Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 8D (Groups)

Identify $A$ with $\{(a,1):a\in A\}$. Componentwise use of the stated multiplication shows that this set contains the identity and is closed under products and inverses, so it is a subgroup. It has the two cosetsand therefore has index two. Every index-two subgroup is normal, so

For $ϵ=-1$,so every element outside $A$ has order two.

The element $(e,-1)$ commutes with $(a,1)$ exactly whenor $a^2=e$. Since $A$ itself is abelian, $D(A)$ is abelian exactly when every $a\in A$ satisfies $a^2=e$.

Suppose $D(A)$ is nonabelian. An element $(a,1)$ is central exactly when $a=a^{-1}$, equivalently $a^2=e$. No element $(a,-1)$ is central, because commuting with every $(b,1)$ would require $b^2=e$ for every $b$, which would make the whole group abelian. Hence

If $Q\in O(2)$, then $Q^{T}Q=I$, soand $\det Q=\pm 1$. If $\det Q=1$, its first column is $(\cos \theta ,\sin \theta {)}^T$ for some $\theta$, and orthonormality plus positive orientation forceThus every element of $SO(2)$ is a rotation.

Let $J=\mathrm{diag}(1,-1)$. If $\det Q=-1$, then $QJ\in SO(2)$ and $Q=(QJ)J\in SO(2)J$. Conversely every matrix in $SO(2)J$ has determinant $-1$. Therefore the union is disjoint and

The standard isomorphisms areand

For $z\in B$, let $R_z$ be the corresponding rotation. The identityshows thatrespects the multiplication of the generalized dihedral group. The decomposition in part (b) makes it bijective. Hence