 Today we get to something very exciting. I want you to consider an isomorphism f from g to h2 groups and we have that g has this binary operation and h has this binary operation so it doesn't necessarily have to be the same but we do have this isomorphism. We've looked at the kernel of f and the kernel of f. Remember that all the elements g elements of g such as the f of g maps to the identity element in h. Very simple. We shown that it's a normal subgroup so now we know that the left cosets and the right cosets are exactly the same. What I want to show though that if we do now have this division of g into these equal sets the kernel and then and then all the cosets I want to show something very special as far as the quotient groups. So remember we have going to have g and we're going to have the kernel of f. Those are the quotient groups. Now I want you to imagine there's a couple of them say there's a and there's b and let's take any arbitrary element and a any arbitrary element and b. You can imagine if there's c d remember how these are constructed from the cosets and I want to show that for all elements all elements a element and a they are going to map to exactly the same element in h element in h. So no matter which of these I take they're all going to map to exactly the same so that in essence this thing maps to you know the same element there. So I'm trying you see where I'm going to go go with this we know there's surjection the surjective mapping but we want to show that that's an injective mapping as well and what do we have then a bijection and then what do we have an isomorphism. So you know can we do that remember that we have this we have this for all elements in the kernel of f that's what we have for this a. Now if I have the f of a composed with n binary operation with m with n remember that's the f of a and then there's binary operation because this is g with n and what is that well that's just that's just f of a composed with or by that binary operation with the e with identity element in h and that is just the f of a. So we've shown that no matter what arbitrary element and a I take what arbitrary element and a I take doesn't matter which one it is I'm always going to map there. Now let's suggest that I take a element of this a and b element of this b and I want to show that they will not map to the same thing but to the contrary let's imagine that they do and that f of a equals the f of b that would be very interesting and I want to show that that is not so so to the contrary let's assume it is so and I'm just going to do a funny little thing I'm going to compose on the right hand side by b inverse b inverse so that's going to be the f of b and the binary operation for h f of b inverse so by the definition of the homomorphism this would be the f of a composed with b inverse that's in g so it's that binary operation and this is the f of b and b inverse and that's nothing other than the identity element in h the identity element in h now this is weird because remember this is the definition of the kernel of f so whatever this a binary operation b inverse is that must be equal to a binary operation that must be equal to n an element of n is an element of that I write the system so let's assume that or just write any arbitrary element n in the kernel of f remember that's all the values in g that map to that map to the identity element in h so we have just shown that you know by definition this was our definition of the kernel of f they all map to the identity element and if I compose both sides b inverse with b that equals that those two is just the identity element so that that I have n is a equals n binary operation b and remember that's nothing other than this so a must be an element of b of b bar one of the cosets so one of these quotient elements so what do we have now so for these two to be the same thing a must be in b so what have we shown really is that we've shown a one-to-one mapping so all the elements are going to go to of this all these elements are going to go to exactly the same and b will go to b and c will go to b and they will be separate in other words if it goes to the same element they must be in the same set here so we have a bijection we know it's a surjective we now know it's an injective so we have a bijection so what we've shown here really is that the quotient group formed by g and this kernel of g is isomorphic to h that's you know if we have an isomorphism such as that we form a quotient group by the kernel of f we say that it is isomorphic to h and that is what is called the first theorem of isomorphisms so what we've shown here really is a very deep something very deep and and it goes by the name and we put an end to it it's the first this first theorem of isomorphism