For $a, b$ in a group and gcd(order of $a$, order of $b$) = 1, prove $(a)\cap (b) =e$ Let $G$ be a group. For $a, b\in G$ and $\gcd(\text{order of }a,\text{order of }b) = 1$, prove that $\{e\} = (a)\cap (b)$, where $e$ is the neutral element, and $(g)$ is the cyclic group generated by $g$.
So, we need to show $\{e\}⊆(a)\cap(b)$ and $(a)\cap(b)\subseteq\{e\}$.
The first part is trivial, and I am wondering how to do the second part.

Let $x\in (a)\cap (b)$.
By Lagrange's theorem, $\text{ord}(x)$ divides both $\text{ord}(a)$ and $\text{ord}(b)$ and so must divide their gcd, which is $1$.
Thus, $\text{ord}(x)=1$ and $x=e$.
