I brute force the solution of order 5 in $Z^*_{36}$ by using the following $a^5 \equiv 1 \mod 36$ and I see that there is no solution for this. However, I don't quite know how to prove this. Can someone help me?
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1$a^5\equiv 1$ has at least one solution no matter what you mod by: consider $a\equiv 1$. – Arthur Oct 20 '19 at 18:03
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1See the duplicates here or here. – Dietrich Burde Oct 20 '19 at 18:22
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The group $\mathbb Z_{36}^*$ has order $12$. Therefore, the order of each of its elements must divide $12$.
José Carlos Santos
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Can you show how you get order 12 in $Z^*_{36}$? Because $(12)^5 \not\equiv 1 \mod 36$ – Algorisum Oct 20 '19 at 18:29
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The order of $\mathbb Z_n^*$ is $\phi(n)$ and $\phi(36)=12$. – José Carlos Santos Oct 20 '19 at 18:33
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But I want to find element of order 5 in $Z^*_{36}$, but not element of order 12. – Algorisum Oct 20 '19 at 18:34
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The order of a group is the number of elements of that group! – José Carlos Santos Oct 20 '19 at 18:37
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So does that mean there is 5 element in the order of 5 in $Z^*_{36}$? – Algorisum Oct 20 '19 at 18:38
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No. It means that there are $\mathbf0$ elements of order $5$. – José Carlos Santos Oct 20 '19 at 19:01
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No. Because the order of an element always divides the order of the group. And $5\nmid12$. – José Carlos Santos Oct 20 '19 at 19:11
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