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Consider the subgroup $G$ of $GL_2(\mathbb C)$ generated by $$a=\begin{bmatrix}i&0\\0&-i\end{bmatrix},\ b=\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$ Find its order and prove that every subgroup of $G$ is normal in $G$.

My guess is that $G=\{1,a,a^2,a^3,b,b^3,ab,a^3b\}$. I checked that these elements are distinct and that these relation hold in $G$: $$b^2=a^2,\ ba=a^3b,\ a^4=b^4=1$$ (maybe some other hold too). Do I need to write out 32 expressions $a^ib^j, b^ia^j$ with $0\le i,j \le 3$ and check that no other elements except those listed above can be obtained? Also, assuming that $G$ has order 8, can I conclude that $G$ is isomorphic to $Q_8$ since there is no relation $a^2=b^2$ in the dihedral group of order 8 (and the group in question is not abelian)?

For the second part, it's a big hassle to list its subgroups and check directly whether they are normal. Is there a quick way to see that all subgroups are normal?

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This is the quaternion group. Clearly, $a$ has order $4$. Also $b^2=a^2 =-I_2$ so $a^2=b^2$ is in the centre of the group. You get $bab^{-1}=a^{-1}$ so that $\left<a\right>$ is normal. But in the quotient $G/\left<a\right>$, the image of $a$ is trivial and that of $b$ has order two, so $|G/\left<a\right>|=2$. Therefore $|G|=8$.

Subgroups of index two are always normal, so subgroups of orders $1$, $4$ and $8$ are normal. One only has to consider subgroups of order $2$, in effect elements of order $2$. The element $a^2$ is central and of order $2$, so that $\left<a^2\right>$ is normal. Are there any other elements of order $2$.

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