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Let $G=GL(2,\mathbb{R})$ and $$H=\Bigg\{{\begin{bmatrix} \cos{\theta} & \sin{\theta}\\ -\sin{\theta} & \cos{\theta} \end{bmatrix}} : \theta \in \mathbb{R}\Bigg\}$$

Prove that $H$ is a subgroup of $G$.

So far I have that the identity element is:\begin{bmatrix} {1} & 0\\ 0 & 1 \end{bmatrix} by multiplying H by \begin{bmatrix} \frac {\cos{2\theta}+1}{2\cos\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta} \end{bmatrix}

but I'm unsure how to show $H \neq \emptyset$ and if $\alpha, \beta \in H$, then $\alpha \beta^{-1} \in H$. (which I'm assuming is the correct direction into proving $H$ is a subgroup of $G$)

Any advice/hints is greatly appreciated.

Chinnapparaj R
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Reety
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    I got the identity as $\pmatrix{1&0\0&1}$. – Angina Seng Nov 02 '18 at 07:34
  • The identity element can not depend on $\theta$. Look for a matrix $E$ such that for every $A\in G, AE=EA=A$. – Yuval Gat Nov 02 '18 at 07:39
  • To prove the subset $H$ being a subgroup of $G$, you want to show that the identity of $G$ is in $H$, so that $H$ is nonempty. – edm Nov 02 '18 at 07:50
  • @LordSharktheUnknown sorry meant to write what I just edited. – Reety Nov 02 '18 at 08:16
  • @edm so all i need to prove is the identity of H is in G which will conclude that H is a subgroup of G? – Reety Nov 02 '18 at 08:20
  • No, of course not. It is not the case that any old subset containing the identity is a subgroup. But surely somewhere you have some notes telling you what needs to be done to prove something is a subgroup? – Gerry Myerson Nov 02 '18 at 08:33
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    I've read so many different things from notes/online I think I'm just confusing myself - from my notes I have the following; to prove that H is a subgroup of G we must show that H is associative, has an identity, has an inverse for every element and is closed under the operation. – Reety Nov 02 '18 at 08:38

1 Answers1

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I haven't used matrices for a while in latex and apologize for any bad formatting. Hopefully someone kind enough can fix it. Also, I am assuming the operation defined on $G$, and thus $H$ is the usual multiplication of matrices.

Note that $H\subset G$ is a subgroup of $G$ $\iff$ for each pair $x,y\in H$, we have $xy^{-1}\in H$

Take $x,y\in H$

I.e. for $\theta_1$, $\theta_2\in \mathbb{R}$, $x$ and $y$ are of the form:

$$x=\begin{bmatrix} \cos(\theta_1) & \sin(\theta_1) \\ -\sin(\theta_1) & \cos(\theta_1) \\ \end{bmatrix} , \ y=\begin{bmatrix} \cos(\theta_2) & \sin(\theta_2) \\ -\sin(\theta_2) & \cos(\theta_2) \\ \end{bmatrix}$$

Noting $\det(y)=\cos^2(\theta_2)+\sin^2(\theta_2)=1$, we see $$y^{-1}=\frac {1}{\det(y)}\begin{bmatrix} \cos(\theta_2) & -\sin(\theta_2) \\ \sin(\theta_2) & \cos(\theta_2) \\ \end{bmatrix}=\begin{bmatrix} \cos(\theta_2) & -\sin(\theta_2) \\ \sin(\theta_2) & \cos(\theta_2) \\ \end{bmatrix}$$

So,

$$xy^{-1}=\begin{bmatrix} \cos(\theta_1) & \sin(\theta_1) \\ -\sin(\theta_1) & \cos(\theta_1) \\ \end{bmatrix}\cdot\begin{bmatrix} \cos(\theta_2) & -\sin(\theta_2) \\ \sin(\theta_2) & \cos(\theta_2) \\ \end{bmatrix}$$

$$=\begin{bmatrix} \cos(\theta_1)\cos(\theta_2)+\sin(\theta_1)\sin(\theta_2) & -\cos(\theta_1)\sin(\theta_2)+\sin(\theta_1)cos(\theta_2) \\ -\sin(\theta_1)\cos(\theta_2)+\cos(\theta_1)\sin(\theta_2) & \sin(\theta_1)\sin(\theta_2)+\cos(\theta_1)\cos(\theta_2) \\ \end{bmatrix}$$

$$=\begin{bmatrix} \cos(\theta_1-\theta_2) & \sin(\theta_1-\theta_2) \\ -\sin(\theta_1-\theta_2) & \cos(\theta_1-\theta_2) \\ \end{bmatrix}$$

Here I used the angle sum formulas.

Clearly $xy^{-1}\in H$ since $\theta_1-\theta_2\in \mathbb{R}$

Thus $H\subset G$ is a subgroup of $G$.

Chinnapparaj R
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Mark Pineau
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