Isomorphism from the multiplicative group U7 Let $H$ be the group $Z_6$ under addition. Find all the isomorphisms from the multiplicative group $U_7$ of units in $Z_7$ to H?
This is one of the practice question for my exam tomorrow.. I dont understand what the author means by $U_7$ of units in $Z_7$ ?
 A: The author means, those elements of ${\bf Z}_7$ that are invertible under multiplication modulo $7$. Do you know what the multiplicative identity is in ${\bf Z}_7$? Knowing that, can you figure out which elements have multiplicative inverses?
A: You are correct to say that every nonzero element of $\mathbb{Z}_7$ has a multiplicative inverse, so the group $\mathbb{Z}_7^*$ has order $6$.  You can also check that $3$ has multiplicative order $6$ in $\mathbb{Z}_7$.  It follows that $\mathbb{Z}_7^*\cong\mathbb{Z}_6$.
Hence, the question reduces to finding all automorphisms of the group $\mathbb{Z}_6$.  Since an automorphism of a cyclic group is determined by where a generator is sent, and since there are $\varphi(6)=2$ generators of $\mathbb{Z}_6$ (where $\varphi$ is Euler's totient function), it follows that there are two automorphisms of $\mathbb{Z}_6$.  Translating this back to answer your question, we have the two maps:
$$
\begin{array}{rcl}f_1:\mathbb{Z}_7^*&\rightarrow &\mathbb{Z}_6\\
3&\mapsto &1
\end{array}$$
$$
\begin{array}{rcl}f_2:\mathbb{Z}_7^*&\rightarrow &\mathbb{Z}_6\\
3&\mapsto &5
\end{array}$$
Now $5$ also has multiplicative order in $\mathbb{Z}_7^*$, so we could have just defined the above maps by deciding where to map $5$.  Suppose we did this to obtain the following isomorphisms
$$
\begin{array}{rcl}g_1:\mathbb{Z}_7^*&\rightarrow &\mathbb{Z}_6\\
5&\mapsto &1
\end{array}$$
$$
\begin{array}{rcl}g_2:\mathbb{Z}_7^*&\rightarrow &\mathbb{Z}_6\\
5&\mapsto &5
\end{array}$$
Then, noticing that $3\equiv5^5\mod 7$, it follows that $g_1(3)=g_1(5^5)=5\cdot g_1(5)=5\cdot1=5$ (remember that the group operation in $\mathbb{Z}_7^*$ is multiplication, while the group operation in $\mathbb{Z}_6$ is addition), so that in fact, $g_1=f_2$.  Similarly, you can show that $g_2=f_1$ and thus we have not defined any new automorphisms.
