Let $a,b\in G$ for a group $G$ with $|a| = m$ and $|b| = n$. Prove that if $(m, n)=1$, then $\langle a\rangle\cap\langle b\rangle = \{e\}$.

Question

Let $a$ and $b$ be elements of a group $G$ with $|a| = m$ and $|b| = n$. Prove that if $m$ and $n$ are relatively prime, then $\langle a\rangle\cap\langle b\rangle = \{e\}$.

Answer 1

Hint:

$$x\in\langle a\rangle \cap\langle b\rangle\implies ord(x)\mid ord(a)\,,\,ord(b)\;\ldots $$

Answer 2

More generally, you could try to prove that if $H,K$ are subgroups of $G$ then so is $H \cap K$, and if $H$ is a subgroup of $G$, then the order of $H$ divides that of $G$.