# Find quantity of elements in group with given order

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?

• The order of $G$ is not $133$, the order of $\Bbb Z/133\Bbb Z$ is $133$. – Hagen von Eitzen Feb 10 at 10:58
• @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? – Midas Hu Feb 10 at 11:13
• 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. – learner Feb 10 at 12:17

Hint: $$( \mathbb { Z } / 133 \mathbb { Z } ) ^ { \times } \cong ( \mathbb { Z } / 7 \mathbb { Z } ) ^ { \times } \times ( \mathbb { Z } / 19 \mathbb { Z } ) ^ { \times } \cong C_6 \times C_{18}$$
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 $$9$$.