Let $a=\begin{pmatrix} \xi & 0\\ 0 & \overline{\xi} \end{pmatrix}$ and $b=\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$ elements of $GL_2(\mathbb{C})$ where $\xi=e^{2\pi i /n}\in \mathbb{C}$.
Prove that $D_{2n}=\left\{1,a,a^2,…,a^{n-1},b,ab,a^2b,…,a^{n-1}b\right\}$ is a subgroup of $GL_2(\mathbb{C})$.
It's clear that $1\in D_{2n}$ is the neutral element and the operation is closed in $D_{2n}$.
About the inverses elements, I have computed that $1^{-1}=I_2=1, (a^i)^{-1}=a^{n-i}, b^{-1}=b$ that are in $D_{2n}$ where $i=1,…,n-1$. But I have problems trying to find the inverse of the elements of the form $a^i b$ where $i=1,…,n-1$.
I think that $(a^i b)^{-1}=ba^{n-i}$. However $ba^{n-i}\notin D_{2n}$. Hence $D_{2n}$ is not a subgroup?
Best Answer
It is a subgroup.
I will use the one-step subgroup test.
Since, indeed, $1\in D_{2n}$, we have $D_{2n}\neq\varnothing$.
Since $a,b\in GL_2(\Bbb C)$, we have $D_{2n}\subseteq GL_2(\Bbb C)$.
Observe that
$$ba^{-1}=ab, b^2=1, a^n=1.\tag{1}$$
Consider arbitrary $a^jb^e, a^kb^f\in D_{2n}$.
Then we have
$$\begin{align} a^jb^e(a^kb^f)^{-1}&=a^jb^e(b^{-f}a^{-k})\quad\text{(as }D_{2n}\subseteq GL_2(\Bbb C)\text{)}\\ &=a^j(b^{e-f}a^{-k})\\ &=a^j(a^kb^{e-f})\\ &=a^{j+k}b^{e-f}, \end{align}$$
where we can take $j+k$ modulo $n$ and $e-f$ modulo $2$ by $(1)$. But then $a^jb^e(a^kb^f)^{-1}=a^{j+k}b^{e-f}\in D_{2n}$.
Hence $D_{2n}\le GL_2(\Bbb C)$.