# Thread:

1. ## Isomorphism

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.

2. Originally Posted by suedenation
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.
Since [ ] is the greatest integer function we have,
$[\, a \, ]=a$ for all integers.

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

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

3)Homomorphism: We need to show that,
$[\, x+y\, ]=[\, x\, ]+_n[\, y\, ]$
Because, $x,y$ are integers we have,
$x+y=x+_ny$. 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.

3. Actually, [a] is the equivalence class of a modulo n; i.e., [a]={k:k in Z and k=a (mod n)}.

4. ## :o

Well, the method is the same as ThePerfectHacker has demonstrated.

For the homomorphism part, check whether $<br /> f(x+y)=[\, x+y\, ]$ is equal to $f(x)+f(y)=[\, x\, ]+[\, y\, ].<br />$ Not hard at all

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

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

Try it.
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