# Relatively prime orders of cyclic groups

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:

1. $\langle a\rangle = \{a^m|m\in \mathbb{Z}\}=G$
2. $\langle b\rangle = \{b^n | n\in \mathbb{Z}\}=G$

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?

• Hint: If $x^m = x^n = e$, then show that $x=e$ (using Bezout's identity) – Prahlad Vaidyanathan Feb 5 '14 at 3:47

## 2 Answers

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 $c=e$.

• Well done, plus one! – Robert Lewis Feb 5 '14 at 3:56

Suppose the order of $a$ and the order of $b$ are relatively prime. The intersection of the two subgroups $\langle a \rangle$ and $\langle b \rangle$ is also a subgroup. By Lagrange's theorem, $\langle a \rangle \cap \langle b \rangle$ has order dividing both $|\langle a \rangle|$ and $|\langle b \rangle|$. Thus $\langle a \rangle \cap \langle b \rangle$ has order 1.