I've been watching your Group Theory playlist from the beginning and so far I'm delighted. Your explanations are crystal clear. So far I grasped everything, even stuff I didn't master in my college algebra course. Keep up the good work!
Excellent presentation! I have one question. You said that all the members of a coset map onto one member of the codomain. For example, two elements from a bar map to some element. Then, in another coset , you say, one cannot have three elements map to some one element in the codomain. Why is that?
Martin, This is a consequence of two important ideas. The first idea is that every coset of a normal subgroup has the same number of elements. We can see this if we remember how cosets are formed: To form a coset, we pick some element that is NOT in our normal subgroup, and we compose (and we'll assume it's a left-composition) that element with EVERY element in our normal subgroup. Therefore, if "a" is our chosen element, the a-bar coset is {ah1, ah2, ah3, ah4..., ahn} where h1, h2, h3, ... hn, are all the members of our normal subgroup H (and there are n of them). Therefore, the coset a-bar will have n elements, where n is the same number of elements as the normal subgroup. We will form b-bar and c-bar, and however many other cosets the group CAN be partitioned into in exactly the same way, which means that the normal subgroup and the a-bar, b-bar, c-bar, etc., cosets will all have the same number of elements: n of them. The second idea is from the video previous to this one, which is that ALL of the elements in the same coset in G will map to the same ELEMENT in G', where G' is the codomain formed by the homomorphism, Φ. Since every element in a-bar, for example maps to the same element in G'. And that element is Φ(a). Every element in b-bar will map to Φ(b), and so on. So now, when we put those two ideas together we get the answer to your question: Every coset has the same number of elements, and every element in the same coset will map to the same element in the co-domain. Therefore every element in the codomain will have the same number of elements from the domain mapped onto it. And in fact, the number of elements from the domain that map to each element in the codomain will be equal to the number of elements in the normal subgroup that we are using to partition the group. In this video, since we are looking at a quotient group formed from the Kernel of the homomorphism, this will also be the number of elements in the Kernel. If there were, for example, 3 elements in the Kernel, then there would also be 3 elements in each coset, and therefore each element in G' would be mapped to from 3 elements in G.
Your small drawings are just the perfect explanations - helped me through my abstract algebra course, thanks!
I've been watching your Group Theory playlist from the beginning and so far I'm delighted. Your explanations are crystal clear. So far I grasped everything, even stuff I didn't master in my college algebra course. Keep up the good work!
Thank you sooo much. This is an incredibly helpful demonstration of the 1st Isomorphism Theorem.
isomorphism from the quotient group of G to the codomain of G
G' (codomain- is where the homomorphism of G go to under the map phi)
Thank you so much for your videos on abstract algebra. They have been essential in my revision for my maths degree.
For myself - this is the Proof of 1st Isomorphism theorem. It's introduced in Part 5. The previous 4 parts are material building up to this theorem.
Thank you so much for all of this. You are saving my algebra class.
Excellent presentation! I have one question. You said that all the members of a coset map onto one member of the codomain. For example, two elements from a bar map to some element. Then, in another coset , you say, one cannot have three elements map to some one element in the codomain. Why is that?
Martin, This is a consequence of two important ideas. The first idea is that every coset of a normal subgroup has the same number of elements. We can see this if we remember how cosets are formed: To form a coset, we pick some element that is NOT in our normal subgroup, and we compose (and we'll assume it's a left-composition) that element with EVERY element in our normal subgroup. Therefore, if "a" is our chosen element, the a-bar coset is {ah1, ah2, ah3, ah4..., ahn} where h1, h2, h3, ... hn, are all the members of our normal subgroup H (and there are n of them). Therefore, the coset a-bar will have n elements, where n is the same number of elements as the normal subgroup. We will form b-bar and c-bar, and however many other cosets the group CAN be partitioned into in exactly the same way, which means that the normal subgroup and the a-bar, b-bar, c-bar, etc., cosets will all have the same number of elements: n of them. The second idea is from the video previous to this one, which is that ALL of the elements in the same coset in G will map to the same ELEMENT in G', where G' is the codomain formed by the homomorphism, Φ. Since every element in a-bar, for example maps to the same element in G'. And that element is Φ(a). Every element in b-bar will map to Φ(b), and so on. So now, when we put those two ideas together we get the answer to your question: Every coset has the same number of elements, and every element in the same coset will map to the same element in the co-domain. Therefore every element in the codomain will have the same number of elements from the domain mapped onto it. And in fact, the number of elements from the domain that map to each element in the codomain will be equal to the number of elements in the normal subgroup that we are using to partition the group. In this video, since we are looking at a quotient group formed from the Kernel of the homomorphism, this will also be the number of elements in the Kernel. If there were, for example, 3 elements in the Kernel, then there would also be 3 elements in each coset, and therefore each element in G' would be mapped to from 3 elements in G.
Thanks, Jeremy, for answering this question!
excellent
I need your e-mail. Please!!!!!!