 Solution

The two diagonal generators are independent involutions, soIt is a Klein four-group. Write the swapping generator as $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Then $J^2=I$, so $H_2=\{I,J\}$ has order two. Moreover,Thus $J$ normalizes $H_1$. The setis a group: its products reduce usingand inverses reduce in the same way. It contains both generating subgroups, so minimality gives $G\subseteq K$; conversely every displayed product is in $G$, giving $G=K$.

There are four diagonal matrices $D$ and four off-diagonal matrices $DJ$, all distinct. ConsequentlyAlso $H_1\cap H_2=\{I\}$, since $J$ is not diagonal, and every element of $G$ has the form $D{J}^{\eta }\in H_1{H}_2$. This proves condition (I), hence the exact factorization of a group. Here $H_1$ is a normal subgroup and $G=H_1⋊H_2$ is a semidirect product. Its matrices are the two-dimensional signed permutation matrices, also realizing the dihedral group of the square.