This the proof we were given as to why a group of order 143 is cyclic. 

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> Let $G$ be a group of order 143. By the Sylow Third Thm., the # of Sylow 11-subgroups of $G$ is $1+11k$ and is a factor of 13, so $k=0$ and there is an element $a\in G$ such that $\mathrm{Ord}(a) = 11$ and $\langle a\rangle$ is normal to $G$. Similarly, the number of Sylow 13-subgroups of $G$ is $1+13k$ and is a factor of 11, so $k=0$ and there is an element $b\in G$ such that $\mathrm{Ord}(b)=13$ and $\langle b\rangle$ is normal to $G$. Since $\langle a\rangle \cap\langle b\rangle = \{e\}$ we see that $ab=ba$, it follows that $\mathrm{Ord}(ab) =143$.
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My question is how do we know that $\langle a\rangle $ and $\langle b\rangle$ are normal to G? What allows us to say that $\langle a\rangle \cap \langle b\rangle = \{e\}$?