I am a math professor at the University of Florida, interested in topology, geometry, data analysis, machine learning, and sensor networks. I am also very interested in teaching, online instruction, and how to make academia more welcoming, more open, more transparent, more accessible, and less intimidating.
Thank you for your demonstration. My question is that how do we assign the arbitrary value between 0 and 1 to each vertex? Earlier it was easier to do that, since I knew that if I chose the vertex to be in the vertex cover, then its value is one, and zero otherwise. But now how does each vertex get assigned their value?
You assign the value between 0 and 1 by solving the relaxed linear programming problem, for example using the simplex algorithm, which is computationally very efficient!
My understanding is that the distributivity axiom does not follow from the other axioms, and that it is therefore necessary to include as an axiom of its own!
Thank you so much for all your hard work, you videos have been tremendously helpful! I just have a quick question, this sounds counterintuitive but I think what I just learned from this video is that any finite abelian group of let's say size 72 is isomorphic to products of cyclic groups (all 6 of them) but those 6 groups are not isomorphic to each other. intuitively it would make sense if the abelian group of size 72 would only be isomorphic to Z/8Z x Z/9Z because 8 and 9 are relatively prime. Also did we not learn if A is isomorphic to B and B is isomorphic to C then A is also isomorphic to C? in here we have an abelian group of size 72 being isomorphic to Z/8Z x Z/9Z and also the same abelian group is isomorphic to Z/2Z x Z/4Z x Z/9Z for example, but Z/8Z x Z/9Z is not isomorphic to Z/2Z x Z/4Z x Z/9Z because the orders are different so how is this possible that the abelian group would be isomorphic to them both? Won't the abelian group of size 72 must have elements of certain orders in which it would be different than some of these products? I am a little confused on this part and apologize for writing so much lol. Thanks in advance.
It is not true that any abelian group of size 72 is isomorphic to Z/8Z x Z/9Z --- that is only one of six possibilities that an abelian group of size 72 could be isomorphic to. Yes, it is true that if A is isomorphic to B and B is isomorphic to C, then A is isomorphic to C. No, it is not the case that Z/8Z x Z/9Z is isomorphic to Z/2Z x Z/4Z x Z/9Z --- these are two different abelian groups of size 72 that are not isomorphic to each other. You seem to be assuming that there is a single abelian group of size 72 --- but this is not correct ---- up to isomorphism there are six different groups of size 72 !
Thank you so much. please i came across this question could you help me with hint on how to solve it. Consider the following group actions (X, G). In each case, describe the set of orbits, and the stabilizer of a chosen point from each orbit. (1) Let G1 = Sn act on the set X1 = {1, . . . , n} by permutations. (2) Let G2∼= (R, +) act on the plane X2 = R2 by horizontal translations. (3) Let X3 be the (boundary of the) unit square in the plane R 2 Let G3 = Sym(X3) be the set of symmetries of the square.
Thanks for posting these videos online they have been helping me (and all my classmates which i told them about lol) a lot... just a quick question, so Homomorphism do not need to be bijective but do they have to be either one to one or onto? (the examples you provided, first one was 1-1and second one was onto) thanks in advance!
Happy to help! No, a group homomorphism does not need to be one-to-one, and a group homomorphism does not need to be surjective. For example, let {id} be the trivial group consisting of only the identity element, and let Z/2Z={0,1} be the unique group with two elements. Then the group homomorphism Z/2Z -> {id} (given by sending 0 to id and 1 to id) is not one-to-one, and the group homomorphism {id} -> Z/2Z (given by sending id to 0) is not onto.
@@HenryAdamsMath thank you so much for responding and thank you for explaining it, I understand now. We’re moving onto Cosets so I’ll be watching the rest of the playlist within the next few days! Looking forward to finishing your videos!
Thanks, @HenryAdamsMath, for the short, simple and marvelous lectures on the Quotient Group 👏 I got a quick idea from you videos 60 and 61 of this series 🤝 I was wondering is it possible to explain or exemplify the quotient group Z^*_m / <p>
but why do you compose them right-to-left? I do it so : (1 2) + (2 3) = (1 2 3) and not the inverse (3 2 1) as it would be for (2 3) + (1 2) = (3 2) + (2 1) or don't u 🤔 ..... your example looks particularly commutative (1 4 3 2) + (1 3) + (2 4) also sums to (1 2 3 4) when (1 4 3 2) is done first
