In Gallian's Contemporary Abstract Algebra, 9th edition, on page 145, he is trying to prove the following: "Let $G$ be a group of order $2p$ , where $p$ is a prime greater than $2$. Then, G is isomorphic to $\mathbb{Z}_{2p}$ or $D_p$". He first proves that G must have an element of order $p$, which he names $a$. The following part of the proof is confusing to me:
Now let $b$ be any element not in $\langle a \rangle$, then by Lagrange's Theorem and the assumption that G does not have an element of order $2p$, we have that $|b|=2$ or p. Because $|\langle a \rangle \cap \langle b \rangle|$ divides $|\langle a \rangle|=p$ and $\langle a \rangle \neq \langle b \rangle$we have that $|\langle a \rangle \cap \langle b \rangle|=1$.
This part I am confused about. How does he conclude that? I can only think that $|\langle a \rangle \cap \langle b \rangle|$ is a subgroup, so its order must divide $2p$.
More generally, if $H$ and $K$ are subgroups of $G$, then $H\cap K$ is a subgroup of $G$. Indeed, let $x,y \in H\cap K$ then $y^{-1} \in H\cap K$ since $y$ is both in $H$ and in $K$ which are subgroups. And thus $xy^{-1} \in H \cap K$ for the same reason.
So in this case, $\langle a \rangle \cap \langle b \rangle$ is a subgroup of $G$, but since it's a subset of $\langle a \rangle$ it's also a subgroup of $\langle a \rangle$, and so its order, which is not $p$, divides $|\langle a \rangle| = p$ which is prime, so it's $1$.