Let $G$ be a group with elements: \begin{align*} \alpha_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \alpha_2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \alpha_3 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \alpha_4 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\ \alpha_5 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \alpha_6 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \alpha_7 = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \alpha_8 = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}. \end{align*} with respect to matrix multiplication.
- Find the generator of $G$.
- Find all subgroups of $G$ and its generator.
The identity element of $G$ is $\alpha_1$. I found that $\alpha_2^2=\alpha_3, \alpha_2^3=\alpha_4, \alpha_2^4=\alpha_1$; that is, the order of $\alpha_2$ is $4$. Next, $\alpha_i^2 = \alpha_1$ for $i=5,6,7,8$, i.e., the order of $\alpha_i,i=5,6,7,8,$ are $2$. Next, I found $\alpha_2 \alpha_5 = \alpha_6, \alpha_2^2 \alpha_5=\alpha_7,$ and $\alpha_2^3 \alpha_5 = \alpha_8$. Hence, I found that the generator of $G$ is $\{\alpha_2, \alpha_5\}$, i.e., $G=\langle \alpha_2, \alpha_5 \rangle$.
Now, for 2, I found that the only subgroups of $G$ are $$M_1=\{\alpha_1\},M_2=\{\alpha_1,\alpha_2,\alpha_4\}, M_3= \{\alpha_1,\alpha_2,\alpha_3,\alpha_4\},G.$$ Thus, $M_1=\langle \alpha_1 \rangle, M_2=\langle \alpha_2,\alpha_3 \rangle, M_3=\langle \alpha_2 \rangle,$ and $G=\langle \alpha_2, \alpha_5 \rangle$.
Am I true? If no, how to handle them correctly? Any ideas? Thanks in advanced.