Homomorphism on modulo groups

  • Thread starter raynard
  • Start date
9
0
I was wondering, if we want to define a morphism from
[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?
 
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_element2006 = 0.

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

f : Z2006 -> Z3008
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
 
Last edited:
9
0
I understand my error in believing every linair function will result in a homomorphism.
However, I don't fully understand how the restriction on a generator (gen2006 = 0) can help in finding function that do result in a morphism. The only possible combination is the trivial solution x |-> x.
 
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 Z6 and Z7 (in which EVERY element is of order 7, as 7 is prime).
 
Last edited:

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