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Deveno

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let G = {e,a,b,ab} where a

and let H = {-1,1}.

define the map φ:G→H by:

φ(e) = 1

φ(a) = 1

φ(b) = -1

φ(ab) = -1.

is φ a homomorphism? let's check:

since φ maps e to 1, we don't have to check any products involving the identity of G (you can, if you really want to) and G is abelian, so we only have to actually check 3 products:

a

φ(a

φ(b

φ(ab) = -1 = (1)(-1) = φ(a)φ(b).

so yes, φ is indeed a homomorphism.

what is ker(φ)? its {g in G : φ(g) = 1}, which is {e,a}, which is not just the identity of G.

- #3

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I understand your reply but I'm looking at {e,a}.

a^2 = 1 so a = 1 I would think and so does e = 1.

Are you saying because the kernal has two elements and they both aren't e that shows the kernal isn't always equal to the identity? I see that but it seems at the end of the day we still wind up with just ones in the kernal (represented by e and a).

Maybe by seeing what I'm struggling with here my delima might be clearer.

Like, is there ever a 2 or a 3 in the kernal?

- #4

Deveno

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let G = Z, the integers under addition, and

let H = Z/5Z, the integers under addition modulo 5.

the homomorphism we have is φ(k) = k (mod 5).

ker(φ) = { 5k : k in Z}, so, for example 5,10,15,-5,-10 are all in ker(φ), as well, of course 0.

this is because 0 + 5Z = 5 + 5Z = 5Z, etc.

the kernel of φ lives in G. φ takes everything in ker(φ), and shrinks it down to the identity of H.

in other words, homomorphisms never make a group "bigger", but they sometimes make a group "smaller". the kernel measures "how much" shrinkage is going on (how much of G we "mod out").

it is a well-known theorem, called the first isomorphism theorem (you may not have covered it yet), that if K = ker(φ), and gK is a coset of K in G, then φ takes the entire coset gK to just a single element, namely φ(g).

- #5

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I'm going to have to spend some time with this to digest it, but I really appreciate it.

- #6

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Thanks again for the help.

- #7

Deveno

Science Advisor

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f

such a set is called a pre-image (under f) of U.

if f:G→H is a homomorphism, in the case of a kernel,

ker(f) = f

(so our "U" in this case is a 1-element set).

- #8

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Edit: I just realized Deveno gave a specific example of this homomorphism already.

- #9

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Thanks. More is always better!

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