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.

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

  • $\begingroup$ Note: the elements of $H_{n}$ are the matrices whose determinants are $n^{th}$ roots of unity. $\endgroup$ – John McGee Aug 20 '14 at 14:24
  • $\begingroup$ @JohnMcGee Yes, the $n$th roots of unity are the kernel of $z\mapsto z^n$. $\endgroup$ – 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, $H_{n} \leq G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.