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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 $\langle x,y \mid x^n=y^2, x^{2n}=1, y^{-1}xy=x^{-1}\rangle.$

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    $\begingroup$ It is easy to prove that the matrices X and Y satisfy the conditions of the group with a given presentation, which will call $ G_{2n} $. Furthermore, we can use the Substitution Test for to ensure the existence of a surjective homomorphism between $ G_{2n} $ and $ Q_{2n} $. This proves that the order of $ G_{2n} $ is greater than or equal to $4n$. But I can not show that the order must be less than or equal to $4n$, which conclude the demonstration exercise.Any help please.Thanks. $\endgroup$
    – user59969
    Commented Feb 16, 2013 at 11:35
  • $\begingroup$ Consider the exact sequence:$1\to \langle X\rangle \to G_{2n} \to \langle Y\rangle \to 1$. This sequence clearly splits, because $G$ is a semi-direct product of the two subgroups. Hence the order is as claimed. I am not sure if this is right, though. $\endgroup$
    – awllower
    Commented Feb 16, 2013 at 11:40
  • $\begingroup$ @awllower there is no such exact sequence: $\langle X\rangle$ is a normal subgroup of order $2n$ while $\langle Y \rangle$ has order $4$, not $2$. The subgroups generated by $X$ and by $Y$ have a nontrivial intersection. $\endgroup$
    – KCd
    Commented Jun 15, 2022 at 12:12

2 Answers 2

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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 $j \in \{0, 1\}$.

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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.

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  • $\begingroup$ Babak (and another Theorem) to the rescue! + 1 $\endgroup$
    – amWhy
    Commented Feb 16, 2013 at 13:00

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