What is the largest order of any element in $U_{900}$? What is the largest order of any element in $U_{900}$?
I found that it is isomorphic to $\mathbb Z _{240}$. So I guess that it should be $240$. Am I correct?
 A: Hint $\ $ The CRT decomposition is $\rm\ \Bbb Z/900 = \Bbb Z/4 \times \Bbb Z/9 \times \Bbb Z/25\ $ so $\rm\: U_{900} = U_{4}\times U_9\times U_{25},\ $ a product of cyclic groups of order $\rm\ \phi(4)=2,\ \phi(9)=6,\ \phi(25)=20,\:$ with lcm $= 60,\:$ not $240$. This implies $\rm\:a^{60} = 1\:$ for all $\rm\:a\in U_{900}.\:$ Indeed, if $\rm\:(a,b,c)\in C(i) \times C(j) \times C(k),\:$ a product of cyclic groups of order $\rm\:i,j,k,\:$ then
$$\rm  \ell = lcm(i,j,k)\mid n\:\Rightarrow\: i,j,k\mid n\:\Rightarrow\:(a,b,c)^n = (a^n,b^n,c^n) = 1$$
So the largest order $\rm\:\color{#c00}{m \le \ell}.\:$ Conversely, taking $\rm\:a,b,c\:$ to be generators 
$$\rm\ 1 = (a,b,c)^n = (a^n,b^n,c^n) \:\Rightarrow\: i,j,k\mid n\:\Rightarrow\ \ell = lcm(i,j,k)\mid n$$
Thus $\rm\: \ell\le m.\:$ Combining $\rm\: \color{#c00}{m \le \ell} \le m,\:$ hence $\rm\:m = \ell.$
Remark $\ $ Look up the Carmichael lambda function to learn more about the general case.
A: You first factorise $900 = 3^2 \cdot 5^2 \cdot 2^2$. Then you determine the maximal order in $U_9$, in $U_{25}$ and in $U_4$ (you should already know the general structure of the groups for prime powers). Finally, the Chinese remainder theorem tells you that $U_{900}$ is the product of these three groups, so the maximal order in $U_{900}$ is the least common multiple of the maximal orders in the three components.
A: That's the largest possible order, but that doesn't mean that there must exist an element with such an order. I ran a Python script and could not find any.
for i in range(900):
    if gcd(900, i) == 1:
        order = 1
        total = i
        while total != 1:
            total = (total * i) % 900
            order += 1
        if order == 240:
            print(i)

EDIT: Today I learned you need <pre> tags.
A: Using Carmichael Function
$\lambda(900)=$lcm$(\phi(2^2),\phi(3^2),\phi(5^2))=$lcm$(2,6,20)=60$
