Prove $H$ is a normal subgroup of $K$ $G$ is a set.
$$G=\left\{\begin{vmatrix}1 &amp; 0 \\ 0  &amp; 1\end{vmatrix}, \begin{vmatrix}0 &amp; 1 \\-1 &amp; 0\end{vmatrix}, \begin{vmatrix}-1 &amp; 0 \\ 0  &amp; -1\end{vmatrix}, \begin{vmatrix}0 &amp; -1 \\ 1  &amp; 0\end{vmatrix}, \begin{vmatrix}i &amp; 0 \\ 0  &amp; i\end{vmatrix}, \begin{vmatrix}0 &amp; i \\ -i  &amp; 0\end{vmatrix}, \begin{vmatrix}-i &amp; 0 \\ 0  &amp; i\end{vmatrix}, \begin{vmatrix}0 &amp; -i \\ i  &amp; 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.
If
$$H=\left\{\begin{vmatrix}1 &amp; 0 \\ 0  &amp; 1\end{vmatrix}, \begin{vmatrix}-1 &amp; 0 \\ 0  &amp; -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 &amp; 0 \\ 0  &amp; 1\end{vmatrix}, \begin{vmatrix}-1 &amp; 0 \\ 0  &amp; -1\end{vmatrix}, \begin{vmatrix}i &amp; 0 \\ 0  &amp; i\end{vmatrix}, \begin{vmatrix}-i &amp; 0 \\ 0  &amp; -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.
My questions are :
a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.
b. What is every element of factor group $G/H?$
as for $H$ in $K$ you could notice that $H$ is of index 2 in K so it is normal.
Longer method than Yotam's: conjugate each subgroup by each element of $K$ and $G$ and show that they are invariant. You'll have to do $4*2 + 8*4 = 40$ different conjugations, but this isn't so large as to make the computation unfeasible to do by hand.