prove there does not exist an onto homomorphism f: Z_n - Z_(n-1)

Question

prove there does not exist an onto homomorphism f: Z_n -> Z_(n-1)

anonymous · Answer

Assume $f$ is a surjective homomorphism. We have the following.
$f(a^n) = f(a)^n = f(0) = 0$

So $f(a)^n = 0$ and therefore the order of $f(a)$ divides $n$. Why is this a contradiction if assuming $f$ is surjective?

moss65 · Answer

this something about isomorphism: ord(a) =ord(f(a)), but it cannot be because ord(a)=n and ord(f(a))=n-1

moss65 · Answer

so f cannot be isomorphic, and it is not one-to-one

anonymous · Answer

Does the order of each element in $Z_{n -1}$ divide $n$?

anonymous · Answer

We are not talking about an isomorphism here. We only know that $f$ is surjective.

moss65 · Answer

no, it doesn't divide n

anonymous · Answer

Well, the order of the identity element (order 1) does divide $n$. However, since this is a surjection, for each $b \in Z_{n - 1}$ we can find some $a \in Z_n$ such that f(a) = b. Therefore it must hold that for each $b \in Z_{n - 1}$ that $ord(b) \mid n$.

anonymous · Answer

Hint: What is the order of the generator in $Z_{n - 1}$?

moss65 · Answer

yes, but it cannot be. is this contradiction you meant?

anonymous · Answer

What is the order of $1$ in $Z_{n−1}$?

anonymous · Answer

We will get our contradiction if you can answer this question. Does $|1_{Z_{n-1}}|$ divide $n$?

moss65 · Answer

Sorry Alchemista, there was some problem with my comp. The order 1 bar in Z_n-1 is n-1, so it definitely doesn't divide n

anonymous · Answer

Exactly, and since we demonstrated above that if $f$ is a surjective homomorphism $\forall x\in Z_{n−1}[|x| \text{ divides } n]$ we have arrived at our contradiction. Since $|1|$ does not divide $n$.

anonymous · Answer

Does that make sense?

moss65 · Answer

absolutely. 
Thank you so much. It was nice to work with you again.