So glad to see you back posting videos again! I think you always explain things in a very clear and direct way (without being too impersonal or lecture-like), and I especially appreciate when you cover more complex topics like group theory/abstract algebra, tensors, higher level differential equations, etc. because it's so difficult to find resources about otherwise. I look forward to the next one! Have a great day! :)
Thank you for a very beautiful and clear explanation! I already know more formal way to prove the first isomorphism theorem, which you mentioned in the beginning. However, I always felt that I lack a more natural approach and some intuition behind this theorem. After watching this video I can finally understand it :) Thank you!
I like this perspective on the first isomorphism theorem, so thanks for sharing! I wouldn't describe the typical proof as using "tricks" though. If one develops a good intuition for what quotient groups mean and also a good intuition for how the kernel relates to injectivity, the typical proof of "making ϕ injective" is pretty straightforward. G/N should be thought of as carefully collapsing G in a way which makes every element of N collapse to the identity. Making N collapse to the identity then forces other elements to collapse to each other. For example, if g = hn for some element n in N, then since n is becoming the identity in this "collapse," after collapsing, we must have g = he = h. So g must collapse to h as well. And similarly, if g = nh, then after the collapse, we have g = eh = h, so g = h. This also explains exactly why we need left- and right cosets to be equal to each other, i.e., why we need a normal subgroup. If one shows, as you showed, that two elements are mapped to the same output if and only if they differ by an element in the kernel, then one sees that, going modulo the kernel, two elements map to the same output if and only if they collapse to same element in the quotient G/kerϕ. So yes, going modulo the kernel "collapses" the non-injectivity of ϕ. Now, it takes some time to develop good intuition for quotients. But what you call "tricks," I really see as "the intuition one should have for quotients." And I think it's not a good idea to disparage good intuition as "tricks".
I share your perspective on the fact that quotienting by the kernel is a natural way to induce injectivity on a homomorphism. In fact, I have a video explaining the intuition behind that approach here: czcams.com/video/ssVIJO5uNeg/video.html The point I wanted to make is that that approach involves intentionally constructing a map that satisfies the desired injectivity and surjectivity conditions. That construction is natural and intuitive, but it is architected by us. We induce bijectivity using our knowledge of how level sets relate to the kernel. In contrast to that approach, I find that these ideas come as a natural consequence of the preimage formulation rather than requiring our insight to impose them. Simply by considering the preimage map, the domain is automatically restricted to the image of ϕ, and the codomain is automatically quotiented by the kernel. The fact that these phenomena arise of their own accord from considering the preimage makes this approach seem more natural to me, even though it is ultimately the same dynamic in either formulation. At the end of the day, which method is preferred is a matter of taste. Point taken on the use of the word "trick"!
@@MuPrimeMath I see what you're saying! And thank you for saying it! I was a bit harsh here in my comment, and I do agree that it's nice seeing the argument you gave in this video. And I think the method used in this video could be used as a potential motivation for quotient groups, as well! Something along the lines of: if we wanted to turn a non-injective homomorphism into an injective one (that essentially does the same thing to the codomain), then we would need to collect elements together according to where they are mapped and "collapse" all elements mapping to the same point. And then we can check that these sets of elements which we have collected together based on having the same image really do have their own group structure. In the end, I don't think I'll use it in the future to motivate quotient groups (I am quite happy with my visual motivation of looking at the symmetries of a square and seeing how they induce permutations of the diagonals and permutations of the perpendicular bisectors of the square, so we can get a group out of collecting elements which do the same thing to the diagonals and which do the same thing to the perpendicular bisectors). But it might be a nice motivation for people seeing abstract algebra for a second time or third time.
Proving that the kernel of phi is a normal subgroup is very straightforward: If k is in the kernel of phi, then phi(k)=e, and phi(gkg^-1) = phi(g)phi(k)phi(g^-1) = phi(g)phi(g)^-1 = e, so gkg^-1 is in the kernel of phi for all g in G.
So glad to see you back posting videos again! I think you always explain things in a very clear and direct way (without being too impersonal or lecture-like), and I especially appreciate when you cover more complex topics like group theory/abstract algebra, tensors, higher level differential equations, etc. because it's so difficult to find resources about otherwise. I look forward to the next one! Have a great day! :)
Very nice explantion and indeed a very natural approach! Thank you very much.
Thank you for a very beautiful and clear explanation! I already know more formal way to prove the first isomorphism theorem, which you mentioned in the beginning. However, I always felt that I lack a more natural approach and some intuition behind this theorem. After watching this video I can finally understand it :) Thank you!
The proof in this video isn't the standard one given in textbooks, but it is still rigorous (i.e. "formal")!
@@MuPrimeMath Exactly! I could have write "more standard proof"
wonderful lecture with grear explanation
Smart and handsome, thanks for the quality content!.
ikr? he's pretty good looking ;p weirdly enough a lot of the math ppl on YT are (3b1b, michael penn, etc)
@ⒶlexInWonderland that's why .... I am mathematician but not on YT hehe...
I like this perspective on the first isomorphism theorem, so thanks for sharing!
I wouldn't describe the typical proof as using "tricks" though. If one develops a good intuition for what quotient groups mean and also a good intuition for how the kernel relates to injectivity, the typical proof of "making ϕ injective" is pretty straightforward.
G/N should be thought of as carefully collapsing G in a way which makes every element of N collapse to the identity. Making N collapse to the identity then forces other elements to collapse to each other. For example, if g = hn for some element n in N, then since n is becoming the identity in this "collapse," after collapsing, we must have g = he = h. So g must collapse to h as well. And similarly, if g = nh, then after the collapse, we have g = eh = h, so g = h. This also explains exactly why we need left- and right cosets to be equal to each other, i.e., why we need a normal subgroup.
If one shows, as you showed, that two elements are mapped to the same output if and only if they differ by an element in the kernel, then one sees that, going modulo the kernel, two elements map to the same output if and only if they collapse to same element in the quotient G/kerϕ. So yes, going modulo the kernel "collapses" the non-injectivity of ϕ.
Now, it takes some time to develop good intuition for quotients. But what you call "tricks," I really see as "the intuition one should have for quotients." And I think it's not a good idea to disparage good intuition as "tricks".
I share your perspective on the fact that quotienting by the kernel is a natural way to induce injectivity on a homomorphism. In fact, I have a video explaining the intuition behind that approach here: czcams.com/video/ssVIJO5uNeg/video.html
The point I wanted to make is that that approach involves intentionally constructing a map that satisfies the desired injectivity and surjectivity conditions. That construction is natural and intuitive, but it is architected by us. We induce bijectivity using our knowledge of how level sets relate to the kernel.
In contrast to that approach, I find that these ideas come as a natural consequence of the preimage formulation rather than requiring our insight to impose them. Simply by considering the preimage map, the domain is automatically restricted to the image of ϕ, and the codomain is automatically quotiented by the kernel. The fact that these phenomena arise of their own accord from considering the preimage makes this approach seem more natural to me, even though it is ultimately the same dynamic in either formulation.
At the end of the day, which method is preferred is a matter of taste. Point taken on the use of the word "trick"!
@@MuPrimeMath I see what you're saying! And thank you for saying it! I was a bit harsh here in my comment, and I do agree that it's nice seeing the argument you gave in this video.
And I think the method used in this video could be used as a potential motivation for quotient groups, as well! Something along the lines of: if we wanted to turn a non-injective homomorphism into an injective one (that essentially does the same thing to the codomain), then we would need to collect elements together according to where they are mapped and "collapse" all elements mapping to the same point. And then we can check that these sets of elements which we have collected together based on having the same image really do have their own group structure.
In the end, I don't think I'll use it in the future to motivate quotient groups (I am quite happy with my visual motivation of looking at the symmetries of a square and seeing how they induce permutations of the diagonals and permutations of the perpendicular bisectors of the square, so we can get a group out of collecting elements which do the same thing to the diagonals and which do the same thing to the perpendicular bisectors). But it might be a nice motivation for people seeing abstract algebra for a second time or third time.
Thank you for making this video! So helpful.
wonderfull
6.6k views and only 17 comments? Let me fix that
@5:50 how is the kernel of phi normal subgroup. Doesn't this fact use this fact to begin with?
Proving that the kernel of phi is a normal subgroup is very straightforward:
If k is in the kernel of phi, then phi(k)=e, and
phi(gkg^-1) = phi(g)phi(k)phi(g^-1) = phi(g)phi(g)^-1 = e,
so gkg^-1 is in the kernel of phi for all g in G.
@@MuPrimeMath ah thanks.