No homework: Task from an old exam (in German): http://docdro.id/GN0ZmvD
$GL_{2}(\mathbb{R})$ is the group of all invertible real $2 \times 2$
matrices and $G:= \left\{ \begin{pmatrix} a & -b\\ b & a
\end{pmatrix} \mid a,b \in \mathbb{R}, a^2+b^2 \neq 0\right\}$Show that $G$ is a subgroup of $GL_{2}(\mathbb{R}).$
I'd like to know how a task like that can be solved correctly, I have no idea 🙁
What also seems strange is the notation $GL_2(\mathbb{R})$, does it have some meaning or is it just the name of the group?
But to $G$ be the subgroup of $GL_2(\mathbb{R})$, I think we firstly need to show that $G$ is a group, and then show it's a subset of $GL_2(\mathbb{R})$. I think it's a subset when it's invertible, so we need to show it is invertible:
It's invertible because its determinant $\neq 0$, thus $G$ is a subset of $GL_2(\mathbb{R})$.
And now I need to show that $G$ is a group. I know I need to show associativity, neutral and inverse element. But how shall I do that with a matrix?
I hope some things I did so far are alright, but I'm very scared of tasks like that in an exam and I hope you can help me understanding them better.
Best Answer
Since we already know that $GL_2(\mathbb{R})$ is a group, we only need to verify that for $A,B\in G$ also $AB$ is in $G$, and $A^{-1}$ is in $G$. This follows easily. For example, $$ \begin{pmatrix} a & -b\\ b & a \end{pmatrix} ^{-1} = \frac{1}{a^2+b^2} \begin{pmatrix} a & b\\ -b & a \end{pmatrix}. $$ We do not need to show associativity, because $A(BC)=(AB)C$ for matrices anyway. Of course, matrices of $G$ have determinant $a^2+b^2\neq 0$, so that $G\subseteq GL_2(\mathbb{R})$. The identity matrix $I_2$ is also in $G$.