# 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?

• Hint: The homomorphism is uniquely defined by the image of a generator. – Tomas Oct 21 '13 at 0:32
• Note, $\langle 3\rangle$ is a group is not an element. You either say $3$ is a generator for $\mathbb{Z}_{17}^{\times}$ or $\langle 3\rangle = \mathbb{Z}_{17}^{\times}$. – Michael Albanese Oct 21 '13 at 0:37

## 2 Answers

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 priori, couldn't there be two different homomorphisms with the same number of elements in their ranges? – Michael Albanese Oct 21 '13 at 0:39
• @Michael: Sure. The hint is merely intended to point the OP in the right direction by eliminating impossibilities. Once you know the possibilities for the kernel and range, the rest is relatively easy. – Brian M. Scott Oct 21 '13 at 0:41
• OK, I thought I was missing something. Thanks. – Michael Albanese Oct 21 '13 at 0:42
• I forgot to look at the order. I naively thought the order was $17$ and $7$, big mistake by me. Thank you very much! – Lays Oct 21 '13 at 0:45
• @Lays: Yes, that would make it a bit difficult. You’re welcome! – Brian M. Scott Oct 21 '13 at 0:46

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.