Finding the elements with order $30$ in group $\mathbb Z^*_{31}$.

Question

The elements of the group $\mathbb Z^*_{31}$ is $1$ to $30$. I found the first element with order $30$ is $11$ by checking the powers of each element by $2,3,5,6,10,15,30$. My professor mentioned that the other elements with order $30$ can be found after finding the first element using a theorem. Could someone please help me with that.

Answer 1

Hint:

Welcome! You've found that every element in $(\mathbf Z/31\mathbf Z)^\times$ is a power of $11$.

Now, if an element $a$ in a group has order $n$, $a^k$ has order $\;\dfrac{n}{\gcd(n,k)}$.

Answer 2

Yes, there is a way to find all other elements of order $30$, once you find one element of order $30$. This is because $31$ is a prime number so $\mathbb{Z}_{31}^{*}$ is cyclic, and once you have found a generator, say $g$, you can find all other elements. The theorem you need to use is that $|g^{s}| = \frac{|g|}{(s,|g|)}$, where $|\cdot|$ denote the order of an element. Then all elements of the form $g^{s}$, with $(s,|g|) = (s,30) = 1$, will have order $30$ by the theorem. After doing the computation, you get the set: $$ \{ 11^1, 11^7 , 11^{11}, 11^{13}, 11^{17} , 11^{19} , 11^{23}, 11^{29} \} = \{ 11 , 13 , 24 , 21 , 3 , 22 , 12 , 17\} \pmod{31} $$