Is $G/\langle x\rangle \cap \langle x \rangle=e$ given the following subgroups have order 11 and 7

Question

I was trying to prove a question in group theory and if I assumed that $G/\langle x\rangle \cap \langle x \rangle=e$ it will solve my question. I believe this is true because I know that the order is as follows, $|G/\langle x\rangle|=11$ and $|\langle x \rangle|=7$. This means that I have two cyclic groups because both are prime and it then follows that their intersection can only be the identity element. But, the identity element of $G/\langle x\rangle$ is $\langle x \rangle$ so I am unsure if I am making a big mistake.

Answer 1

Actually $G / \langle x \rangle \cap \langle x \rangle = \emptyset$ since $G / \langle x \rangle$ has elements that are cosets and $\langle x \rangle$ contains elements of $G$.