$\varphi_n (z) = z^n$ find all subgroups between $\ker \varphi _3$ and $ \ker \varphi _{12}$ let $C^*$ be the group of complex numbers excluding zero with * operation.
I need to show the following - 
1) $\varphi_n : C^* \to C^*$  such that $\varphi_n (z) = z^n$ is homomorphism.
2) $\ker \varphi _3 \subseteq \ker \varphi _{12}$
3) find all subgroups between the two groups from question two.
What I did -  I think I solved 1 and 2 but Im not confident about it
And with question 3 I don't know how to solve. 
so for 1 i said that
$\varphi_n (z_1*z_2) = (z_1 * z_2)^n = z_1^n * z_2^n = \varphi_n (z_1) * \varphi_n (z_2)$
is that correct? 
for two i said that if $z^3 = 1$ then $z^{3*4} = 1 = z^{12}$
not sure about that one also . 
I need help in 3 . 
any help will be appreciated
 A: The answer of 3 is $Ker \varphi_6$.
Proof: $Ker \varphi_{12}$ have $12$ elements:$\{e^{2\pi i{\frac{k}{12}}}, 0\leq k\leq 11\} $ and the $Ker \varphi_3$ have $3$ elements : $\{e^{2\pi i{\frac{k}{3}}}, 0\leq k\leq 2\} $ .
Now the only subgroup of $Ker \varphi_{12}$ that contains $Ker \varphi_3$, is $Ker \varphi_{6}=\{e^{2\pi i{\frac{k}{6}}}, 0\leq k\leq 5\}$.
A: Your work on (1) and (2) are correct, assuming you know that the kernel of a homomorphism is always a subgroup of the domain. It will help if you recognize that $z^{n}=1$ only if $z$ lies on the unit circle in the complex plane.
A: The multiplicative group $\;\ker\varphi_n$, also known as $U_n$, is cyclic, isomorphic to the additive group $\mathbf Z/n\mathbf Z$. Now for any cyclic group of order $n$, generated by $\zeta$, say, and any divisor $d$ of $n$, there exists exactly one subgroup of order $d$, generated by $\zeta^{\tfrac nd}$.
In the present case ($n=12$), we're looking for subgroups of order a divisor of $12$, which is a multiple of $3$. In addition to the trivial subgroups, the only one is the subgroup of order $6$, and since $\zeta=\mathrm e^{\tfrac{\mathrm i\pi}6}$, it is generated by 
$$\mathrm e^{\tfrac{\mathrm 2i\pi}6}=\mathrm e^{\tfrac{\mathrm i\pi}3}.$$
