 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? Sigma says, I don't care what I'm handed, I spit out e. 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. Here, let's call them little f and little f1 and little f2. 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. The typical game plan is write down a bunch of examples of that structure. We've done a bunch of examples of homomorphisms. 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 what a group was and then we wrote down a bunch of properties that groups necessarily have to have. 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. So we've written down a bunch of examples. 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