I already know that
$\left(\mathbb{Z}/10\mathbb{Z}\right)^{*}=\left(\mathbb{Z}/2\mathbb{Z}\right)^{*}\times\left(\mathbb{Z}/5\mathbb{Z}\right)^{*}$
Theses groups have order 1 and 4 so the group is cyclic of order 4 isomorphic to $\mathbb{F_5^{*}}$
Is that right?
Can anybody explain me how to do with $\left(\mathbb{Z}/9\mathbb{Z}\right)^{*}$ and $\left(\mathbb{Z}/16\mathbb{Z}\right)^{*}$ to know their orders and if they are cyclic?
Thank you!
If you're new to these groups, I'd recommend determining whether or not $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic for small, composite $n$ by brute force. Roughly speaking, you could follow the following algorithm: - Determine the subset $S$ of $\{1,\ldots,n-1\}$ that is a complete family of representatives for the residue classes of $(\mathbb{Z}/n\mathbb{Z})^\times$ - Take the smallest integer $m > 1$ in $S$ and compute $m^2, m^3,\ldots$ modulo $n$, stopping at the first $k$ for which $m^k = 1$. - If $\{1,m,\ldots,m^{k-1}\} = S$, then $m$ generates the group. Otherwise, choose the smallest $m' \in S\setminus\{1,m,\ldots,m^{k-1}\}$ and repeat these steps. - If you never arrive at a generator, then the group is not cyclic. The other advantage to this is that it produces an explicit generator of your group if there is one.
For example, for $n = 9$ one has $$ S = \{1,2,4,5,7,8\}. $$ The smallest $m \in S$ apart from $1$ is $m = 2$. Now we compute $2^2 = 4$, $2^3 = 8$, $2^4 = 7$, $2^5 = 5$, and now we can stop computing powers since this already shows $2$ generates $(\mathbb{Z}/9\mathbb{Z})^\times$.
As another example, let $n = 8$. Then $$ S = \{1,3,5,7\}. $$ Following the above algorithm, we'd note $3^2 = 1$, then $5^2 = 1$, and finally $7^2 = 1$, which together show that $(\mathbb{Z}/8\mathbb{Z})^\times$ is not cyclic.
Of course, there are more efficient ways to do this, and there are other algorithms that compute the isomorphism class of $(\mathbb{Z}/n\mathbb{Z})^\times$ along the way (including modifications of the above algorithm). I recommend a brute force approach here, though, because it will give you a good sense of the structure of these groups and improve your comfort level with manipulating them.
A more thorough approach that will express $(\mathbb{Z}/n\mathbb{Z})^\times$ as a product of cyclic groups can be found here.