Step in a proof that $G:=\langle x,y \mid x^n,y^2, (xy)^2\rangle$ is isomorphic to the dihedral group $D_n$

Question

I’m reading a proof of the fact that the group given by the presentation $G:=\langle x,y \mid x^n,y^2, (xy)^2\rangle$ is isomorphic to the dihedral group $D_n$. It begins like this:

Let $G=\langle \bar{x},\bar{y}\rangle$ where $\bar{x}$, $\bar{y}$ are the classes
of the elements $x$, $y$ modulo the relations. Then the relations
imply that the subgroup $H:=\langle \bar{x}\rangle$, whose cardinal is $\leq n$, is
normal in $G$.

So far so good. But then I don’t understand the next assertion:

We have $G/H \subseteq \{H, \bar{y}H\}$ then $|G| \leq 2n$.

My group theory might be a bit rusty right now and I don’t see why $G/H \subseteq \{H, \bar{y}H\}$ and how $|G| \leq 2n$ follows from that. Any help would be appreciated.

Answer 1

$G/H=\langle x,y\mid x^n,y^2, (xy)^2,x\rangle=\langle y\mid y^2\rangle=\{H, \bar yH\}$

and

$|G|=|G/H||H|.$