Isomorphism of a quotient group of C* with C*

Question

Let for $n\ge 1, \mu_n \subset \Bbb C^{\times}$ be such that $\mu_n:=\{z\in\Bbb C^{\times}\;|\;z^n=1\}$.

First I need to prove that the homomorphism $\Bbb C^{\times}\to\Bbb C^{\times}$ given by $z\mapsto z^n$ induces an isomorphism $\Bbb C^{\times}/\mu_n ≅ \Bbb C^{\times}$

Then I'm left to prove that if $H \subset \Bbb C^{\times}$ is a finite subgroup with $|H| = n$ then $H = \mu_n$.

To prove the isomorphism I thought I'd use the first isomorphism theorem. As a mapping $f$ defined by $z\mapsto z^n$ gives $\ker(f) = \mu_n$ and the mapping also gives us $f(\Bbb C^{\times}) =\Bbb C^{\times}$ which concludes the isomorphism.

Now the second part seemed very weird to me, I don't know how to do this. Please feel free to correct me with the isomorphism as well.

Now I've been thinking a bit and my idea for the second part was to take an element $h∈H$ because $ord(h)|ord(H)$ we must have $ord(h)|n$ which implies $h^n = 1$..?

Answer 1

For the first part, make sure to prove that $f$ is indeed surjective.

For the second part, you need two things:

• if $H$ is a group of order $n$, then for each $a \in H$ one has $a^{n} = 1$ (this is a consequence of Lagrange's theorem), and
• over a field, a polynomial of degree $n$ can have at most $n$ roots.

Now consider the polynomial $x^{n} - 1$.