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Let $G = ( \mathbb { Z } / 133 \mathbb { Z } ) ^ { \times }$ be the group of units of the ring $\mathbb { Z } / 133 \mathbb { Z }$ . Find the number of elements of $G$ of order $9 .$

133 cannot be divided by 9. So what is the solution to the problem? Or my consideration is too simple to realize the problem precisely?

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    $\begingroup$ The order of $G$ is not $133$, the order of $\Bbb Z/133\Bbb Z$ is $133$. $\endgroup$ – Hagen von Eitzen Feb 10 at 10:58
  • $\begingroup$ @HagenvonEitzen I am sorry but I wonder 1.Why $G$ has 108 elements as mentioned in the answer below? 2. Is there a universal solution for problem like this? $\endgroup$ – Midas Hu Feb 10 at 11:13
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    $\begingroup$ Not every element of $\Bbb Z/\Bbb 133\Bbb Z$ is a unit. The units are those coprime to $133$ (inverse of $k$ mod $n$ exists iff $k$ is coprime to $n$). The order of the group of units is given by $\phi(133)=\phi(7\times 19)=6\times 18=108$ where $\phi$ is the Euler totient function. $\endgroup$ – learner Feb 10 at 12:17
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Hint: $ ( \mathbb { Z } / 133 \mathbb { Z } ) ^ { \times } \cong ( \mathbb { Z } / 7 \mathbb { Z } ) ^ { \times } \times ( \mathbb { Z } / 19 \mathbb { Z } ) ^ { \times } \cong C_6 \times C_{18} $

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The fact that $9\nmid133$ is irrelevant here, since the group $\mathbb{Z}_{133}^\times$ has $108$ elements. Since $9\mid108$, Lagrange's theorem is not an obstacle to the existence of elements of order $9$.

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