Surjective homomorphism on roots of unity

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Let $\displaystyle G = \{ z \in \mathbb{C} |z^n = 1 \text{ for some n } \in \mathbb{Z} ^+ \}$. Prove that for any fixed integer k > 1 the map from G to itself defined by $\displaystyle z \mapsto z^k$ is a surjective homomorphism but is not injective.

So let

$\displaystyle \phi _k : G \to G : g \mapsto g^k$.

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 $\displaystyle x^k = y^k$ 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, $\displaystyle (-1)^k = (1)^k$ 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

$\displaystyle \phi _k : G \to G : g \mapsto g^k$.

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 $\displaystyle x^k = y^k$ 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 $\displaystyle z^n= 1$ for some n then there exist a complex number w such that $\displaystyle w^n= 1$ such that w^k= z for this specific z.

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My other question about this is do we need to find an x, y for each k? For example, $\displaystyle (-1)^k = (1)^k$ is only true for even k.

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

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

Thanks!

-Dan

Re: Surjective homomorphism on roots of unity

1. Surjective: if $\displaystyle z^n=1$ then $\displaystyle z=e^{2mi\pi/n}$ for some $\displaystyle m$ with $\displaystyle 0\leq m<n$. Can't you find $\displaystyle w$ with $\displaystyle w^{kn}=1$ and $\displaystyle w^k=z$?

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