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.

  • 2
    $\begingroup$ Just a comment on the name $GL_2 (\mathbb{R})$ stands for 'General Linear 2x2 matrices over $\mathbb{R}$'. That is, they are all the invertible 2x2 matrices with entries in $\mathbb{R}$. Edit: the 'linear' is because the columns/rows are linearly independent because it is invertible $\endgroup$ – TheMathsGeek Mar 15 '17 at 11:53

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 $G$.

  • $\begingroup$ Thank you for answer :) Can you say why we need to show that $AB$ is in $G$? $\endgroup$ – cnmesr Mar 15 '17 at 12:37
  • $\begingroup$ Because that is one of the two subgroup axioms: a nonempty subset $S$ of a group $G$ is a subgroup if (S1) with $a,b\in S$ also $ab\in S$, and $(S2)$ with $a\in S$ also $a^{-1}\in S$. $\endgroup$ – Dietrich Burde Mar 15 '17 at 12:40
  • $\begingroup$ $G$ is a subgroup of $GL_2(\mathbb{R})$ if 1. $G$ is a group 2. $G$ is a subset of $GL_2(\mathbb{R})$ Or did I understand something wrong and these aren't all axioms? What are the axioms you refer to? $\endgroup$ – cnmesr Mar 15 '17 at 12:47
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    $\begingroup$ Well, you are right. This can be proved from what you say, and then its called subgroup criterion. Anyway, you do not need to verify all group axioms, since $GL(2)$ is already a group. It is enough to verify $(S1)$ and $(S2)$. $\endgroup$ – Dietrich Burde Mar 15 '17 at 12:49

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