- #1

Ledger

- 11

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Ledger
- Start date

- #1

Ledger

- 11

- 0

- #2

Deveno

Science Advisor

Gold Member

MHB

- 2,725

- 6

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

Ledger

- 11

- 0

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

Science Advisor

Gold Member

MHB

- 2,725

- 6

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

Ledger

- 11

- 0

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

- #6

Ledger

- 11

- 0

Thanks again for the help.

- #7

Deveno

Science Advisor

Gold Member

MHB

- 2,725

- 6

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

Theorem.

- 237

- 5

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

- #9

Ledger

- 11

- 0

Thanks. More is always better!

Share:

- Last Post

- Replies
- 14

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 474

- Last Post

- Replies
- 0

- Views
- 347

- Last Post

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 932

- Last Post

- Replies
- 1

- Views
- 822

- Last Post

- Replies
- 5

- Views
- 2K

MHB
Subgroup nesting

- Last Post

- Replies
- 3

- Views
- 1K