Does the group $G$ have an element of order $6$? Is there any easy way to check is there any element of order $6$ in a group $(G,\cdot),$ where $$G=\left\{A_{x,y}=
\begin{bmatrix}
 x+7y & 5y \\
 -10y & x-7y \\
\end{bmatrix},
\in M_2(\mathbb{R}) \ \middle|\ x^2+y^2\neq0\right\}$$
or do I have to find $A_{x,y}^6$ and solve the equation $A_{x,y}^6=I$?
 A: Your group is the set of the matrices of the form $xI + yA$ (subject to the constraint $x^2 + y^2 \neq 0$, where $I$ denotes the identity matrix and
$$
A = \pmatrix{7&5\\ -10&-7}.
$$
Note that $A^2 = -I$. With that, we can deduce that the map $\phi:(G,\cdot) \to (\Bbb C^*,\cdot)$ given by
$$
\phi(xI + yA) = x + yi
$$
is an isomorphism. Thus, it suffices to solve the equation $z^6 = 1$ for $z \in \Bbb C^*$.  In particular, with De Moivre's Theorem, we can see that taking $x = \cos(2 \pi/6)$ and $y = \sin(2 \pi /6)$ gives us an element of order $6$.
A: The intuition:
Note that $\det A_{x, y} = x^2 + y^2$. For $A_{x, y}^6 = I$, we must have $x^2 + y^2 = 1$. This suggests us the substitution $(x, y) = (\sin \theta, \cos \theta)$.

For $\theta \in \Bbb R$, define $A_\theta$ as
$$A_\theta := A_{\sin\theta, \cos\theta}.$$
A simple calculation shows us that
$$A_\theta\cdot A_\varphi = A_{\theta + \varphi}.$$
Thus, we see that $$(A_\theta)^n = A_{n\theta}$$ for any $n \in \Bbb Z$.
It can be seen that $\theta = \pi/3$ then does the job. That is,
$$A_{\pi/3}^6 = I$$
and $$A_{\pi/3}^n \neq I$$
for any $1 \le n < 6$.
