Prove that $(\Bbb{Z}/8)^∗\cong \Bbb{Z}/2×\Bbb{Z}/2$ and $(\Bbb{Z}/9)^∗\cong \Bbb{Z}/6$ Prove that $$(\Bbb{Z}/8)^∗\cong \Bbb{Z}/2×\Bbb{Z}/2$$ and $$(\Bbb{Z}/9)^∗\cong \Bbb{Z}/6$$
Is there any way to do this using the Automorphism group or otherwise?
 A: Otherwise:


*

*$(\mathbf Z/8\mathbf Z)^\times$ has order $4$ and is not cyclic since all its elements have order $2$  (except $1$).

*$(\mathbf Z/9\mathbf Z)^\times$ has order $6$ and is cyclic: $2$ has order $6$ (and also $2^5=5$). 

A: The group $(\mathbb{Z}/8)^{\ast}$ has $\phi(8)=4$ elements. Since we know that it is not cyclic, e.g.
Why multiplicative group $\mathbb{Z}_n^*$ is not cyclic for $n = 2^k$ and $k \ge 3$
it must be the other group of order $4$, namely $\mathbb{Z}/2\times \mathbb{Z}/2$. The second group has $\phi(9)=6$ elements and is abelian, hence it must be $\mathbb{Z}/6$. Actually, we know exactly when these abelian groups are cyclic:
When is the group of units in $\mathbb{Z}_n$ cyclic?
A: If you have to work for specific small number like $8$ and $9$ one can enumerate  the numbers coprime to them.
IN the case $n=8$, the units  are $\{1,3,5,7\}$ which is a group for multiplication modulo $8$ with every element having $1$ as square mod 8. ANd the product of two elements from 3,5,7 is the third. Then it is clear that it is the Klein's 4-group.
You can work out the case $n=9$ and verify the units form a cyclic group of order $6$. If you want a general theory  you can consult, for example Ireland and Rosen's Invitation to Number Theory.
A: Since we are dealing with "small" groups, we can solve this problem with a bit of brute force.
Let's start making a multiplication table for $\Bbb Z_8^\times$:
$$\begin{array}{|c|c|c|c|c|}
\hline
&1&3&5&7\\
\hline
1&1&3&5&7\\
\hline
3&3&1&7&5\\
\hline
5&5&7&1&3\\
\hline
7&7&5&3&1\\
\hline
\end{array}$$
Note that there are no elements of order $4$. You can define
$$f:(\Bbb Z_8^\times,\cdot)\to(\Bbb Z_2\times\Bbb Z_2,+)$$
in the following way (it is not unique): $f(1)=(0,0)$, $f(3)=(1,0)$, $f(5)=(0,1)$, $f(7)=(1,1)$.
It is easy to see that this function is an isomorphism.
Can you try it for the other pair of groups?
