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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- Mar 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. :( - Mar 27th 2006, 02:07 PMThePerfectHackerQuote:

Originally Posted by**suedenation**

$\displaystyle [\, a \, ]=a$ for all integers.

1)Onto: We need to show that for all $\displaystyle x\in\mathbb{Z}_n$ we can find such a $\displaystyle y\in \mathbb{Z}$ such as, $\displaystyle [\, y\, ]=x$ which is true if you take $\displaystyle x=y$.

2)One-to-One: We need to show that if $\displaystyle [\, x \, ]=[\, y \, ]$ then, $\displaystyle x=y$. Based on the first paragraph that $\displaystyle x,y$ are integers, we have that $\displaystyle x=y$

3)Homomorphism: We need to show that,

$\displaystyle [\, x+y\, ]=[\, x\, ]+_n[\, y\, ]$

Because, $\displaystyle x,y$ are integers we have,

$\displaystyle x+y=x+_ny$. 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. - Apr 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)}.

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

For the homomorphism part, check whether $\displaystyle

f(x+y)=[\, x+y\, ]$ is equal to $\displaystyle f(x)+f(y)=[\, x\, ]+[\, y\, ].

$ Not hard at all :)

Ifr $\displaystyle f$ were to be one-to-one, you would have had to show that $\displaystyle x\neq y \Rightarrow f(x)\neq f(y).$ Try testing this, with integers like $\displaystyle a$ and $\displaystyle 2a$.

The onto part is also not hard. Just consider an equivalence class $\displaystyle [a]$, 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.