$x^{20}=1$ for all $x\in U(100)$ Show that $x^{20}=1$ for all $x\in U(100)$.
In the book that I'm reading, it says "Since $U(100)\thickapprox Z_2\oplus\,Z_{20}$ we see that $x^{20}=1$ for all $x$ in $U(100)$".I think maybe that remark is due to this theorem which states "The order of an element of a direct product of a finite groups is the least common multiple of the orders of the components of the elements". i.e $$|(g_1,g_2,.\,.\,.,g_n)|=lcm(|g_1|,|g_2|,.\,.\,.,|g_n|).$$
We define $U(n)$ to be the set of all positive integers less than $n$ and relatively prime to $n$ with the operation  multiplication modulo $n$
 A: Using Carmichael Function, $$\lambda(100)=(\lambda(25),\lambda(4))=(20,2)=20$$
So, for any integer $x,(x,100)=1\iff(x,2)=(x,5)=1,$
$$x^{20}\equiv1\pmod{100}$$ 
A: 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,  $x^{20} = 1$.
A: It's easy to verify that the least common multiple of a divisor in $2$ (the order of $\mathbb Z_2$) and a divisor in $20$ the order of $\mathbb Z_{20}$ is a divisor of $20$. And that $u^{20}=1$ for all $u\in \mathbb Z_2\oplus \mathbb Z_{20}$ follows from that, so basically you're right. I don't recall what $U(100)$ might be, so I can't tell if that actually transfers to $U(100)$.
