 So let's do an example of a quotient group and I'm not going to be any fancier than most of the textbooks are. Most of the textbooks you'll see your group is going to be the integers under addition and my normal subgroup of g is going to be 3 is going to equal 3 times z, the 3 times the integers. So what does what does the set look like that makes up 3 times the integers? Well that is remember that it's just going to be say 3 times z such that z is an element of z something like that. Remember if you were to look at that that would be say minus 9 minus 6 minus 3 0 3 6 9 etc. So that's how you can see it. I'm calling that I'm calling that my normal subgroup it is stable you can check it out it is stable under conjugation. So what I'm now looking for is this quotient group of this quotient group integers over 3 times over 3 times the integers. So if this is my you know this is my n in other words remember so that would just be my n and this is my g so the quotient group g over n. So now let's make the cosets because we've got to you know it's all the integers negative infinity positive infinity I've got to make you know I've got my normal group now what about the cosets. Now since this is a normal group you can check that out we have to have we know that the left and right cosets are the same so let's stick to the left cosets. So my first coset is I'm going to let I'm going to let a equals 1 remember where a is an element of my bigger group g so what do I have here now I have I have 1z an in other words my first left coset an 1z and that is just going to be remember my binary operation here is addition so I'm just going to have this set 1 plus x such that x is an element of of n so what is that going to look like so my first coset 1 integers that is just going to be I'm going to add 1 to each of these elements and n and this n remember is three times I'm using those two interchangeably so that's going to be minus 8 and that is going to be minus 5 and it's going to be minus 2 it's going to be 1 4 7 etc so that's one let's choose a equals 2 so I'm now looking at 2n 2h or remember ah an the next coset well that is just going to be 2 plus x such that x is an element this of this n my normal subgroup and what is that going to look like well that's minus 7 and minus 4 and minus 1 and where are we 3 where am I let's not let's not make too many mistakes here and in being so hasty so I'm just going to take each of these and I'm going to add 2 to that so that's all the way negative 7 and I add 2 to that that's minus 4 and I add 2 to that's minus 1 and I add 2 to that and that is 2 and I add 2 to that that's 5 and I add 2 to that that's 8 etc and what we'll now find is that we have partitioned into equivalence classes basically the whole of my original the set that makes up my original group because if I chose a equals 3 I'm just land up straight back within so what I have here now what I have here now is the following that I have this question group and that is going to consist of my elements 0 1 and 2 and we've just proven that that is indeed a group and we can have this binary operation of addition on these two and I can have 0 1 and 2 and 0 1 and 2 and remember this is going to be my identity element because if I add anyone to that means I take a representative one here say 0 why not and I take one in there which is 1 0 plus 1 is 1 where is it it is in this remember this is this this is this okay so I just end up there here I'm going to end up here and then I'm going to end up there 1 and 1 is going to be 2 because if I added you know 1 plus 1 is 2 2 is in this is in this subclass 1 and 2 is 3 but 3 is back in in which was that 2 and 1 is 3 which ends up in that and 2 and 2 is 4 and if you see 4 ends us up back in 1 so just as we have you know we showed Cayley's theorem there these are all unique these are all unique no problem I indeed have a group and I can find the identity element I can find the inverses and I can find associativity and of course this closure so that is a beautiful example of just cementing this understanding of the very simple in the end concept of a quotient group one more thing one more thing that reminds me so if I what do what do we have here we have a quotient group with one two we have a group and so that's going to be my group now and I see that it contains a prime number at this contains a prime number of elements so this is really going to be isomorphic and this is what we're leading to we're going to look at homomorphisms again this leads us to the fact that this is isomorphic to the cyclic group with three elements and you can see because I can take any one of these so what is that going to that way is that going to leave me well this element one if we just raise that to the zero of power that is just going to give me this to itself and then and and you know one of them plus another one then is two of them and one plus two of them is three which lands me up at zero so there we are that is indeed I can start with any one of these I'm going to get the cyclic group so this is indeed isomorphic to the cyclic group in three elements