$x^{20}=1$ for all $x\in U(100)$

Solution 1:

Using Carmichael Function, $$\lambda(100)=(\lambda(25),\lambda(4))=(20,2)=20$$

So, for any integer $x,(x,100)=1\iff(x,2)=(x,5)=1,$

$$x^{20}\equiv1\pmod{100}$$

Solution 2:

Since $U_{100} \thickapprox \mathbb Z_2\oplus\,\mathbb Z_{20}$, then there exists a group isomorphism $f : U_{100} \to \mathbb Z_2\oplus\,\mathbb Z_{20}$.

Note that $U_{100}$ is a multiplicative group while $\mathbb Z_2$ and $\mathbb Z_{20}$ are additive groups. So the identity element of $U_{100}$ is $1$ while the identity element of $\mathbb Z_2$ and $\mathbb Z_{20}$ is $0$.

Let $x \in U_{100}$. Then $f(x) = (y,z)$ for some $y \in \mathbb Z_2$ and some $z \in \mathbb Z_{20}$.

The smallest positive integer, n, such that, for all $y \in \mathbb Z_2, \; ny = 0$ is $n = 2$.
The smallest positive integer, n, such that, for all $z \in \mathbb Z_{20}, \; nz = 0$ is $n = 20$.

Since $20$ is the least common multiple of $2$ and $20$ it follows that the smallest positive integer, n, such that, for all $x \in \mathbb Z_2$ and $y \in \mathbb Z_{20}, \; ny =0$ and $nz = 0$ is $n = 20$.

Since $f$ is an isomorphism, $f(x^{20}) = (20y, 20z) = (0, 0) = f(1)$.

Since $f$ is an isomorphism, $x^{20} = 1$.