Let n be a positive integer. Define f : Z ! Zn by f(a) = [a]. Is f a homomorphism? One-to-one? Onto? An isomorphism? Explain.

Can any one help? Thanks very much. :(

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- March 27th 2006, 12:21 PMsuedenationIsomorphism
Let n be a positive integer. Define f : Z ! Zn by f(a) = [a]. Is f a homomorphism? One-to-one? Onto? An isomorphism? Explain.

Can any one help? Thanks very much. :( - March 27th 2006, 02:07 PMThePerfectHackerQuote:

Originally Posted by**suedenation**

for all integers.

1)Onto: We need to show that for all we can find such a such as, which is true if you take .

2)One-to-One: We need to show that if then, . Based on the first paragraph that are integers, we have that

3)Homomorphism: We need to show that,

Because, are integers we have,

. Which is definitely not true.

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Thus, this map is not an isomorphism. Think about it, how can you have an isomorphsim between a finite and an infinite set? Impossible. - April 1st 2006, 07:14 PMTreadstone 71
Actually, [a] is the equivalence class of a modulo n; i.e., [a]={k:k in Z and k=a (mod n)}.

- April 5th 2006, 09:13 AMRebesques:o
Well, the method is the same as ThePerfectHacker has demonstrated.

For the homomorphism part, check whether is equal to Not hard at all :)

Ifr were to be one-to-one, you would have had to show that Try testing this, with integers like and .

The onto part is also not hard. Just consider an equivalence class , and find an integer (obvious!) to map to this. :)

And, since an isomorphism needs to be one-to-one, this (is?/is not?) an isomorphism. :eek:

Try it.