R_+,R_-)$ which result from the attachment of the tree $T_1$ to  the first copy of $T$ and the trees $T_2$ and $C_1$ to the second copy of $T$. Similarly, $\mbox{(C)}$ contains all the pairs of branches of $(R_+,R_-)$ which result from the attachment of the tree $T_3$ to  the first copy of $T$ and the trees $T_4$ and $C_2$ to the second copy of $T$.

Now, the first and last pairs of branches of $(R_+,R_-)$ show that $\pi_{ab}(g)=(c,d)$ as required. Let $H=\langle f,g\rangle$. It suffices to prove that $H$ contains the derived subgroup of $F$. First, note that the pairs of branches $w100\to w100$ and $w10100\to w1010$ in $\mbox{(B)}$ show that $g$ fixes the dyadic fraction $\alpha=.w101=.w1001^{\mathbb{N}}$ and that $g'(\alpha^-)=1$ and $g'(\alpha^+)=2$. Hence, $H$ satisfies condition (2) of Theorem \ref{gen}. Therefore, to prove that $H$ contains $[F,F]$ it suffices to prove that $\mathrm{Cl}(H)$ contains $[F,F]$. For that, we will make use of Lemma \ref{lem:tree}. Let us denote the equivalence relation $\sim_H$ by $\sim$. To prove that $\mathrm{Cl}(H)$ contains the derived subgroup of $F$ it suffices to prove that for the tree $T$ (with branches $u_1,\dots,u_n$) and the word $w$, conditions (1)-(4) of Lemma \ref{lem:tree} hold.

First, note that
 since $f$ or $f^{-1}$ has the pairs of branches $u\to v$ and $v\to w$, we have $u\sim v\sim w$.
Now, let us consider the pairs of branches of $g$. The pairs of branches $u_i\to u_{i+1}$, $2\leq i\leq k-2$
imply that $u_2\sim u_3\sim\cdots\sim u_{k-1}$. Recall that $u,v0$ and $v1$ are branches of $T$ (distinct from $u_1$) which precede $u_k\equiv w0$.
As such, $u,v0,v1\in\{u_2,\dots,u_{k-1}\}$, so that $u\sim v0\sim v1$. Since $u\sim v$, we get that $v\sim v0\sim v1$. It follows that every descendant of $v$ is $\sim$-equivalent to $v$. Since $w\sim v$, every descendant of $w$ is $\sim$-equivalent to $w$ (and as such, Condition (1) of Lemma \ref{lem:tree} holds).
  In particular, since $u_k\equiv w0, u_{k+1}\equiv w10$ and $u_{k+2}\equiv w11$, we have $u_k\sim u_{k+1}\sim u_{k+2}\sim w$. Since $u_2\sim\cdots\sim u_{k-1}\sim u\sim w$, we have $u_2\sim\cdots\sim u_{k-1}\sim u_k\sim u_{k+1}\sim  u_{k+2}\sim w$.

Now, the pair of branches $u_{k+3}\to w111$ in $\mbox{(C)}$ implies that $u_{k+3}\sim w111\sim w$. The pairs of branches $u_i\to u_{i-1}$ for $k+4\leq i\leq n-1$ imply that $u_{k+3}\sim u_{k+4}\sim\cdots\sim u_{n-1}$. Since $u_{k+3}\sim w$, we have that $u_2\sim \cdots u_{k+2}\sim u_{k+3}\sim\cdots\sim u_{n-1}\sim w$. That is, for every $2\leq i\leq n-1$ we have $u_i\sim w$. Hence, Condition (2) of Lemma \ref{lem:tree} holds.

Next, we show that Condition (3) of Lemma \ref{lem:tree} holds. That is, we claim that for every $i\geq 0$, we have $u_10^i1\sim w$.
Indeed, for $i=0$ it holds since the pair of branches  $u_11\to u_2$ of the element $g$ implies that $u_11\sim u_2\sim w$. Moreover, since $u_11\sim  w$, every descendant of $u_11$ is  $\sim$-equivalent to $w$.
 Hence, the pair of branches $u_101\to u_11^c$ of $g$ implies that $u_101\sim u_11^c\sim w$, so the claim holds for $i=1$ as well.
Now, the pairs of branches $u_10^{c+1-j}1\to u_11^j0$ for $1\leq j\leq c-1$ of $g$ imply that for all  $1\leq j\leq c-1$ we have $u_10^{c+1-j}1\sim w$. Letting $i=c+1-j$, we get that for all $i=2,\dots,c$ we have $u_10^i1\sim w$. Hence, the claim holds for every $i=0,\dots,c$. We prove by induction that the claim also holds for every $i>c$. Indeed, let $i\geq c+1$. Then the pair of branches $u_10^{c+1}\to u_10$ of $g$ implies that
$$u_10^i\equiv u_10^{c+1}0^{i-(c+1)}\sim
u_100^{i-(c+1)}\equiv u_10^{i-c}.$$
That implies that $u_10^i1\sim u_10^{i-c}1\sim w$, by induction. Hence, for all $i\geq 0$, we have $u_10^i1\sim w$ and Condition (3) of Lemma \ref{lem:tree} holds.

Note that in the proof of Condition (3) we have made use of branches in $\mbox{(A)}$ (more specifically, of the branches written in the first $4$ rows in $\mbox{(A)}$). In an almost identical manner, using branches of $\mbox{(C)}$ (more specifically, the branches written in the last $4$ rows of $\mbox{(C)}$), one can show that Condition (4) from Lemma \ref{lem:tree} holds. It follows that $\mathrm{Cl}(H)$ contains the derived subgroup of $F$. Hence, by Theorem \ref{gen}, $H=\langle f,g\rangle$ contains the derived subgroup of $F$, as required.

It remains to note that one can modify the proof for the case where $c>0$ and $d<0$ as follows. First, wherever $d$ appears in the above proof, we replace it by $|d|$. Second, in the construction of $R_+$ and $R_-$, instead of performing operation $(3a)$ on the first copy of $T$ (used in the construction of $R_+$) and operations $(4b)$ and $(5b)$ on the second copy of $T$ (used in the construction of $R_-$), we perform operation $(3a)$ on the second copy of $T$ and operations $(4b)$ and $(5b)$ on the first copy of $T$. The result is a tree-diagram $(R'_+,R'_-)$, whose pairs of branches coincide with the pairs of branches of $(R_+,R_-)$, other than the ones listed in $\mbox{(C)}$: for every pair of branches $p\to q$ in $\mbox{(C)}$, the tree-diagram $(R'_+,R'_-)$ has the ``opposite'' pair of branches $q\to p$. Note that this change, does not affect the proof  that the subgroup generated by $f$ and the element represented by $(R'_+,R'_-)$ contains the derived subgroup of $F$, as the equivalence relation $\sim$ is not affected.

(2) The proof is similar to the proof of (1).
First, let $u,v$ and $w$ be as in part (1). By assumption, the element $f$ has non-trivial slope at $1^-$. Replacing $f$ by $f^{-1}$ if necessary, we can assume that $f'(1^-)>1$. Hence, there exist $m>\ell$ in $\mathbb{N}$ such that $f$ has the pair of branches $1^{m}\to 1^{m-\ell}$.

Let $T'$ be a finite binary tree such that $u,v0,v1,w0,w10,w11$ are branches of $T'$. Let $T$ be the tree obtained from $T'$ by attaching the minimal binary tree with branch $1^m$ to the right-most leaf of $T'$.
Let $r$ be the length of the last branch of $T'$ and note that $T$ has the branches $1^r0,1^{r+1}0,\dots,1^{r+m-1}0,1^{r+m}$. (In particular, since $m\geq 2$, there are at least three branches in $T$ after the branch $w11$.)

 Let $(S_+,S_-)$, $T_1$, $T_2$, $C_1$ and $C_2$ be as in part (1).
Let $T_3$ be the minimal binary tree
with branch $00$ and $T_4$ be a tree which consists of a single caret.

Let $u_1,\dots,u_n$ be the branches of $T$.	 We use two copies of the tree $T$ to construct a new tree-diagram $(R_+,R_-)$ in $F$, by performing the
operations $(1a)-(3a)$ on the first copy of $T$ and $(1b)-(5b)$ on the second copy of $T$, as in the proof of part $(1)$. Note that there exists $5\leq k\leq n-5$ such that the branch $u_k\equiv w0$. By construction, the tree-diagram $(R_+,R_-)$ has the sets $\mbox{(A)}$ and $\mbox{(B)}$ of pairs of branches as in the proof of part (1) as well as the following set
\begin{align*}
	\mbox{(C')}& \begin{cases}
		w11\to w110& \\
		u_{k+3}	\to w111\\
		u_{i} \to u_{i-1} & \mbox{for } k+4\leq i\leq n-1\\
		u_n00\to u_{n-1}\\
		u_n01\to u_n0\\
		u_n1\to u_n 1\\
	\end{cases}
\end{align*}

Let $g\in F$ be the element represented by $(R_+,R_-)$. Then $\pi_{ab}(g)=(c,0)$. Let $H=\langle f,g\rangle$. The proof that $H$ contains the derived subgroup of $F$ is almost identical to the proof in part (1) (note that the first $3$ rows in $\mbox{(C')}$ also coincide with the first three rows in $\mbox{(C)}$). The only difference is in the proof that Condition (4) of Lemma \ref{lem:tree} holds for $H$ (with  the tree $T$ and the word $w$).
Let us prove that the condition holds. That is, we claim that for every $i\geq 0$, we have $u_n1^i0\sim w$ (where $\sim$ stands for $\sim_H$).
Recall that by construction $u_n\equiv 1^{r+m}$. Hence, we need to prove that for every $i\geq 0$, we have $1^{r+m+i}0\sim w$. Clearly, it suffices to prove that for every $i\geq 0$, we have $1^{r+i}0\sim w$.  By Condition (1) of Lemma \ref{lem:tree} (which holds here, as in part (1)), for every $j\in\{2,\dots,n-1\}$ we have $u_j\sim w$. Since $1^r0,1^{r+1}0,\dots,1^{r+m-1}0$ are branches of $T$ (which are clearly not the first nor the last branch of $T$), for every $0\leq i\leq m-1$, we have $1^{r+i}0\sim w$. We prove by induction that the claim also holds for every $i\geq m$. Indeed, let $i\geq m$. Since $f$ has the pair of branches $1^m\to 1^{m-\ell}$, we have $1^m\sim 1^{m-\ell}$. It follows that $1^{r+i}0\sim 1^{r+(i-\ell)}0$. Since $i-\ell<i$, we are done by induction. Hence, Condition (4) of Lemma \ref{lem:tree} holds for $H$, as required.
\end{proof}

Theorem \ref{thm:main intro} is a corollary of Proposition \ref{prop:main}. Recall the statement of the theorem.

\begin{Theorem2}\label{thm:main}
	Let $(a,b),(c,d)\in\mathbb{Z}^2$ be such that  $\{a,c\}\neq\{0\}$ and $\{b,d\}\neq\{0\}$. Let $f\in F$ be a non-trivial element such that $\pi_{ab}(f)=(a,b)$. Then there exists an element $g\in F$ such that $\pi_{ab}(g)=(c,d)$ and such that $\langle f,g\rangle=\pi_{ab}^{-1}(\langle (a,b),(c,d)\rangle)$.
\end{Theorem2}

\begin{proof}
	It suffices to prove that there exists an element $g\in F$ such that $\pi_{ab}(g)=(c,d)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$. Indeed, in that case, $\pi_{ab}(\langle f,g\rangle)=\langle (a,b),(c,d)\rangle$ and since $\langle f,g\rangle$ contains the derived subgroup of $F$, we have that $$\langle f,g\rangle=\pi_{ab}^{-1}(\langle(a,b),(c,d)\rangle),$$ as necessary.

	If $c,d\neq 0$, then by Proposition \ref{prop:main}(1), there is an element $g\in F$ such that $\pi_{ab}(g)=(c,d)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$, as required.

	If $c\neq 0$ and $d=0$, then by assumption $b\neq 0$. Hence, $f$ has non-trivial slope at $1^-$. Then by Proposition \ref{prop:main}(2), there is an element $g\in F$ such that $\pi_{ab}(g)=(c,0)=(c,d)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$, as necessary.

	If $c=0$ and $d\neq 0$, we are done in a similar way, using Proposition \ref{prop:main}(3).

	Finally, if $c=d=0$, then $a\neq 0$ and $b\neq 0$. Then $f$ has non-trivial slope both at $0^+$ and at $1^-$. Hence, by Proposition \ref{prop:main}(4), there is an element $g\in F$ such that $\pi_{ab}(g)=(0,0)=(c,d)$ and such that $\langle f,g\rangle$ contains the derived subgroup of $F$, as necessary.
\end{proof}

Finally, Theorem \ref{thm:almost3/2} is a corollary of Theorem \ref{thm:main intro}.
 Indeed, let $f\in F$ be such that $\pi_{ab}(f)=(a,b)$ forms part of a generating pair of $\mathbb{Z}^2$. Then there exists $(c,d)\in\mathbb{Z}^2$ such that $\{(a,b),(c,d)\}$ is a generating set of $\mathbb{Z}^2$. Since $(a,b)$ and $(c,d)$ generate $\mathbb{Z}^2$, we have that $\{a,c\}\neq \{0\}$ and $\{b,d\}\neq \{0\}$. Hence, by Theorem \ref{thm:main intro}, there is an element $g\in F$ such that $\pi_{ab}(g)=(c,d)$ and such that
$$	\langle f,g\rangle=\pi_{ab}^{-1}(\langle (a,b),(c,d)\rangle)=\pi_{ab}^{-1}(\mathbb{Z}^2)=F.$$
Hence, $f$ is part of a $2$-generating set of $F$.
\qed

redpajama_set_name: RedPajamaArXiv


Card number one made by Jenine using the Christmas Cocoa Specialty DSP and Christmas Punches.
This card made by Charlene using the Christmas Punch as well.
Renne made this card using the Hostess Level One set Patterned Pines from the Holiday Mini and the Merry Moments DSP.


Sandwiches of turkey, roast beef and sopresata are made with hand-roasted meats and homemade condiments at Hoagitos in Asbury Park.
What you should know before you indulge in one of the cute, homemade, artisanal $5 sandwiches on the Asbury Park boardwalk: The owners of Hoagitos beat out 30 other vendors to be there.
In 2012, Mark Spagnuolo, his wife, Emma, and brother-in-law Terence McGill were toying around with what they believed was a great business idea: sandwiches served up like tapas; sandwiches made with hand-roasted meats (turkey, pork, beef) and homemade condiments; sandwiches served on homemade rolls commissioned daily from Del Ponte's Bakery in Bradley Beach. The sandwiches at Hoagitos are more a small plate rather than a full meal, bigger than a slider but just as adorable.
Hoagitos managed to impress the team at Madison Marquette, the property management group that oversees boardwalk vendors. They brought in sandwiches. It was a green light.
The shop opened last summer and sold 800 sandwiches on its busiest day.
Hoagitos offers a tight menu, with few choices. Each sandwich is thoughtfully constructed. The roast beef, for example, comes with roasted garlic mayo and enough pickled onions to add bite, plus a bit of peppery arugula.
Turkey is brightened by a cilantro-lime mayo, and the pork tenderloin dressed up in creole mayo. One sandwich is a snack; two or three is a meal, says Mark Spagnuolo. Mix and match.
One never knows where Mr. Hoagito will pop up.
Additional whimsy comes from the company logo, Mr. Hoagito, a personified sandwich with a mustache and sunglasses. His likeness pops up all over town as part of the company's social media campaign. People spot Mr. Hoagito at The Stone Pony, for example, then post a Facebook announcement, an Instagram photo or a tweet.
"We just hand out stickers and see where he ends up stuck," Spagnuolo says.
This year, the shop introduced Kobe and portobello options to its menu, as well as breakfast sandwiches.
Hoagitos is open daily and stays open on weekends until 3 a.m.


Brick Town Talk: Another Church! - LEGO Town, Architecture, Building Tips, Inspiration Ideas, and more!
The really steep roof is probably my favorite part of this model. It has a very well thought out look and carries across that elegance and style of a real church.Nice job on this one.


Load your family into the 2016 Hyundai Tucson. Under the hood you'll find a 4 cylinder engine with more than 150 horsepower, and for added security, dynamic Stability Control supplements the drivetrain. Top features include a split folding rear seat, a rear window wiper, adjustable headrests in all seating positions, fully automatic headlights, remote keyless entry, cruise control, an overhead console, and power windows. You and your passengers will enjoy the stereo system, which includes a CD player with MP3 capability, steering wheel mounted audio controls, and 6 well positioned speakers. Hyundai also prioritized safety and security by including: dual front impact airbags with occupant sensing airbag, front side impact airbags, traction control, brake assist, anti-whiplash front head restraints, a panic alarm, and 4 wheel disc brakes with ABS. When road conditions become unpredictable, rely on all wheel drive to maintain outstanding control. This vehicle has achieved Certified Pre-Owned status, by passing Hyundai's comprehensive certification process, including a rigorous 150 point inspection! Our sales reps are knowledgeable and professional. We'd be happy to answer any questions that you may have. We are here to help you.