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.

Suppose $x\in (a)\cap (b)$. Then $x=a^k=b^l$ for some non-negative integers $k<\text{ord}(a)$ and $l<\text{ord}(b)$. Then $x^{\text{ord}(a)} = e = b^{\text{ord}(a)l}$, and similarly $x^{\text{ord}(b)} = e = a^{\text{ord}(b)k}$.
Thus $a^{\text{ord}(b)k} = b^{\text{ord}(a)l}=e$, which shows that $\text{ord}(a)|\text{ord}(b)k$, and since $\gcd(\text{ord}(a),\text{ord}(b))=1$, we have $\text{ord}(a)|k$. Since $k<\text{ord}(a)$, we must have $k=0$, but then $x=e$.
