Let $G=(\mathbb{Z}/540\mathbb{Z})^{\times} \cong C_2 \times C_{18} \times C_4 $.
Find elements in $G$ so that $ord(a)=2$, $ord(b)=18$, $ord(c)=4$ and $\{e\}=\langle a\rangle \cap \langle b,c\rangle = \langle b\rangle \cap \langle a,c\rangle = \langle c\rangle \cap \langle a,b\rangle$
The first part is easy but is there also an easy way to check if the intersection is trivial? What properties must $a,b,c$ have for that to happen?
My approach was to pick $a=161$, $b=11$, $c=107$. Then $\langle a \rangle \cap \langle b \rangle \cap \langle c \rangle = \{e\}$ and they have the order I want.
Then I have 3 cases to check. The easiest is
$\{1\}=\langle 11\rangle \cap \langle 161,107\rangle \Rightarrow \{1,11,11^2,...,11^{17}\}\cap\{1,161,107,107^2,107^3,161\cdot107,161\cdot107^2,...\}\overset{!}{=}\{1\}$
I know that $11$ is prime and the numbers in the second set must have a prime factorization of $(7\cdot 23)^x\cdot 107^y, x,y\in \mathbb{Z}$. But this doesn't guarantee anything, since the remainder can still be divisible by $11$.
So I'd like to have some advice how I can proceed from here.