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For example, to find 4 elements of order 47 in U(187), is there a simpler way to do it instead of finding 2 more numbers except 1 and 186 by going through all possibilities from 2 to 185?

user9826208
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    The order of any element of a group must divide the order of the group. If your $U(m)$ is the group of units (invertible elements) of $\mathbb{Z}/m\mathbb{Z}$, then, in your case, the order must divide $\phi(187)=160$. Your $47$ does not. – metamorphy Oct 28 '18 at 19:07

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$187 = 11 \times 17$ so $\phi(187) = 10 \cdot 16 = 160$.

The order of the group of units $U(187)$ is thus 160.

The order of an element of $U(187)$ thus divides $160$, so no element of order $47$ can exist.

Henno Brandsma
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