Determine all group homomorphisms from $\mathbb{Z}_{17}^{\times}$ into $\mathbb{Z}_7^{\times}$. 
Determine all the group homomorphisms from $\mathbb{Z}_{17}^{\times}$ into $\mathbb{Z}_7^{\times}$.

I noticed that the group is cyclic and found that the generator of both groups is $\langle 3 \rangle$ but how can I find all the group homomorphism? 
 A: HINT: $\Bbb Z_{17}^\times$ is a group of order $16$, and $\Bbb Z_7^\times$ is a group of order $6$. If $\varphi:\Bbb Z_{17}^\times\to\Bbb Z_7^\times$ is a homomorphism, the order of $\ker\varphi$ must divide $16$, and the order of $\varphi[\Bbb Z_{17}^\times]$ must divide $6$. Moreover, $\Bbb Z_{17}^\times/\ker\varphi\cong\varphi[\Bbb Z_7^\times]$, so the product of these two orders is ... what?
A: Pick a generator $x$ of $\mathbb{Z}_{17}^{\times} \cong \mathbb{Z}_{16}$, then $x$ has order $16$. Let $\varphi : \mathbb{Z}_{17}^{\times} \to \mathbb{Z}_7^{\times}$ be a group homomorphism. Note that the entire homomorphism is specified by $\varphi(x)$ because $\mathbb{Z}_{17}^{\times}$ is cyclic. As $\varphi(x^n) = \varphi(x)^n$, $\varphi(x)^{16} = 1$. For any choice of $\varphi(x)$ in $\mathbb{Z}_7^{\times}$ satisfying $\varphi(x)^{16}$, we have a genuine group homomorphism. Therefore, the number of group homomorphisms is equal to the number of choices for $\varphi(x)$. As $\mathbb{Z}_7^{\times} \cong \mathbb{Z}_6$, there are only six possibilities to check. In general though, you could use the condition $\varphi(x)^{16} = 1$ to determine the possible orders of $\varphi(x)$ and find all the elements which had one of the admissible orders.
