Generalized Quaternion Group

Question

Let $w = e^{\Large\frac{i\pi}{n}} \in \mathbb{C}.$ Prove that the matrices $X=\left( \begin{array}{cc} w & 0 \\ 0 & \overline{w} \\ \end{array} \right)$ and $Y = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$ generate a subgroup $Q_{2n}$ of order $4n$ in $\operatorname{GL}(2, \mathbb{C})$, with presentation $\langle x,y \mid x^n=y^2, x^{2n}=1, y^{-1}xy=x^{-1}\rangle.$

Answer 1

Hint: Consider the group which is presented by $$G=\langle X,Y\mid X^n=Y^2,X^{2n}=1,Y^{-1}XY=X^{-1}\rangle$$ and then use the von Dyck's Theorem.

Answer 2

Consider the group $G = \langle x,y \mid x^n=y^2, x^{2n}=1, y^{-1}xy=x^{-1}\rangle$. As noted by @AgenorAndrade, this group has order at least $4 n$, as it has your group as a homomorphic image.

To show that it as at most $4n$ elements, just note that every element of $G$ can be written as $x^{i} y^{j}$, for $0 \le i < 2n$, and $j \in \{0, 1\}$. In fact, in an arbitrary product of $x$ and $y$ and their inverses, the third relation $$ y^{-1} x = x^{-1} y^{-1}, \quad\text{or}\quad y x = x^{-1} y $$ allows you to move all $y$ to the right, so you end up with $$ x^i y^j, $$ and then the second relation gives you $0 \le i < 2n$, and the first one tells you you just need to take $j \in \{0, 1\}$.