- #1

- 9

- 0

[tex]\mathbb{Z}[/tex]

_{2006}to, lets say [tex]\mathbb{Z}[/tex]

_{3008}.

Obviously, all linear functions like [tex]$ x \rightarrow a\cdot x$[/tex] will do, but are there any other functions which can result in a homomorphism?

- Thread starter raynard
- Start date

- #1

- 9

- 0

[tex]\mathbb{Z}[/tex]

Obviously, all linear functions like [tex]$ x \rightarrow a\cdot x$[/tex] will do, but are there any other functions which can result in a homomorphism?

- #2

- 124

- 0

Well, in the case of this cyclic group an homomorphism depends on the generating element of the group. When you have the image of your generating element you have the image of every other element of the group. The only restriction is

image_generating_element^{2006} = 0.

In your case the claim that all homom. will do is FALSE. Example, define

f : Z_{2006} -> Z_{3008}

x |-> 20x.

Apply f to 2006 = 1 + 1 + ... + 1:

1016 = 40120 = 20 + 20 + ... + 20_{[2006 times]}= f(1) + f(1) + ... + f(1) _{[2006 times]}= f(1 + 1 + ... + 1_{ [2006 times]}) = f(2006) = f(0) = 0

CONTRADICTION

image_generating_element

In your case the claim that all homom. will do is FALSE. Example, define

f : Z

x |-> 20x.

Apply f to 2006 = 1 + 1 + ... + 1:

1016 = 40120 = 20 + 20 + ... + 20

CONTRADICTION

Last edited:

- #3

- 9

- 0

However, I don't fully understand how the restriction on a generator (gen

- #4

- 124

- 0

Well, sometimes a homomorphism doesn't even exists (except for the trivial homom. which maps everything to 0), for example, try to find a homom. between Z_{6 }and Z_{7} (in which EVERY element is of order 7, as 7 is prime).

Last edited:

- Last Post

- Replies
- 7

- Views
- 5K

- Replies
- 1

- Views
- 772

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 3K

- Last Post

- Replies
- 9

- Views
- 1K

- Replies
- 2

- Views
- 6K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 9

- Views
- 311

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 67

- Views
- 8K