If $a$ and $b$ are elements of a group whose orders are relatively prime, what can you say about $\langle a\rangle\cap \langle b\rangle$?
Let the order of $a$ be $m$ and the order of $b$ be $n$. Since $m$ and $n$ are relatively prime, we know that $\gcd(m,n)=Id$.
I know:
- $\langle a\rangle = \{a^m|m\in \mathbb{Z}\}=G$
- $\langle b\rangle = \{b^n | n\in \mathbb{Z}\}=G$
And I'm pretty sure that $\langle a\rangle\cap \langle b\rangle = Id.$, but I'm not sure how to show this.
Any suggestions?
Best Answer
If $c\in\langle a\rangle\cap\langle b\rangle$ then the order $|c|$ of $c$ complies: $$|c|\quad \mbox{divides both $m$ and $n$}.$$
But $m,n$ are relatively prime, then $|c|=1$, so $c=e$.