# Show $H_n$ is a subgroup of $G$ for any $n\in\mathbb N$

Let $G = GL_2(\mathbb C)$ be the group of invertible complex $2 \times 2$ matrices and for each $n\in\mathbb N$ consider the subset:

$$H_n = \{\,A \in G : (\det A)^n= 1\,\}.$$

I know to prove if it is a subgroup of $G$ is by verifying:

1. $H$ is closed under the operation in $G$ and

2. Every element in $H$ has and inverse in $H$.

However, I am a bit confused how to prove this specific case, any help please.

• Is $A\in G$ a matrix? Is $\det$ the usual determinant of matrices? (sorry for poor understanding) – The Great Seo Aug 20 '14 at 14:17
• @TheGreatSeo Sorry, just edited. – user164945 Aug 20 '14 at 14:21

$\det\colon G\to\mathbb C^\times$ is a group homomorphism, and so is $\mathbb C^\times\to\mathbb C^\times$, $z\mapsto z^n$. Your $H_n$ is the kernel of the combination of these two, hence is a subgroup (and in fact normal).
• Note: the elements of $H_{n}$ are the matrices whose determinants are $n^{th}$ roots of unity. – John McGee Aug 20 '14 at 14:24
• @JohnMcGee Yes, the $n$th roots of unity are the kernel of $z\mapsto z^n$. – Hagen von Eitzen Aug 20 '14 at 14:53
$H_{n} \neq \emptyset$ because $\det (I_{2}) = 1$.
Choose $A, B \in H_{n}$. We look at the element $AB^{-1} \in G$. Since for matrices $X, Y$, we have $\det(XY) = \det(X)\det(Y)$ and $\det(X^{-1}) = \frac{1}{\det(X)}$, it follows that $\left(\det(AB^{-1})\right)^{n} = \frac{ \left(\det(A)\right)^{n}}{\left(\det(B)\right)^{n}} = 1 \implies AB^{-1} \in H_{N}$. Therefore, $H_{n} \leq G$.