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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.

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  • $\begingroup$ You needed to check only $6, 10, $ and $15$, not $2, 3, $ and $5$ $\endgroup$ – J. W. Tanner Apr 5 at 23:31
  • $\begingroup$ Hi and welcome! I think you can find the answer to your question here $\endgroup$ – Menezio Apr 5 at 23:42
  • $\begingroup$ @J.W.Tanner , Could you please explain why not 2,3 and 5. Because, i found the order of 2 in the group as 5 (2^5 = 1(mod 31). perhaps, you mean for the elements after 11? $\endgroup$ – Art Apr 6 at 0:20
  • $\begingroup$ you would have found $2$ not to be a primitive root when you checked $2^{15}$ $\endgroup$ – J. W. Tanner Apr 6 at 0:25
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    $\begingroup$ @J.W Tanner Yes.Understood. Thanks $\endgroup$ – Art Apr 6 at 0:32
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Since $\Bbb Z_{31}^*\cong\Bbb Z_{30}$ and $\varphi(30)=8$, there are $8$ with order $30$. They are $1,7,11,13,17,19,23,29$, each of which is coprime with $30$.

The theorem your professor is referring to is $|g^k|=|g|/\operatorname{gcd}(|g|,k)$.

You can use any generator, say $11$, to construct an isomorphism $\psi$ between the multiplicative group $\Bbb Z_{31}^*$ and the additive group $\Bbb Z_{30}$, given by $\psi(11^n)=n$.

So, under the isomorphism, we get $\{11,11^7,11^{11},11^{13},11^{17},11^{19},11^{23},11^{29}\}$ as the generators.

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    $\begingroup$ $27$ is not coprime to $30$. $(27,30) = 3$ $\endgroup$ – user100101212 Apr 5 at 23:44
  • $\begingroup$ Oops. Thanks for catching that. @user100101212 $\endgroup$ – Chris Custer Apr 5 at 23:57
  • $\begingroup$ You should give some details on the isomorphism: one is a multiplicative group, the other an additive group, and it may be not completely trivial to every reader. $\endgroup$ – Bernard Apr 6 at 13:46
  • $\begingroup$ @Bernard I've added something on that. $\endgroup$ – Chris Custer Apr 6 at 15:37
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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: $$ \{ 11^1, 11^7 , 11^{11}, 11^{13}, 11^{17} , 11^{19} , 11^{23}, 11^{29} \} = \{ 11 , 13 , 24 , 21 , 3 , 22 , 12 , 17\} \pmod{31} $$

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  • $\begingroup$ $24 \equiv 11^{11} \pmod{31}$ , and $22 \equiv 11^{19} \pmod{31}$. $\endgroup$ – user100101212 Apr 5 at 23:42
  • $\begingroup$ No, those are exponents of the generator. $\endgroup$ – user100101212 Apr 5 at 23:44
  • $\begingroup$ I'm sorry. Under the isomorphism you would be correct. $\endgroup$ – Chris Custer Apr 6 at 0:09
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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 $\;\dfrac{n}{\gcd(n,k)}$.

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