First Isomorphism Theorem of Groups -- Abstract Algebra 14 

43:15 a much easier version, which works the same for GL_n(IR) so general nxn matrices with real entries: For the preimage of x take the diagonal matrix with 1's on the diagonal except for one entry (maybe the first) which is x. [ x 0 0 __ 0 0 0 1 0 __ 0 0 0 0 1 __ 0 0 | 0 0 0 __ 1 0 0 0 0 __ 0 1 ] The detrminant is x*1*1*...*1 = x which is non-zero, so the matrix is in GL_n(IR)

If you start the last example with "prove thatZ_6/ is isomorphic to Z_3", I would have serious problems to imagine that I had to create phi from Z_6 to Z_15 given by phi(n) = 5n.

For the first warmup exercise... I don't know if I can solve it. By that I mean I think there are 2 different homorphisms that satisfy the conditions. You gave me phi(5)=1 I know that the identity in the domain maps to the identity in the co-domain phi(1 (in a multiplicative group)) = 0 (in an additive group) I can find values for the values in the domain which are generated by 5: phi(5²)=phi((5)\*(5))=phi(5)+phi(5)=1+1=2 I can keep doing that... I got: phi(1)=0, phi(5)=1, phi(25)=2, phi(17)=3, phi(13)=4, phi(29)=5 (note that 125=36\*3+17=17 mod 36, and so on) 5 has order 6 in U36 I know how to map these values, but I don't know how to map others, like 7 Let's say phi(7)=x, and try to get the values generated by 7: phi(1)=0x, phi(7)=1x, phi(13)=2x, phi(19)=3x, phi(25)=4x, phi(31)=5x We have 13 and 25 in common, so we have 2x=4 mod 6 4x=2 mod 6 both have x=2 mod 3... or x=2,5 mod 6 And that is my problem: While I know exactly the values of a few inputs: phi(1)=0, phi(5)=1, phi(25)=2, phi(17)=3, phi(13)=4, phi(29)=5 For the other values, I can either have: phi(7)=2, phi(35)=3, phi(31)=4, phi(11)=5, phi(19)=0, phi(23)=1 Or: phi(7)=5, phi(35)=0, phi(31)=1, phi(11)=2, phi(19)=3, phi(23)=4 So either {1, 19} is the kernel of phi, or {1, 35} is. I guess at some point it is "obvious", because the U36 has 12 elements(namely: {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35}) while Z6 has 6, if the image of phi is Z6(which in our case it is... because phi()=Z6), then we need a kernel of size 2 so that U36/kernel(phi)... which must be isomorphic to Z6(and, consequently, have the same cardinality), so it has to have 6 elements.

Awesome

American hero.

3:37 don't forget that A needs to be nonempty (which it is in the applications to (1) and (2)) 23:39 since you're using left cosets you should really write x^(-1)y ϵ ker(phi), but in this case it's okay because H := ker(phi) is normal (although it can cause confusion for situations where H is not normal) 39:58 this is a mistake, although we can't blame you towards the end of an hour-long lecture. the _congruence class_ n-1 equals the _congruence class_ 1, so n-1 ≡ 1 (mod n) hence -1 ≡ 1 ⇒ 2 ≡ 0 (mod n) and therefore n | 2, giving not only n = 2 but also n = 1. so at 41:55 it should be "if n > 2" instead of "if n ≠ 2".

if N

For the proof, don't we still need to show that psi is unique?