Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 9

Assume (I). Existence of a decomposition is part of the hypothesis. If $h_1{h}_2=k_1{k}_2$ are two decompositions, multiply on the left by $k_1^{-1}$ and on the right by $h_2^{-1}$ to obtainThe left side belongs to $H_1$ and the right side to $H_2$, since they are subgroups. Their common value belongs to $H_1\cap H_2=\{e\}$, so $h_1=k_1$ and $h_2=k_2$. This proves uniqueness, hence (II).

Conversely, assume (II). Existence immediately gives $G=H_1{H}_2$. If $a\in H_1\cap H_2$, thenare two allowed factorizations, because the identity is in each subgroup. Uniqueness makes their first components equal, so $a=e$. Thus $H_1\cap H_2=\{e\}$ and (I) follows.

Therefore (I) and (II) are equivalent. This is an exact factorization of a group. For finite groups it gives $|G|=|H_1||H_2|$, but it does not by itself identify $G$ with a direct product of groups; multiplication between the two factors can be noncommutative.

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.