Surjective homomorphism on roots of unity

Quote:

Let

. Prove that for any fixed integer k > 1 the map from G to itself defined by

is a surjective homomorphism but is not injective.

So let

.

Proving the homomorphism is trivial. It's the surjection part that's getting to me. (I haven't tackled the injection yet. Please leave that out of the replies for now.) What I am looking at is the following...I want to show that there is at least one pair of x, y in G such that is true when x is not equal to y. My other question about this is do we need to find an x, y for each k? For example, is only true for even k.

I'm just not wrapping this one around in my head today.

Thanks!

-Dan

Re: Surjective homomorphism on roots of unity

Quote:

Originally Posted by

**topsquark** So let

.

Proving the homomorphism is trivial. It's the surjection part that's getting to me. (I haven't tackled the injection yet. Please leave that out of the replies for now.) What I am looking at is the following...I want to show that there is at least one pair of x, y in G such that

is true when x is not equal to y.

No, you don't! That would be showing this is a NOT an injection. It would not be showing this is a surjection. To show this is a surjection you need to show that if z is a complex number such that for some n then there exist a complex number w such that such that w^k= z for this specific z.

Quote:

My other question about this is do we need to find an x, y for each k? For example,

is only true for even k.

For z= -1, take w= -1.

Quote:

I'm just not wrapping this one around in my head today.

Thanks!

-Dan

Re: Surjective homomorphism on roots of unity

1. Surjective: if then for some with . Can't you find with and ?

2. Injective: Think about the kernel of the homomorphism.

Re: Surjective homomorphism on roots of unity

Sorry I didn't get back sooner. I just finished the problem, both surjective and injective. I appreciate the help.

-Dan