For starters, $|U(100)| = \varphi(100) = 40$. So by Lagrange, $\forall \hat{x} \in U(100), ord(\hat{x}) \mid 40$. Now, how can one prove there aren't any generators in U(100) without computing the order of any element?
2026-04-02 23:28:55.1775172535
How does one prove U(100) is not cyclic without computing the order of any element?
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As recommended in the comments, I'm posting here the answer I was looking for:
$U(100) = U(25⋅4) \cong U(25) \oplus U(4)$. By Gauss, $U(25) \oplus U(4) \cong \mathbb{Z}_{20} \oplus \mathbb{Z}_{2}$. And since the order of any element in $\mathbb{Z}_{20} \oplus \mathbb{Z}_{2}$ is at most lcm(20, 2) = 20, there are no elements of order 40 in U(100).
Thus, U(100) is not cyclic.