First Isomorphism Theorem of Groups -- Abstract Algebra 14

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MathMajor

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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)

SeeTv.

If you start the last example with "prove thatZ_6/<3> 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.

matematicacommarcospaulo

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:

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(<5>)=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.

matheusjahnke

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".

schweinmachtbree

if N<G1 and N is normal then will phi(N) always be a *normal* subgroup of G2 or will it be normal only some conditions or we can't say anything?

atifachaudhry

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

MadocComadrin

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