Hey, yep, we're on What is up my exchange family from all over the world and thank you for tuning into another episode of chiefs chat My name is Chief Massar and Kevin Osby, and I'm your senior listed advisor for the Army and Air Force exchange service Before we get started with our guests today. I would like to introduce my lovely and they told me to stress lovely lovely Matthews and Julie Mitchell. How y'all ladies doing? Gosh, hi chief doing good. Hi chief. We are good and we do you want to say that your predecessor never referred to us as lovely So thank you. Thank you for that. We appreciate it. I'm gonna change the game So today we have a very special guest that's been doing some outstanding work for our veteran Julie Do you mind introducing today's guests? Not at all chief You're right. We are very honored to have today's guest with us He served in the Army National Guard for 27 years including two tours in Iraq and after his retirement last year He was unanimously elected as national commander for disabled American veterans Please join us in welcoming commander Stephen butch whitehead Thanks guys Butch thanks so much for joining us and taking time out of your day We really appreciate it and for everybody watching. Thank you as well drop a note in the comments Let us know where you're tuning in from and if you have any questions for butch will be reading those live throughout the broadcast Now's a good time to start your watch party to enjoy this live interview with your friends And if you're not already following us, you should because cheap chats are every Tuesday and Thursday. So then you'll know who's coming up next Awesome well butch. Thank you for joining us today. It's great having you with us Uh, I first want to start off asking about your your name butch It's a very unique name. It probably comes with a lot of misconceptions So Can you please give us a little insight on on your name? I sure can chief, you know, and it's actually first of all, thank you guys for having me on I was truly an honor to be joining you chief and your talk here. It's great to be part of it The nickname though chief that's it is very unique. I actually had it ever since I was a little baby You know, I think I was the first name my Mom and dad's friends gave me was butch and it stuck with me all through my life all childhood all the way to today So it's very unique because the fact of it's actually I don't know where my friends my mom parents friends came up with the name butch So I asked I always asked it was because of short hair. I had did I not have hair? What was this whole thing that got me butch? I never got a true answer out of anybody about how I got the word butch So it's just but it's stuck I I think the dog in in tom and jerry's name was butch too. Uh, that could be maybe I was the cartoon guy, you know It could be that exactly that Awesome. Awesome. Well, I told I was telling my co-host uh yesterday I should change my name to butch because then that that'll our our complaints will probably decline a lot if I have to go to chief butch There you go That's awesome. So where you calling us from today? Well chief. I'm actually on here this great state of minnesota Uh, and I'm talking to the host this morning. We finally we flipped the page a little bit And it's gorgeous weather for us right now. Um, I think at today's high it's supposed to be maybe 78 70 degrees low humidity It's gorgeous up here minnesota. That's awesome. When is the when does the switch flip up there? Well, I'm sure we'll get a nice couple more a couple weeks of hot weather humid weather But the the weather the slide I would say july was very very hot for us I mean, I think we're in the 90s and then with the heat index into the hundreds for several days in july So it was very hot and here in minnesota with all of the lovely 10 000 plus lakes that we have the humidity Thanks up pretty high pretty quickly when you get those warm days. Yeah Absolutely. Okay. Thank you So butch can you tell us a little bit about your military career and kind of what led you to a life of service? You know the military for me. It was kind of a family-driven thing. Uh, both my grandparents my father I'm the bkb of seven and all three of my brothers joined the military as well Um, everybody thought I was going to be joining the air force chief. I'm sure you'd like to hear that Uh, my oldest brother was army next brother was navy next brother was marine So that just left either the coast guard air force. So all of my aunt's uncle So I guess butch is going to air force and here I went army And it was one of those things that I was actually going to enlist in the marines My brother that was in marines. He was uh, when I turned 17 He was actually deployed to kawai for the gulf war And uh, he was one of my idols and I looked up to him, uh, obviously he and I were the closest in age Uh, but he had the power of calling back to talking to mom dad the same day actually He set up a meeting with the marine recruiter to cover my house and he said do not let him join the marines I'm so baffled by that, you know But he wanted me to go off to be a college. Uh, so I joined the army national guard at the age of 17 And uh, one thing that's changed from the bio that you read there, uh, Julie is I'm actually still serving I actually didn't fully retire. I took a leave of absence when I became commander and instead of going into the full retirement, I stayed in and now I am also the Division 34th infantry division commands aren't major for the 34th infantry division up here in minnesota One of the one of eight army national guard Divisions. Oh, that's great. I did not know that. Thank you for I'm sorry. I got it wrong. Um, I Is the bio that was out there last year, you know when I was when I was elected, that's the true thing, you know But later on, um, I decided to keep the uniform on I never fully retired Um and uh continue to serve today Oh, that's that's awesome. And so just to give you a little uh history on myself I started off in the marine corps. So, uh, okay. Yeah. Yeah. And so nobody gave me that briefing. Don't join the marine corps I just went well the recruiter showed up at my house and then the rest was history. So that story behind that but Must be some of the marine recruiter's a little bit more aggressive than the army the air force marine recruits Must be a little bit more aggressive when it comes to that. Yeah, I think I was 16 when they started Started nearly Wow, and so butch now you're serving still with the disabled american veterans or DAV So can you tell us a little bit more about DAV? What's your mission? Who do you serve and then what resources do you offer? You know, I you know Leah, that's a great question because you know one thing that really thrives really well with me with the DAV and why I'm so proud to be you know part of it and not just because I'm a national commander but Just being a member of the organization is it's truly my passion that is taking care of those that You know fight for our our country and when I came home from my deployment to iraq um We had a bunch of vietnam veterans there that met us when we got off the bus and uh the message that they told me is that We'll take care of you we have your back and I truly it really sunk in really hard with me because of the fact of I didn't know what I was getting into now. You know what a lot of things have changed life does not stop When you deploy and we're in the military life continues and on a deployment I was on a 18 month long deployment to iraq and a lot of change when I came home knowing that somebody was there for me and The dv was that so I was able to lean on the dv through some struggles that I had They got me to identify the resources that needed to continue to move forward and not You know fall into deeper into a darker hole, but the dv was definitely there for me Awesome, so okay, which can you tell the viewers about the benefits of having a dv membership and how can you know you know chief you're When you're in the military you're part of a family right and when you wear that uniform Your brothers and sisters all around you and being a member of the dv Is just that too, you know, none of us have a uniform on we take the uniform off Southern country now we put a different uniform on that's our civilian uniform But the dv is still a family and some of the perks about getting a dv member is we are very tight We're very connected, but some of the you know special perks, you know, we have great benefits that our members They can save on their prescriptions for their medication. We get discounts when you purchase, you know More for the company is a huge huge support of ours and being a member They give us a little perk for when you're buying new vehicles and stuff like that um When we rent vehicles, you know aphids budgets and stuff like that So there's all these little miscellaneous small perks, but I would say the biggest benefit we have is being it We have another family that we can lean on through some tough times that we don't we're not left out there by ourselves That's awesome So you took you mentioned this um Little bit before but you became national commander Last year 2020 has been pretty crazy so far So how has the pandemic affected dav and veterans in general? You know, I think the biggest uh Well, I want to issue the biggest hurdle that we've had across julie when it comes to the kovat and what it's done to the veterans it's really You know put a hinder on our veterans getting the health care, right?  It was only outpatient only But it also it forced us to expand, you know Making sure the va's looking at other resources and that has been a great resource to The kovat has helped us become more telehealth right now We have veterans understanding the telehealth and how that's also still being able to get you the health care You need you know Seeing a doctor and things like that. So that's been a big thing for us. Uh pushing for that advocating for that Um, the kovat also is affected, you know a lot of our this is a hundred year We are actually celebrating a hundred year centennial this year And we've had to cancel a lot of our events. Well all of our events actually so far since my last travel was in february I haven't traveled since and We had a lot of trips planned in between there. So while it's a good thing I've been able to stay home longer, but it's also I look as a downfall for the fact of I wanted that you know The country needs to know that we made a hundred years and here's why we made the hundred years You know, we're doing such wonderful things taking care of our veterans out there that fought for us and But we're so continued to do that. Uh, but it's we just had to change the way how we do it Oh, well, happy anniversary. Happy 100th to you guys. That's that's quite a milestone and I'm sorry that you guys that it's kind of been More subdued this year. That's that's disappointing for sure But I hope that you guys are able to you know, find other ways to connect whether that's virtually or find other ways to celebrate their Great work that you guys are doing Julie, that's a great point. I'm glad you brought that up because you know one thing I want to highlight Through this pandemic stuff this covered what it really did is we were still able to help Thousands of veterans, you know with their benefits, you know, we really shall we're still able to do because of technology We didn't stop helping veterans through any of this covid time and that was a huge win for us because As we all know the covid-19 affected so many people when it came to Unappointments and loss of revenue and things like that And our veterans rely on the disability and if that's only income they have coming in or maybe we're waiting on it We're able to still help them through that as well as we we started up a Grant process that veterans could apply for Small grants if they were a disabled veteran that we were able to help give a supplement income as well That's great. That's great that you're still taking care of family Just like you have done for 100 years. You didn't let the pandemic stop that so hats off to you for sure Yes, and uh, I got a happy anniversary To da v and all the viewers out there So earlier this year the the exchange was honored to provide a new benefit to veterans All veterans with service connected disabilities were welcome back to their exchange with in-store shopping privileges In your view, what's the feedback that you get get from the veteran community on that benefit? It's been long overdue as a vote up in here. You know, they were so happy We've been advocating for that change for several years and we were so happy and our members Were so happy the disabled veterans were happy to have that change You know, as we all know a lot of our veterans They have a tendency to live really close to Through the military base. So that's just the life they knew that's where they tend to go back to and be close to the military bases and Have the ability to go in and take part and be you know The a fees and the great service that you guys provide is a great tool It's a great thing that the veterans really are proud to be part of and they wanted to use So they were very very excited to be able to come in and use the the benefits of a fees. Yes, they absolutely deserve it So we're glad that they're you know back back on bases and uh and shopping with a fee That's great in butch You've been with da v for a while Including as the adjudant for the department of minnesota. Can you talk about how you first became involved with the organization? Yeah, lia, you know, as I said a little earlier, you know When I got off that bus and that vietnam veteran who happened to be a da v member You know when that that organization itself reached out to me to make sure that they were there for me That really excited me it fell in love Made me fall in love with and kind of drive my passion and you know My wife has been a huge support of that and I think that's what kind of drove me to be part of the da v Is she always said when that phone rang it would be one of my soldiers I would literally drop what I was doing and be there for that soldier and that's literally what Got me excited got me through my tough times as well. And that's what the da v did for me and I wanted to continue doing that So here up in minnesota being the adjudant uh being part of the Setting up program of services at the local level that national does but even expanding them a little further For instance here in minnesota, we have a wonderful, you know, obviously I said earlier I talked about the 10 000 lakes, right? We're a huge outdoor community up here and you know taking disabled veterans and getting them acting the outdoors When you're disabled not knowing that well, I suppose I can't do that anymore because i'm disabled And we're showing these disabled veterans that even with a disability They can still get back out enjoy the activity. So that's one of the great passion about here in minnesota being the adjudant And at the national level continue to spread the word and let the disabled veterans Know that they're not alone and that there's somebody there to help them through any struggles He or she may be going through that there's people out there that care and want to help them And that's so important Especially right now when people are feeling more confined and feeling that they can't go out and do things to be able to Show them that you can go outside and you you can still have a life and you do still matter and you We are all part of a big connected family. That's terrific. Absolutely terrific Exactly and yeah to still feel connected to people Yeah, absolutely, you know the one of the great programs at the national level that we partner with veterans affairs I want to highlight there's two great programs that we do One is called the winter sports clinic, which is out in colorado and that's um A program where we're showing these disabled veterans that do know there's downhill skiing and it's uh sledding it's cross-country skiing Rock climbing we bring in hundreds of disabled veterans all across the country and A year bonding with other veterans maybe have a similar injury or illness that you have but we're also teaching them like I said earlier that You can still do things and then the other event is the t tournament It's partnered with the va as well. That's held in iowa iowa city and I think about 80 of those veterans that Just do we lose them? Yeah, I'm not sure if we lost them or not Which I think you may be frozen if we might have lost yeah, we might have lost you Let's cross our fingers that you come back Can you hear me now? Welcome back to chief It's not about that. I don't think I'd love this technology. Um, I know right Boom it's gone. It's okay So but now what I was saying is about the t tournament and how 80 of the participants are legally blind And we're all golfing three or four days, you know legally blind person and showing them that they can still golf You know and that was when you see it you get you know choked up you just get emotional yourself You're like oh my god. I'm glad I'm part of an organization that's doing this for veterans. Yeah, and I'm sure that can still beat me in golf So you became the national commander last year and then you were talking about how this year there wasn't a convention So you kind of have an additional what you have an additional year to serve an additional year to get things done Um, so what are some of your goals and any specific initiatives that you're focusing on, you know With it being a one-year commitment and like you're saying we didn't have our convention So I did get extended one more year, which I'm honored to serve, you know, I'm very honored to serve One additional year But one of the goals I'm still very pushing for is the advocacy for veterans and making sure our legislators are, you know Doing what there should be to take care of those that serve, you know A couple big things that we'll continue to push for Is the caregivers, you know and people forget about they don't forget But they have a tendency to not remember that when that service member is wearing that uniform There's usually a caregiver on the back side, you know a spouse and things like that And when the service member becomes a disabled veteran That caregiver sometimes has to make some sacrifices here. She is well And we want to make sure that they're taken care of so that's a big initiative of mine And then as well as I want to make sure we continue to uh advocate, you know For the burn pit and toxic exposure. It's another big issue that I think Our service members are being exposed to uh, we were calling it very similar to the Agent Orange that are vietnam veterans Are exposed to so the advocacy is very important to me But also the on the other side is continuing to make sure that all of our veterans are being taken care of Awesome and and yeah those burn pits. I know a lot of uh, you know fellow fellow service members that that uh were in charge of, you know stirring the the burn pit and and being, you know, uh at lower ranks doing those type of duties So i'm glad you're you're advocating uh for their health, uh because it does have long-term effects it does And it's and so we find out what they are.  Yeah So and you mentioned about the uh convention being cancelled uh last year and congrats on your new term So, uh, and I know kovic you just changed a whole bunch of stuff So, uh, how are you guys gonna reach out to the veterans despite not having a annual event this year? You know, we're we're actually doing a virtual uh Update here actually next week. We're actually going to do a virtual Update with all of our members we sent out So we're still going to do our reports. It just will be no elections. It's just going to be all informational only So we're doing that virtual Events as well as we're continuing to rely on our our local chapters And departments to continue to get out and do the outreach and just maybe smaller gatherings We can't do these big gatherings obviously because the cdc and this covet keeping our our members safe is very important to us But we can still get out there and research not research But reach our disabled veterans to make sure that they're being taken care of And i'll tell you what we're using today the zoom meeting That's a great tool And who would have thought you know five years ago This is going to be the way we're holding meetings conducting interviews But i'll tell you it's been a godsend to be able to still do our work and still be able to reach out to veterans Excellent So butch just want to pause for a second and take a look at the live feed Wanted to share that people are tuning in they're watching from really all over the world As far as germany right now it looks like Christina says so very honored to be of service to our military. I couldn't join and so for me This is such a great way to serve Oh awesome. Thank you. I'm not sure what capacity Christina is serving in but thanks to Christina and butch. Thank you for your service as well Thank you Yes, and thank you for your time this morning and then before we wrap up Can you tell us where can veterans go to find out more information about da v and all of the resources available to them? You know our website has a lot of great information on it You know the da v.org and i'll say that again da v.org um You can go there and if you need help or And have some questions about a particular filing for benefits or anything with the va You can click on and type in your zip code and it'll tell pull up the closest resource for you But as well as if you want to volunteer Our website has a great link to show you where you can sign up to be a volunteer as well So da v.org is a great great tool Got it da v.org Last year liya and i were honored to attend the convention in orlando We had a great time and we really enjoyed talking with yes It was it was a good weekend We got to talk with some of your members and I remember liya We talked to a 92 year old world war two in korean war navy veteran who was going to have Be welcomed back with his exchange shopping privileges in store in january and he and his wife They were overjoyed. They lived in columbia south carolina They were going to be able to go to fort jackson to shop in person and he was just blown away He was one of the most memorable people i've ever met in in my life. Um, it was such a great i'm so honored to have been there with y'all and to Be able to talk to him and um help make you know a difference to to those who have served And our great nation so it was great You know, it's a great tool and we got to keep advocating and let make sure everybody knows it's out there I still think we have veterans that are not aware that this Policy changed and we got to keep pushing for that to make sure that again It's such a great great tool a great resource for our disabled veterans to be able to take advantage of We got to make sure that they know about it Awesome. Absolutely. Yeah, well butch man. Thank you so much for for just your dedicated service of In uniform and out of uniform you've been doing A ton for for our country for for a lot of years. So thank you for your Unwavering service to our our country. Thank you for taking some time to chat with us today You shared a bunch of wonderful valuable information out there. So hopefully people get on dav.org Maybe we can post it in our comment section So people have the link this means so much to the airmen soldiers sailors marines and coasties We wish you all the best and thank you for taking care of our veterans Thank you chief and lia. Julie. Thank you. I was truly an honor to be part of it And I really enjoyed my time chatting with you guys today Yeah, and if you don't mind staying uh staying after the live feed, uh, I gotta get some information from you Absolutely It's our pleasure. Thanks, butch. Thanks everybody. Take care. All right. Thank you chat out


function that sort of, I don't want you to think of it as some sort of trivial function that regards to what goes in, e comes out. Then sigma is a group homomorphism. Then sigma is a homomorphism, a homomorphism of groups. I'm going to start abbreviating homomorphism just by hom. Let's check that it is. It's not too bad to do. Let's see, take any two elements of g. So pick x and y and g and compute each of the two expressions. If I compute phi of x, y, folks, if I compute sigma of x, y, what does sigma do?  That was easy because I don't care what's in the window. It doesn't matter. E always comes out. But what's sigma of x, sigma of y, e times e, which is e. So the two expressions are the same. We usually call this the trivial homomorphism. It's no big deal. All right. More examples. Interesting ones here. Example. Let g be this group, the group of invertible n-by-n matrices with entries from the reals. So what does this mean? It means n-by-n matrices fixed n, n-by-n matrices with non-zero determinant. So that's a group. And here's what I want you to do. Pick any element in there. Any, let's call it, I don't know, capital A. So A is some matrix. You've written it down and it's not going to change throughout the discussion. Now we're going to write down a function. Define, I'm going to call it phi sub-capital A. It's going to be a function from g to itself. You're starting to maybe see that a lot of homomorphisms turn out to be homomorphisms from the group to itself that certainly doesn't have to be the case, as in the example from SN to Z2, but it turns out to be an important case. This is going to be one of them. As follows, phi sub-A of any element in the group, well the group consists of matrices, so it's not unreasonable to call a generic element in the group M equals A inverse MA. So regardless of what you throw into this function, what comes out is what you get when you multiply on the left by A inverse and what you multiply on the right by A. If you've got an element in the group, it's got non-zero determinant. Well, if you've picked something in the group, it's got non-zero determinant, so its inverse exists, it's got non-zero determinant, so all this stuff winds up back in g. So the fact that this is a function from g to g is not that big an issue, but it turns out to be a homomorphism. Then phi sub-A is a homomorphism. Let's check it. Check. Let's see, why is that? We need to take any two things in the domain group. Well, in the general definition, I called those things x and y, but you should call them whatever is representative of the things in the domain group. Here are their matrices, so let's call them, I don't know, M and M prime or something like that, or M and N. Let's do the computation. If I do the first computation, the first computation says go ahead and combine the two things inside the group and then run the product through the function. Well, folks, here's what this function does. It takes whatever it's handed, it multiplies it on the left by A inverse and on the right by A. On the other hand, if I compute this expression, that's the second thing I need to compute. I need to take the function, run each of the individual group elements through it individually, and then combine, well, let's see. The definition of this is you multiply on the left by A inverse. You take whatever's in the window and you multiply on the right by A. The definition of this, you multiply on the left by A inverse, you multiply. Oh, and you see what's going to happen here? It's a trick that we've seen before. That is the identity matrix, A inverse M identity and A. Yeah, but if you multiply by the identity, it doesn't change anything. The two are equal, equal. So we've got a homomorphism. In fact, it turns out here, folks, that there's nothing really special. Megan, question? Nope. There's nothing really special about the fact that I happen to be working with matrices. In fact, you can define this sort of function from any group to itself. Take a group, pick your favorite element in the group, call it A, and define something called phi sub A from the group to itself simply by setting phi sub A of the element to be A inverse times M times A. Now, if we were to do this, but instead of picking GLNR, we were to pick something like the complex numbers under multiplication in a billion group, then this sort of operation, I mean, if the group's a billion, then this just slides across that and you get A inverse times A, you just get M. So a lot of times this type of homomorphism turns out to just be the identity homomorphism. In fact, it's not too hard to show that if you pick something called A and you can slide A past all the elements in the group, maybe if the group's a billion or if you start with the identity element or something like that, then it turns out that this homomorphism might be sort of uninteresting. It might be one of these trivial homomorphisms, but in a situation like this, this turns out to be at least non-trivial because what's coming out is not necessarily what is not necessarily the original matrix. Alright, let's do some more. More examples. Example, I'm going to hand you two groups, G1 and G2, and I'm going to ask you to form their direct product so that gives me a group. So it makes sense to talk about a function from this group to other groups. Well, how about G1? I can do the same thing with G2 in a minute. Here's the definition of the function. What I'm asking you to input are ordered pairs, and what I'm going to ask you to spit out is just tell me what's in the first coordinate. Bless you. So all this is, folks. I mean, it's something that you're familiar with in linear algebra as well. It's simply the function that's typically called the projection function, the projection onto the first coordinate. Here, all you're doing is taking the function that picks off the first coordinate and fees the homomorphism. In fact, let me call it piece of one is a homomorphism. The one, because what I'm asking you to do is pick off whatever's in the first coordinate here. Reason, it's easy to do. In fact, I'll leave it to the reader here because it simply boils down to how you combine things inside the appropriate product. So I'll let you pound out the appropriate equation here. Let's look at another example of a homomorphism example. Yeah, let's do two more here. If I look at, so this is the collection of polynomials, polynomials, coefficients in R, and the operation is addition. So this is a group that we looked at once, way back when we were looking at a lot of examples of groups, but haven't really pulled this one back up yet. So the elements of this particular group are simply polynomials, you know, a0 plus a1x plus a2x squared, et cetera, et cetera, et cetera, and you simply add them together. It's no big deal. The zero polynomial is the identity element. If you have me a polynomial, you just throw negative signs in front of all the coefficients and you get the negative of it. Well, here's a function from this group to itself, define phi from g to g. In fact, instead of calling it phi, let's call it something a little more suggestive. This is Greek letter delta, and the definition of delta is take delta of something in this group. Well, the things in that group are polynomials, so it's not unreasonable to call a polynomial that, and what I want it to spit out is the derivative of the polynomial. It makes perfect sense. I know what the derivative of a polynomial is, and it's another polynomial. So at least I've got a function from polynomials to polynomials, and it turns out this particular function is a homomorphism reason. Let's see, what do we have to do? Show delta is a homomorphism. Greek letter delta because this stands for derivative, so it's the letter d here. So we have to pick two things inside the domain. You know, in general we call them x and y. Sometimes we call them m and n. Sometimes we call them g1. It doesn't matter.  How about delta of f1 of x plus f2 of x is you take the derivative of it. So it's the derivative of x. That's the definition of the function. Just take the derivative of whatever you've been handed. Yeah, but equals property of what, calculus result? If you have two functions in the atom together and you take the derivative, it's the derivative of the first plus the derivative of the second. It's a one there that's prime. So you proved in your COC-1 course that, I mean it's just properties of derivatives here. But on the other hand, if I do delta of f1 plus delta of f2, that's exactly what I get. f1 prime of x plus f2 prime of x. Check. So the point is that the derivative function is a homomorphism from the collection of polynomials to itself. And if you took linear algebra from me, that was a strong example that just all the time has an example of a linear transformation from a vector space to itself. It turns out that this thing happens to be a vector space with coefficients in the real. But all I'm worried about is the addition operation. The addition operation makes it into a group. All right, final example before we move on and look at some properties of homomorphisms. And the final example is, oh yeah, this one. Let's call it v from the group of non-zero complex numbers to the group of positive real numbers. Both of these of course under multiplication. And the definition of the function v is take v of some complex number. And what I want you to spit out is the length of that complex number. I mean it's absolute value if you want to think in terms of Pythagorean theorem or find that complex number in the complex plane and then just tell me how far it is from the origin. All right, so there's a function. Let's see if I hand you a non-zero complex number, the length of a non-zero complex number is always a positive real number. So at least it lands inside this group. So the function is a legit one and now the question is can we show it's a group homomorphism and it turns out to be, well let's see what's the function v of alpha beta is the length of alpha beta, which is property of complex numbers. Of complex numbers that's the same as the length of alpha times the length of beta as the length function preserves multiplication. But on the other hand v of alpha times v of beta is the beta and we're done. The two are equal. So there's another example of homomorphism. So there are homomorphisms everywhere. You've got to be careful just because you write down a function there's certainly no guarantee it's going to be a homomorphism. So I tried to keep you a little bit honest on Monday by giving you that example. I mean here's a perfectly good way of getting from the integers to the integers. Just add four to everything sort of shifting by four and it turns out that's not a homomorphism. We gave a specific counter example to show that. What we're about to do and folks this is what you always do when you write down a new sort of structure.   Then show that if you somehow know that you've got one of these things what necessarily follows. What's necessarily true about this thing? Just like we do with groups.  We wrote down subgroups. We wrote down properties that subgroups necessarily have to have like Lagrange's theorem or something. Now we've written down a new notion, homomorphism.  Here are some properties. So here are some properties of homomorphisms, of homomorphisms. So the hypothesis is proposition assume that you've got a homomorphism from a group to a group. Be a homomorphism, homomorphism. Then here's the first thing that's true. Here's the first thing that's true. This property isn't given in the definition but it turns out if you can convince me that you've got a homomorphism then necessarily the following properties have to hold. If, for example, you choose to plug in the identity element of the group here what you get out is the identity element of the range group. So I don't think I need to write out what this notation means. E is the identity element of G, E prime is the identity element of G prime. It turns out if you have a homomorphism necessarily the identity element of G has to land on the identity element of G prime. Oh, here's another thing that's true about homomorphisms. For each element in the group, let's call it A, A in G, if you throw in the inverse of A into the homomorphism, in fact what you get out is if you had thrown in A into the homomorphism written down whatever the output is over in G prime and then told me what its inverse is. So the verbiage is homomorphisms preserve identity elements and homomorphisms preserve inverses. That shouldn't be surprising because homomorphisms preserve the group structures. That's the sort of philosophy or the intuition behind what a homomorphism does. Again, this isn't part of the definition and it's not something that we just hope to be the case. In fact it's something that we'll be able to prove is true. So let's prove these proof. Well, I'm going to haul, I'm going to haul something out for you. This is a good one. Remember you prove for homework the following property. That if you hand me a group there's exactly one element in the group that is what's called item potent. There's exactly one element in each group with the property that when you combine it with itself you get itself back. As we saw in the exam if you're not in a group structure there might be more than one. As we saw in the exam if you're in a group structure there is only one that came from a homework problem. So we know that to be true because we proved it for homework problem. So I'm going to use that property of groups. So folks if I want to convince you that this is E prime well this is an element of that group just because phi is a function from g to g prime. So this is an element of g prime if I can convince you that this is an item potent element of g prime. Folks there's only one item potent element in g prime. It's the identity of g prime in other words a d prime. So we use an old homework problem. Old homework problem. Problem. The result of which says there is only one element in g prime having the property that when you combine it with itself I'll call it a star a equals a it's e prime. So the point is so if we can show that this particular element phi of e combined with itself is itself in other words if I can show that this specific element of g prime has the property that when you combine it with itself to get itself back then we can conclude that phi of e is the identity element of g prime. It's nice we did some work for homework that we can now use here in a really nice way and this is easy to do easy. So I'm going to show you this let's see phi of e star with phi of e and maybe I should be a little more I mean this is the the star prime operation but I'm already getting a little bit sloppy on the notation is why wait a minute phi is a homomorphism so if you take phi of something star with phi of something else it's phi of because phi is a homomorphism might say well you've written it down backwards folks it's an equation I don't care whether you write this equals this or this equals this it's the same information here we're sort of starting from the right hand side and trading it in for the left hand side that's fine oh but wait a minute what's e star e? e star e is just e check that was easy so I've just shown that phi of e star phi of e is phi of e so the conclusion is that phi of e z prime okay property two same idea let's see how do you show that something is the inverse of something else you convince me that if you combine the two you get the identity element of whatever group it sits in so here's all we need to do we need only show the following is true that if we take this thing phi of a inverse and we combine it with phi of a that we get the identity element of the group in which these things live here is a thing I'm going to claim that it's the inverse of that yeah being the inverse of something means that when you combine it with the other thing that you get the identity let's see if it happens but that's easy to do phi of a inverse star phi of a oh phi is a homomorphism that's phi of a inverse star a which is what phi of e which is what we just proved what phi of e is we just proved in part one if you drop the identity element of the group g into this function phi and phi is a homomorphism that e prime comes out check so we're done with that one too so these homomorphisms really do as advertised they somehow preserve much of the group structure that's involved the identity elements have to go where they need to the appropriate the appropriate inverses go where they need to quick observation what we've just shown is that if you have a homomorphism then necessarily it takes the identity element of the domain group to the identity element of the output group so here's a quick proof that the function that i wrote down on monday the function from the integers to the integers that sent phi of z equal to z plus four in other words the add four function can't be a homomorphism because what happens to zero zero if you add four to it goes to four so it can't be a homomorphism because every homomorphism has the property that if you throw in the identity element that the identity element has to come out all right so we get some sort of easy check marks yet that particular map can't be a homomorphism so here's some properties now just as when you looked at linear transformations in your linear algebra course um you talked about various subspaces of the input space or the output space you talked about the kernel of the linear transformation you talked about the range of linear transformation if the linear transformation happened to be a matrix multiplication that boiled down to looking at row spaces or column spaces or something like that it turns out similar quantities are around in the more general context of group homomorphism from one group to another you can talk about well let's see what is the kernel of a linear transformation it's the collection of things in the input space that wind up going to zero in the output space that's what the kernel is well we're going to talk about the same sort of idea here and it turns out we can talk about the range of linear transformation so uh if uh phi from g to g prime is again homomorphism then we make the following definitions define the kernel of phi and here it is the kernel of phi here's the notation for it k e r phi sometimes we use the prens and sometimes we don't it's the intuitively it's the collection of things that you put in that wind up going to the identity element when they come out so it's the collection of elements in the input group with the property that when you run the thing through the function that what comes out is the identity element of the output group let's do a bunch of examples uh no yeah let's do examples first and then we'll look at some properties example well we can run through some of them like uh how about the function that we wrote down sigma from s n to z two the one we looked at last time this was you know sigma what do we call this one phi i guess we called it phi sorry phi of sigma is it's the parity function it's the zero if sigma is even and one if sigma is odd so the question is what's the kernel of this homomorphism well the kernel by definition is the collection of things inside the domain so it's a subset of the domain it's those things that have the property that when you run them through the function that the identity element of the range or the codomain or the target group gets spit out so the question that we're asking is what are the inputs that have the property that well let's see the identity element over here is zero so what are the things that spit out zero answer the even ones so the things inside here that get mapped to zero are precisely the even permutations and we had a name for that subset it was called a sub n so here the kernel of phi is the even permutations which just coincidentally happens to be given a name and as we'll see not so coincidentally not only is it a subset of g this one happens to be a subgroup of g let's do another example how about yeah let's do this one oh yeah um this thing we call delta from r bracket x to itself the derivative function turned out to be a homomorphism what's its kernel it's the collection of things in here that have the property that when you run them through this function that zero comes out so rephrase that it's the collection of things in here with the property that the derivative is zero well what polynomials have derivative equal to zero the constant functions is the constant only polynomials that have derivative equal to zero all right that's sort of an interesting example let's try one more example this one what i call it v from the non-zero complex numbers to the positive real numbers again both groups having operation multiplication v was the length function question what's the kernel of v it's the same question you're asking every time if you want the kernel you simply look inside here and you ask which things in the domain spit out or get taken to the identity function in the range so the first thing you want to do is identify what the identity element in the range is well remember this group is under multiplication so the identity element over here is one that's the operations multiplication so the question we're asking is what complex numbers have the property that their length is one and the answer is those are simply the complex numbers that sit on the unit circle the things that are length one away from the origin so kernel v is what we call the unit circle if you want to draw these as complex numbers if this is the complex plane then the kernel looks like those things that are one unit away from the origin there's a picture of the kernel and we can write down the kernel for other examples i mean sometimes it's the case folks that the kernel is just zero so let's do one more example of a situation where we've got a homomorphism and the kernel is just well the example let's see the homomorphism that we looked at last time uh phi from z to z the multiplication by four homomorphism phi of z equals four times z you know the observation at the beginning of tonight was nothing special about four just do it what's the kernel of phi well it's the same question that you ask every time what things in here spit out the identity element here well identify the identity element over here zero so the question is what things here spit out zero what things here have the property that when you multiply by four you get zero just zero that's the only thing that has the property that when you multiply it by four zero gets kicked out so it's possible let's see it's possible that the kernel of a homomorphism is just well it's just one element is zero now make an observation and then we'll write down a proposition that you all know how to attack we'll probably prove it because it's important enough in here for what we're going to do later on but note let's see i have a property of homomorphisms it's the property that i just erased property one we know that phi of e is e prime for any homomorphism for any homomorphism phi that was property one of the proposition wait a minute what's the kernel of homomorphism the kernel of the homomorphism is the things that you plug in that spit out e prime well folks it's always the case that if i plug in e that e prime gets spit out that's what we've proved in proposition one so it's always the case so in particular we always have kernel of phi anytime we have a homomorphism so there's always something in there the identity proposition phi from g to g prime a homomorphism and folks when i make this statement of course there's lots of other window dressing i could put in but we don't need to do that as soon as i tell you i've got a homomorphism this automatically implies that the two things i'm looking at are groups with whatever operation you've got it also doesn't preclude the possibility that g prime is g but that's right then the punchline is this then let's see the kernel of phi well the kernel of phi is a subset of g in fact not surprisingly is a subgroup of g the kernel of any homomorphism from g to any other group you want i don't care what g prime is you always get a subgroup the original group guess how we prove this subgroup theorem exactly right okay theorem in fact let's knock through all three steps quickly step one subgroup theorem pick two things in the subset show that the corresponding binary operation on those things is in the subset so pick x and y in the kernel of phi show that x y is in the kernel of phi all right so we can all do that in our sleep now let's see what that boils down to in this particular situation the definition of kernel of phi means what does it mean to say that x is in the kernel of phi it means that phi of x is e prime all i've done is rephrased what this inclusion means by definition and phi of y is also e prime that's what it means to be in the kernel what do we have to do we have to show that this thing is in the kernel what does it mean to be in the kernel it means if you drop it into the function that e prime comes out so we have to show that if you drop that thing into the function that e prime comes out that's the task and hopefully you see what's going to happen here well but phi of x y is what phi is the homomorphism folks so if you have phi of phi of x phi of y and as promised the notation has gotten a little bit sloppy but we're familiar enough with it or comfortable enough with it technically this should be x star y that's the operation going on in g technically this should be star prime because this is the operation happening inside g prime but as soon as we know where everything is it's understood what the operation is so i don't have to keep putting in the stars in the star primes oh but now we're in good shape phi of x is e prime and phi of y is e prime and e prime star e prime is e prime check that was easy and just in the interest of time i'll wrap things up and say well you have to show the identity elements in there done there it is check you have to show that the inverse of any element in there is in there and i'm not going to prove it for you but i'll let you think about it at home three follows directly from from property two of the proposition that we proved above of previous proposition i'll let you prove that how does that follow well think about it part two of the previous proposition says that if you take the inverse of an element and you run it through the homomorphism you get the inverse of what would have come out so if you take something such that phi of a is e prime and you want to prove that phi of a inverse is also e prime well phi of a inverse is phi of a inverse but phi of a is e prime so it's the inverse of e prime but the inverse of the identity is just the identity all right so we get a subgroup and this subgroup is not just any old garden variety subgroup these types of subgroups will play an unbelievably important role when we try to build a new type of group based on what's going on inside subgroups so you might think well these are pretty what's a good property these are pretty ubiquitous that's a good word these should be everywhere because these subgroups are allowed to be the kernels for any homomorphism that you want so start with any homomorphism the kernel is always a subgroup but it turns out the kernels of homomorphisms are actually very special kinds of subgroups okay the first thing we're going to show about kernels is that they can determine whether or not the homomorphism is a one-to-one function which is a pretty surprising result and the second thing we're going to determine or show about kernels is that kernels have a very special property that if you have a subgroup that happens to be the kernel of a homomorphism then necessarily every left coset of such a subgroup is also a right coset such a subgroup and in the homework that you turned in tonight you wrote down examples of subgroups in other groups that didn't have that property so the kernels of homomorphisms will in sort of a in overall sense be important because they have this property and after you think well why is that property important that's a great question to ask and we will see the answer to that by the middle of next Monday okay properties kernels of homomorphisms have some really nice properties here's the first one proposition and because the proposition it sounds like proposition the kernel of the homomorphism determines whether or not or not the function is one-to-one here's the specific statement specifically the only time a function viewed as a group homomorphism is one-to-one is when the kernel is exactly this subgroup if and only a fee is a one-to-one function so if you've got a group homomorphism and if it's the case that the kernel of the group homomorphism is just the identity element of the group i mean it always has to contain at least the identity element of the group because the kernel is a subgroup the point is if that's the only thing in the kernel then necessarily fee as a function is one-to-one and conversely if fee is one-to-one then that's the only thing in the current and this will be a really nice property to have around it'll allow you to determine when it's the case that a function is one-to-one just by examining well the things that spit out the identity element of the target group here's the proof it's not too bad so we'll do the the harder direction first so what do we want to do we want to show that if fee of x is fee of y then x equals y that's what it means for a function to be one-to-one that's the definition of being a one-to-one function if you hand me any two inputs if they kicked out the same output then in fact they would the same input to begin with okay but let's see i'm starting with this equation fee of x equals fee of y fee of x is fee of y that's the given information i eventually want to conclude that x is y but folks this isn't just happening in arbitrary sets this is happening inside some sort of group so the hypothesis and i should probably have written this out again the standing hypothesis in all that we do tonight is this that fee from g to g prime is a homomorphism so let me just write that down here just so that we're talking the same language is a homomorphism all right so all this is happening inside g prime this is a group element that's a group element so what i can choose to do is multiply quote-unquote both sides by the inverse of either one of these doesn't really matter i'm going to choose to multiply by the inverse of this so fee of x combined with fee of y inverse equals fee of y combined with fee of y so i've simply chosen because i can pull any element out of the group g prime i want i've chosen to combine each side of this equation by the specific element fee of y inverse now the reason i want to do that is the following first of all this is just an element combined with its inverse so it is not an issue or any sort of property of homomorphisms that allows me to simplify this expression as just the identity element of the group that all this is happening in which happens to be called e product so that's a non issue what is of interest though is let's see we prove property to that proposition that if you have the inverse a fee of something then that's the same as this so this is used property two of the proposition the inverse of fee of something is fee of the inverse of that thing oh but wait a minute this is fee of something combined with fee of something else so since fee is a homomorphism this is the same as fee of x y inverse fee of one thing combined with fee of another is fee of combined the two things within the prens here yep is e prime sorry all right anybody want to tell me what the next step should be what does this equation tell me yeah only e goes to e prime right the hypothesis is that the kernel of fee is e and the point is that this equation tells me by definition let me write in one more line this says that x y inverse is in the kernel of fee because x y inverse when you run it through the function spits out e prime the kernel of fee consists of just one element there it is so by hypothesis hypothesis this is sort of nice x y inverse is e again the hypothesis is that the only thing in the kernel is e itself well i've just identified something that's in the kernel so the conclusion is it has to be the one thing that i know is in there and now simply multiply both sides on the right by y so x y inverse y is e y these then become e so we get x is y and we're done that's really nice so if the kernel is just the identity then fee is a one to one function and the converse i won't spend too much time proving because the converse folks is really just a statement about functions it's not a statement about group homomorphisms at all if the functions one to one is it the case that the kernel of fee is just the identity yeah because you can't have two different things going to the same element fee of e is e prime could you have fee of something else being e prime no because the functions one to one so follows just from properties from the definition of being a one to one function so this is a nice property of kernels kernels of homomorphisms determine or dictate whether or not the function that you've got your hands on is actually one to one function or not so for example the function from z to z that you get by multiplying by four is a one to one function all right second important property of kernels turns out to be the property that i just mentioned uh here is here is arguably the most important property most important property so here's the idea i'm gonna call this a theorem now i just call it a proposition proposition says this so the hypothesis standing all day is that fee is a group homomorphism a homomorphism so that means g and g prime are groups fee is a function from g to g prime that satisfies the requisite property g might be g prime doesn't matter what i want you to do is look at the following subgroup let capital k denote kernel of fee so i'm going to put in parentheses so k is a subgroup of g that's what we just proved in this proposition by the subgroup theorem i want you to look at this subgroup then the punchline is this then every left coset of k is also a right coset in other words if you've written down a subgroup and that subgroup happens to be the kernel of a homomorphism from the group g to any other group i don't care what g prime is at all the role of g prime is completely irrelevant here but if you've written down a subgroup that can be viewed as the kernel of some group homomorphism if you go ahead and you write down all of the left cosets of that subgroup and then you go back and you write down all the right cosets of that subgroup it turns out the two sets that you'll be looking at are identical remember in class we gave a couple of examples of subgroups where when you look at the left cosets and then you go ahead and look at the right cosets that you're not looking at the same subsets so this property definitely does not hold for any arbitrary subgroup of a group but this property does hold for subgroups that happen to come up or arise as kernels of homomorphisms all right any suggestions on how i might prove this proof i'll give you a hint let's see you did a problem for homework that said what that if you have a subgroup and the subgroup has the property that let's see what what were the letters that were used in the problem with the property that g inverse h g is back in the subgroup then you can conclude that the subgroup has the property that every left coset to right coset so if this is the conclusion i want to draw then i can do so simply by showing you that this subgroup has the property indicated in the homework problem what was that problem 28 or something like that okay so here's the proof we use problem 28 in section what section 10 that's exactly why problem 28 section 10 was assigned let's see we show that the subgroup k has the property that for every element in the group let's call it little g in capital g and every let's call it little k in capital k that if we compute g inverse k g that we get something back in k because we proved in problem 28 of section 10 that if the subgroup that you've got your hands on has this property then you can conclude that the subgroup that you have your hands on has the property that every left coset to right coset it turns out this property is going to be relatively easy to prove knowing that k is the kernel of the homomorphism here's how you do it well let's see so what have i done i pick any g in the group i pick any k in the subgroup well remember what the subgroup is it's the kernel of phi so we're assuming that k is in k so what does that mean ie that phi of k is the identity element in g prime that's what it means to be in the kernel it means if you run the element through the function that the identity element of the target group comes out and what do we have to do we have to show well the goal is to show that this element is in k but what does it mean to be in k it means that when you run it through phi that the identity is supposed to come out we have to show that if we run g inverse k g through the function that the identity comes out but that's what it means to be in k because k by definition is the kernel of phi can we do it not too bad what is phi of g inverse k g well phi is a homomorphism so anytime you have phi of the product of two things you can split it up so this is i could do it all in one step but i'll go ahead and do it in two steps for you so i'm asking you to combine that with that so think there's a little dot in there yeah there's also a dot there but for now just think of it as that star that so it's phi of the first thing star phi of the second thing because phi is a homomorphism and now i'm going to use the fact that phi is a homomorphism again that's phi of g inverse but what's this it's phi of something combined with something else so it's phi of the second one phi of the third one and again the notation is a little bit sloppy because technically this is all happening in g prime so there should be star primes here in star primes but we know where this is happening because it's the image of phi so necessarily this thing lives in g prime now for what it's worth folks in the future you don't need to do two intermediate steps if you have the product of three or more things inside the window here and phi is a homomorphism it breaks up over as many pieces as you want so i mean this is sort of approved by induction and that's the first step of it but i'll let you use that in the future all right so what well here's the so what we know something about this t prime so i just substitute because that's what i know about v of k well that's totally convenient because now i'm multiplying by e prime so that's not going to change anything so that's phi of g inverse phi of g which is what well that's i mean there's a couple ways i could do this let's just go ahead and use property two of the proposition that's phi of g inverse phi of g because that's what the proposition says if you run the inverse of a function or of an element through the function and you get the what you would have gotten if you ran the original element through and then take its inverse yeah so then i got something inverse times the thing which is e prime check so here's what we've just done we've started with a subgroup that happens to be the kernel of a homomorphism and we've just shown that if you take any element in the group and any element in that subgroup and you compute or you combine or you perform the expression g inverse k g that the results not only always back in the group but in fact is back in the subgroup and once we have that property established for this subgroup then the result of problem 28 in section 10 says therefore every left coset is a right coset so conclude by the problem 28 section 10 number 28 that every left coset okay also a right coset so that's sort of interesting okay i got 10 minutes so i'm going to free associate for 10 minutes this is good at this stage you're thinking oh so what why do i care at all about this property that every left coset is a right coset but great question and here is a glimpse towards the future here's what we're going to do with first of all that particular property one thing we'll be able to say is that there's some symmetry here if you start with the kernel of a homomorphism then every left coset is a right coset the converse will be true as well if you happen to have a subgroup and the subgroup happens to have the property that every left coset is a right coset then necessarily that subgroup is the kernel of some homomorphism from the group to some other group that'll wind up being sort of the the cherry on top of the the sunday that we'll get to about two weeks from now so it turns out that that sort of connection between left cosets being right cosets and being kernels of homomorphisms turns out to be a very tight relationship but what will be of significantly more interest to us surrounding this property of every left coset being right coset is the following okay what we're going to do starting monday is we're going to start building new groups like well we've done that before we built groups out of matrices and out of functions and out of permutations and out of thises and that's the way we're going to build these new types of groups is as follows we're going to take a group something that we know to be a group we're going to write down whatever subgroup you want tell me which one it is that's fine call it h and the new group that we're going to form is going to have as its elements subsets of the original group seems a little bit weird who said again the elements of this new group that we're going to form the elements each are going to be subsets so what we're eventually going to teach you how to do is take a subset of the group and combine it with another subset of the group to get yet another subset of the group so we're going to have a binary operation on the collection of certain subsets of the group that's going to turn this collection of subsets into a group in its own right now i'm not just going to ask you to sort of willy-nilly pick whatever subsets you want to do this the way you're going to choose or write down those subsets is you're going to take the subgroup that i've handed you and you're going to list out all of its left cosets warm place in my heart for things on the left and i say that sort of decisively because if you say well what if we did things on the right well the point's going to be this folks the situations that are going to be of most interest are subgroups with the property that every left coset is a right coset so if instead you had said well i wanted to choose all of the right cosets you and i would be at least talking the same language here now what i'm going to ask you to do is come up with a way of taking two cosets and producing from them another coset because the overall goal is to somehow construct a binary operation where the things in the set happen to be sets in their own right happen to be cosets of this subgroup now surface well actually there's there's maybe more than one way of thinking how you might do that if i hand you a coset maybe it looks like a h and i hand you another left coset maybe it looks like b h and i ask you to combine those two cosets to produce another coset how might you do that well here's one possible way i mean this is just a collection of things in the group and this is just a collection of things in the group they're both just subsets so one way to produce another subset of the group is just you know multiply everything in sight multiply everything in here times everything in here so do that times that and that times that and that times that and that times that you know just do all the possible combinations well there's a reasonably good way of combining all the things in here with all the things in here the problem with that though is that in general what you get when you do that operation or that process might not be another coset it's too bad because if you're trying to describe a binary operation on the collection of cosets you want it to be the case that when you take two cosets and combine them you get another coset if you do it in the manner that you might think is the most natural one just sort of combine everything in here with everything in here you might not get the coset too bad all right well let me give you another possible way of trying to describe a binary operation on the collection of cosets if this one's called a h and this one's called b h here's a way of combining those to get another coset right down the coset a b h just combine the two elements that are producing the cosets and say all right there's the coset i want you to spit out that's a good idea too it's just the issue is and this is where we're going to wind up going back to the stuff we talked about in days one and two in here the issue with that binary operation might not be well defined might not be well defined because the issue with cosets is there's lots of different ways of describing the same coset there's lots of different names for the same coset so here's the big picture of where we're headed and we'll start doing all the details on monday big picture of why this is important of why the property every left coset is a right coset is important important take any group g and any subgroup h g so i don't care what group you start with and i don't care what subgroup you start with then this set the collection of left cosets is a set is a set just list them all out in fact i can tell you how many there are if you start with a finite group and you tell me how big the subgroup is the number of left cosets is this number that we call the index if you hand me for instance a group with eight elements and a subgroup of two elements then i'll tell you how many left cosets there are there's eight divided by two or four of them write them down now here's what i'm going to ask you to do we attempt or we seek or we hope to define a binary operation on this collection on the collection of cosets and that already seems weird i'm going to teach you how to take two left cosets and combine them to produce another left coset and once it's all said and done once i teach you how to do that the collection of left cosets is going to wind up being your group it's just we have to get this process off the ground and it turns out that turns out there are at least two at least two but i'm going to focus on two possible ways of doing this natural ways to do this ways to do this but in either of these ways but to make these work make these work we need this property the every left coset is a right coset property every left coset is a right coset in fact let me modify this slightly we'll get out of here there are two possible natural ways to try to do this there are two natural ways of trying to define a binary operation on the set of left cosets and i just described them in words and i will write out the details of those two possibilities on monday the first is just multiply everything in sight the second is define something like a h combined with b h to be a b h it turns out that if you don't have this property there are issues with both of those things but if you do have this property on the subgroup then both of those ways will give you a perfectly good binary operation on the collection of left cosets and not only that they actually give the same binary operation and not only that once you've got the binary operation defined in either of those ways then in fact you wind up getting a group all right