Proving $H$ is a subgroup of $GL(2,\Bbb{R})$ 
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.
 A: 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 $G$.
