Mate thanks for the help and aspiration , i passed the algebraic structures subject and i am waiting the results of number theory one, you where a great reason for it and of understanding better the fascinating world of algebra. Thanks again keep going !!!
6:14 im trying to follow along but this abstract stuff is hard for me... i dont understand the statement here: "Thus, when we construct quotient groups of a group G, not only are we constructing homomorphic images of G, but we are constructing all the homomorphic images of G (up to isomorphism)" I am confused about this because quotient groups of G are sets of cosets of H, whereas the fundamental theorem isnt a statement about the cosets of H, but rather the cosets of the kernel of the thing that gets you from G to H.. so i dont see how those theorems are related or how we know we are getting all the homomorphic images from that. i think i understood the theorem up to that point but i didnt understand that implication.
Thanks for the well thought out question! I'm going to use some different letters here to avoid muddying the point due to a clash of notation. Say f is a homomorphism of A onto C. Like you said, quotient groups of A are sets of cosets of B in A where B is a normal subgroup. In the FHT, it is the kernel k of f which plays the role of B - the normal subgroup. [Kernels are normal subgroups. we prove that in my lesson introducing kernels I think, but also in this shorter yet to be released video on specifically that result: th-cam.com/video/NaaATC_th58/w-d-xo.html] And so in the context of the FHT, the kernel K of the thing that takes us from G to the homomorphic image H is itself a normal subgroup, and thus we can construct a quotient group G/K with it. The theorem, then, is that this G/K is in fact isomorphic to the homomorphic image H. So where there is a homomorphic image, there is also a group isomorphic to that image made up entirely of the cosets of the homomorphism's kernel, cosets which are in the original domain group G. Is that helpful?
@@WrathofMath Thanks! i think that helped. But im still having trouble understanding how we know that that accounts for **all** homomorphic images of G (up to isomorphism)? we know G/K doesnt add anything new, but what about G/X or G/Y, or any homomorphic image that isnt a quotient group? Most likely im just misunderstanding what you meant in the video or or caught up on the wording of something, but im not sure what.
We know it account for all homomorphic images of G because in the proof we begin with an arbitrary homomorphic image! Recall a homomorphic image of G is a group H such that there exists a homomorphism from G onto H. In the proof we begin with an arbitrary onto homomorphism, which is the same as beginning with an arbitrary homomorphic image. To perhaps belabor this point... In the proof, we say "Let f be a homomorphism of G onto H", which emphasizes the homomorphism more than the image, but we just as well could have said "Let H be a homomorphic image of G. Then there exists a homomorphism f from G onto H." And proceeded from there. It doesn't matter what the homomorphic image H is, most likely it will not be a quotient group since that's a very particular type of group, it will just be some other homomorphic image. But regardless, we could ask if we can construct a quotient group of G that is isomorphic to H, and the theorem shows us that YES, we can ALWAYS construct such a quotient group. How? By taking the quotient group of G by the kernel of the homomorphism in question, which maps G onto H. Every homomorphic image MUST have a homomorphism mapping G onto H, and every homomorphism has a kernel, and every quotient group by that kernel is isomorphic to the corresponding homomorphic image. That is the FHT. Side tangent: TH-cam allows creators to post guidelines for their comments sections now. No use for this has occurred to me until your perfectly formatted question. I get lots of comments, and while I respond to lots of them, the serious math questions are difficult to respond to as they take the longest and often do not provide context. Viewers often, understandably, ask their question as if I just watched the video too and will know what they're talking about - but at this point some of these videos, and indeed even some of the subject matter, I haven't looked at in 5+ years. Your format - timestamp with quote, made responding to your comment convenient and as easy as it could possibly be for me! I will definitely include this in comment guidelines so people can maximize their chance of a useful response. Thanks for sparking the idea!
@@WrathofMath is it correct to say that not every quotient group we construct is necessarily a homomorphic image, since it has to be the kernel of a homomorphism for that to be the case? or can we say that any image A is necessarily the kernel of some homomorphism, and thus G/A is really G/K in disguise for some kernel K?
To be a homomorphic image a group need not be a kernel of a homomorphism, it only need be the codomain of a surjective homomorphism. Any quotient group G/A MUST be isomorphic to a homomorphic image of G, and the kernel of that forced homomorphism is A.
![“อ.เบียร์ คนตื่นธรรม” มาตามคำเรียกร้อง แหกศาสตร์ฮวงจุ้ย เตือนสติอย่างมงาย | แฉ 7 พ.ย. 67 [1/3]](http://i.ytimg.com/vi/5fc4S3xBNfc/mqdefault.jpg)
So helpful! Had to wrap my head around this theorem and quotient groups... your teaching style is very soothing and nice to follow 🙂
So glad it helped - thanks for watching!
I needed this a few years ago 😢. Good work! I love supporting math videos!
Thank you! Better late than never!
Thank you so much for the whole series! You are a blessing from heaven haha
Mate thanks for the help and aspiration , i passed the algebraic structures subject and i am waiting the results of number theory one, you where a great reason for it and of understanding better the fascinating world of algebra. Thanks again keep going !!!
Thanks so much, glad to help!
Thank you so much 🙏 great video
My pleasure, thanks for watching!
6:14 im trying to follow along but this abstract stuff is hard for me... i dont understand the statement here: "Thus, when we construct quotient groups of a group G, not only are we constructing homomorphic images of G, but we are constructing all the homomorphic images of G (up to isomorphism)"
I am confused about this because quotient groups of G are sets of cosets of H, whereas the fundamental theorem isnt a statement about the cosets of H, but rather the cosets of the kernel of the thing that gets you from G to H.. so i dont see how those theorems are related or how we know we are getting all the homomorphic images from that. i think i understood the theorem up to that point but i didnt understand that implication.
Thanks for the well thought out question!
I'm going to use some different letters here to avoid muddying the point due to a clash of notation. Say f is a homomorphism of A onto C. Like you said, quotient groups of A are sets of cosets of B in A where B is a normal subgroup. In the FHT, it is the kernel k of f which plays the role of B - the normal subgroup.
[Kernels are normal subgroups. we prove that in my lesson introducing kernels I think, but also in this shorter yet to be released video on specifically that result: th-cam.com/video/NaaATC_th58/w-d-xo.html]
And so in the context of the FHT, the kernel K of the thing that takes us from G to the homomorphic image H is itself a normal subgroup, and thus we can construct a quotient group G/K with it. The theorem, then, is that this G/K is in fact isomorphic to the homomorphic image H. So where there is a homomorphic image, there is also a group isomorphic to that image made up entirely of the cosets of the homomorphism's kernel, cosets which are in the original domain group G. Is that helpful?
@@WrathofMath Thanks! i think that helped. But im still having trouble understanding how we know that that accounts for **all** homomorphic images of G (up to isomorphism)? we know G/K doesnt add anything new, but what about G/X or G/Y, or any homomorphic image that isnt a quotient group? Most likely im just misunderstanding what you meant in the video or or caught up on the wording of something, but im not sure what.
We know it account for all homomorphic images of G because in the proof we begin with an arbitrary homomorphic image! Recall a homomorphic image of G is a group H such that there exists a homomorphism from G onto H. In the proof we begin with an arbitrary onto homomorphism, which is the same as beginning with an arbitrary homomorphic image. To perhaps belabor this point...
In the proof, we say "Let f be a homomorphism of G onto H", which emphasizes the homomorphism more than the image, but we just as well could have said "Let H be a homomorphic image of G. Then there exists a homomorphism f from G onto H." And proceeded from there.
It doesn't matter what the homomorphic image H is, most likely it will not be a quotient group since that's a very particular type of group, it will just be some other homomorphic image. But regardless, we could ask if we can construct a quotient group of G that is isomorphic to H, and the theorem shows us that YES, we can ALWAYS construct such a quotient group. How? By taking the quotient group of G by the kernel of the homomorphism in question, which maps G onto H. Every homomorphic image MUST have a homomorphism mapping G onto H, and every homomorphism has a kernel, and every quotient group by that kernel is isomorphic to the corresponding homomorphic image. That is the FHT.
Side tangent: TH-cam allows creators to post guidelines for their comments sections now. No use for this has occurred to me until your perfectly formatted question. I get lots of comments, and while I respond to lots of them, the serious math questions are difficult to respond to as they take the longest and often do not provide context. Viewers often, understandably, ask their question as if I just watched the video too and will know what they're talking about - but at this point some of these videos, and indeed even some of the subject matter, I haven't looked at in 5+ years. Your format - timestamp with quote, made responding to your comment convenient and as easy as it could possibly be for me! I will definitely include this in comment guidelines so people can maximize their chance of a useful response. Thanks for sparking the idea!
@@WrathofMath is it correct to say that not every quotient group we construct is necessarily a homomorphic image, since it has to be the kernel of a homomorphism for that to be the case? or can we say that any image A is necessarily the kernel of some homomorphism, and thus G/A is really G/K in disguise for some kernel K?
To be a homomorphic image a group need not be a kernel of a homomorphism, it only need be the codomain of a surjective homomorphism. Any quotient group G/A MUST be isomorphic to a homomorphic image of G, and the kernel of that forced homomorphism is A.
Waiting to see this theorem "put into action". I guess you need some more Patreon subscribers, eh?
It certainly helps!
th-cam.com/video/Go9vBI-syis/w-d-xo.html
When you prove that G/H is a homomorphic image of G, doesn't H have to be a normal subgroup?
I believe in his video about quotient group he defined G/H when H is a normal subgroup of G so writing G/H is implicitly saying that
Also, the kernel of a homomorphism from G into H is necessarily a normal subgroup of G - this is a structural theorem of homomorphisms