Mont Gum Did you know you can slow down the video playback? Click on the little "Gear" icon and try Speed 0.5. I sound a little funny but the pace is easier to keep up with :)
Hello few litle errors in Example A quotient of S4 at 4min52 Now New H =list([]) ,list([1 2],[3 4]),list([1 3],[2 4]),list([1 4],[2 3]) New G =list([1 2]),list([1 3]),list([1 4])),list([1 2 3]),list([1 3 2]) add to this new H all element of G on the left
i like your videos a lot, but the speed at which you talk is so fast that i can't keep up. but still you're my favorite so far. just wish i wasn't having a panic attack the whole time.
It has to be fast. This video covers a 50 minute lecture in 11 minutes 36 seconds. If you find this video ‘fast,’ please consider... 1. Watch this video multiple times. Its structure, incomprehensible in the first viewing, becomes comprehensible in subsequent viewings (Gestalt). 2. Do a quotient groups search: www.google.com/search?q=quotient+groups Recommend Millersville: www.millersville.edu/~bikenaga/abstract-algebra-1/normal/normal.html Also recommend its Abstract Algebra 1 series: www.millersville.edu/~bikenaga/abstract-algebra-1/abstract-algebra-1-notes.html
In the "Quotient Mapping Theorem", you refer to an epimorphism from G to G/N, but since we aren't necessarily assuming that G/N is a group (since N is not necessarily assumed to be normal), it seems like we should not be using the word "epimorphism".
Joanne Koratich Thanks! And you're right, taking (1,1) as a generator in Z2xZ3 shows that that group is in fact isomorphic to Z6, so they shouldn't be separate (except insofar as to make this point).
So at first I figured that quotient groups are very similar to groups but different in that the elements are not sets but cosets (I also originally wondered why they aren't called cosubgroups haha). Now if my interpretation is correct I see the quotient group to still be a set of elements namely for Z mod 12 { {0,3,6,9},{1,4,7,10},{2,5,8,11} }. In set theory these sets (the cosets) are valid elements of a set. So we strip the subgroup of its binary op and use it and the other cosets for the set.
" the kernel of any mapping out of G is isomorphic to a normal subgroup of G " Better: the kernel of any homomorphism out of G *is* a normal subgroup of G, not merely up to isomorphism.
Okay, I think I get it :) Thanks Professor. And on another note I think you should find I updated the notes pdf in dropbox. I am not used to dropbox, so I hope I did that correctly.
5 years late, but perhaps for other viewers: they're both common functions in AA1 (since this is AA2). The image of a function is the codomain. The kernel of a function consists of all elements in the domain that are sent to the identity element in the codomain.
@@ThefamousMrcroissant the image is everything you can get to using the function, not necessarily the whole of the codomain. The range of a function is not equal to the codomain in general, only if the function is onto. But elsewhere in Group theory we have that Im(phi) is a subgroup in the codomain.
Actually I think I understand this! So since there is a kernel to the mapping of a homormorphism out of G that is contained in every codomain group and every kernel is a normal subgroup, the kernel of any mapping out of G is isomorphic to a normal subgroup of G okay okay I got it! That is the 1-1 correspondence part Wow so any homomorphism you want and you have associated a normal subgroup in your codomain..quite powerful
Hello Professor Salomone, would you mind providing some references for the proof of the Fourth Isomorphism Theorem? It seems that this proof can not be found in Gallian's book. Is there any other text book recommended? Thanks so much.
Mont Gum Did you know you can slow down the video playback? Click on the little "Gear" icon and try Speed 0.5. I sound a little funny but the pace is easier to keep up with :)
It sounds like a drunkard teaching algebra
:)
Hello
few litle errors in Example A quotient of S4 at 4min52
Now New H =list([]) ,list([1 2],[3 4]),list([1 3],[2 4]),list([1 4],[2 3])
New G =list([1 2]),list([1 3]),list([1 4])),list([1 2 3]),list([1 3 2])
add to this new H all element of G on the left
( 1 2 )*H
SS:BACD S:2134 Sequence:( 1 2 )
SS:ABDC S:1243 Sequence:( 1 )( 3 4 )
SS:CDBA S:3421 Sequence:( 1 4 2 3 )
SS:DCAB S:4312 Sequence:( 1 3 2 4 )
( 1 3 )*H
SS:CBAD S:3214 Sequence:( 1 3 )
SS:BCDA S:2341 Sequence:( 1 4 3 2 ) not(2 4) !
SS:ADCB S:1432 Sequence:( 1 )( 2 4 )
SS:DABC S:4123 Sequence:( 1 2 3 4 )
( 1 4 )*H
SS:DBCA S:4231 Sequence:( 1 4 )
SS:BDAC S:2413 Sequence:( 1 3 4 2 )
SS:CADB S:3142 Sequence:( 1 2 4 3 )
SS:ACBD S:1324 Sequence:( 1 )( 2 3 )
( 1 2 3 )*H
SS:CABD S:3124 Sequence:( 1 2 3 )
SS:ACDB S:1342 Sequence:( 1 )( 2 4 3 )
SS:BDCA S:2431 Sequence:( 1 4 2 )
SS:DBAC S:4213 Sequence:( 1 3 4 )
( 1 3 2 )*H
SS:BCAD S:2314 Sequence:( 1 3 2 )
SS:CBDA S:3241 Sequence:( 1 4 3 )
SS:ADBC S:1423 Sequence:( 1 )( 2 3 4 )
SS:DACB S:4132 Sequence:( 1 2 4 )
you can check with Scilab scripts here
fileexchange.scilab.org/toolboxes/498000
2:48 Cayley Table :) These videos are very helpful, thank you for posting them.
These lessons are amazing, thanks. Theorem 40 possibly has a typo, the epimorphism is named psi but the kernel is over phi.
I was incredibly confused with this slide too. Almost certainly a typo
i like your videos a lot, but the speed at which you talk is so fast that i can't keep up. but still you're my favorite so far. just wish i wasn't having a panic attack the whole time.
These videos are real helpful!!!
It has to be fast. This video covers a 50 minute lecture in 11 minutes 36 seconds. If you find this video ‘fast,’ please consider...
1. Watch this video multiple times. Its structure, incomprehensible in the first viewing, becomes comprehensible in subsequent viewings (Gestalt).
2. Do a quotient groups search:
www.google.com/search?q=quotient+groups
Recommend Millersville:
www.millersville.edu/~bikenaga/abstract-algebra-1/normal/normal.html
Also recommend its Abstract Algebra 1 series:
www.millersville.edu/~bikenaga/abstract-algebra-1/abstract-algebra-1-notes.html
+kstahmer He also suggests slowing the speed down do 0.5, though I've been unable to find that option. All I see is font options.
In the "Quotient Mapping Theorem", you refer to an epimorphism from G to G/N, but since we aren't necessarily assuming that G/N is a group (since N is not necessarily assumed to be normal), it seems like we should not be using the word "epimorphism".
At ~5:54, Z6 is isomorphic to Z2xZ3, so I am confused as to why you list them separately. ? Otherwise I love your clean, clear, colorful approach.
Joanne Koratich Thanks! And you're right, taking (1,1) as a generator in Z2xZ3 shows that that group is in fact isomorphic to Z6, so they shouldn't be separate (except insofar as to make this point).
Why would you want to do arithmetic on cosets anyway? What's the intuitive meaning / real-word examples of the concept?
Nice and neat lecture thanks for free help !!!
So at first I figured that quotient groups are very similar to groups but different in that the elements are not sets but cosets (I also originally wondered why they aren't called cosubgroups haha). Now if my interpretation is correct I see the quotient group to still be a set of elements namely for Z mod 12 { {0,3,6,9},{1,4,7,10},{2,5,8,11} }. In set theory these sets (the cosets) are valid elements of a set. So we strip the subgroup of its binary op and use it and the other cosets for the set.
" the kernel of any mapping out of G is isomorphic to a normal subgroup of G " Better: the kernel of any homomorphism out of G *is* a normal subgroup of G, not merely up to isomorphism.
Okay, I think I get it :) Thanks Professor. And on another note I think you should find I updated the notes pdf in dropbox. I am not used to dropbox, so I hope I did that correctly.
Quotient? More like "Quite excellent!" 👍
06:09 OK you've lost me here :q Is there a video that I'm missing here that explained what is the "kernel" and "image"?
5 years late, but perhaps for other viewers: they're both common functions in AA1 (since this is AA2). The image of a function is the codomain. The kernel of a function consists of all elements in the domain that are sent to the identity element in the codomain.
@@ThefamousMrcroissant the image is everything you can get to using the function, not necessarily the whole of the codomain. The range of a function is not equal to the codomain in general, only if the function is onto. But elsewhere in Group theory we have that Im(phi) is a subgroup in the codomain.
How can compute the image of (2,3,5) in R=Z/3Z xZ/4Z x Z/11Z
Actually I think I understand this! So since there is a kernel to the mapping of a homormorphism out of G that is contained in every codomain group and every kernel is a normal subgroup, the kernel of any mapping out of G is isomorphic to a normal subgroup of G okay okay I got it! That is the 1-1 correspondence part Wow so any homomorphism you want and you have associated a normal subgroup in your codomain..quite powerful
Hello Professor Salomone, would you mind providing some references for the proof of the Fourth Isomorphism Theorem? It seems that this proof can not be found in Gallian's book. Is there any other text book recommended? Thanks so much.
how to find the order of quotient group?
You look like Ted Mosby.
He looks way better than Ted Mosby.
More like Émilien (Frédéric Diefenthal) from the movie "Taxi" ;)
How can find an integer m such that R=Z/mZ? How to computer the ring homomorphism and the inverse