Past exam of the groups and representations course of the University of Oxford / 2019 / 1

A subgroup $N\le G$ is a normal subgroup if $gN{g}^{-1}=N$ for every $g\in G$. A group homomorphism $f:G\to \tilde{G}$ satisfies $f(gh)=f(g)f(h)$ for every $g,h\in G$. In particular $f(e)=\tilde{e}$ and $f(g^{-1})=f(g{)}^{-1}$. A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$ into the general linear group of invertible linear maps on $V$.

For the kernel of a group homomorphism, put $K=\ker f=\{g:f(g)=\tilde{e}\}$. It contains $e$, and if $k,\ell \in K$ then $f(k{\ell }^{-1})=f(k)f(\ell {)}^{-1}=\tilde{e}$, so $K$ is a subgroup. For $g\in G$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Applying this inclusion with $g^{-1}$ gives the reverse inclusion, proving that $\ker f$ is a normal subgroup.

For a normal subgroup $N$, define multiplication of left cosets byTo check independence of representatives, replace $g,h$ by $gn,hm$, with $n,m\in N$. Thenbecause normality gives $h^{-1}nh\in N$. Hence the operation is well-defined. Associativity follows from $(gh)k=g(hk)$ in $G$; the identity is $N=eN$; and the inverse of $gN$ is $g^{-1}N$. These verify the group axioms, giving the quotient group $G/N$. The map $g\mapsto gN$ is a surjective group homomorphism with kernel of a group homomorphism equal to $N$.

Nonzero scalar matrices form a subgroup $N$ of the general linear group, since products and inverses remain nonzero scalar matrices. Moreover, for every invertible $g$,Thus $N$ lies in the center of a group and is a normal subgroup. The quotient group is the complex projective linear group ${\mathrm{PGL}}_n(\mathbb{C})$, whose elements identify invertible matrices differing by a nonzero scalar.

The canonical group homomorphism $\pi :g\mapsto [g]$ obeys $[g][h]=[gh]$. Composing it with the given group representation givesConsequentlyThis is the projective representation associated with $R$: a nonzero scalar acts trivially on each line in the projective space of $V$. Its kernel of a group homomorphism consists of those $g$ for which $R(g)$ is scalar, which can be larger than the kernel of a group homomorphism of $R$ itself.

Write $A=g_1$, $B=g_2$ and $C=AB$. Direct multiplication givesThe relation $BA=-AB$ moves all occurrences of $A$ to the left of all occurrences of $B$. Removing even powers reduces every product of the generators to $\pm A^a{B}^b$ with $a,b\in \{0,1\}$. The resulting eight matrices are distinct, soExplicitly, the four pairs areSince $AB=-BA$ and $AB\ne 0$, the group is noncommutative. More precisely, $C$ has order four, $A$ has order two and $ACA=C^{-1}$, identifying $G$ with the dihedral group of a square. This fixes the convention unambiguously: it is the dihedral group of group order eight, irrespective of whether it is denoted $D_4$ or $D_8$.

Use the matrices of solution. Their projective classes satisfy $[-M]=[M]$, since $-I_2\in N$. The group homomorphism is therefore explicitlyThese four classes are distinct: the two diagonal representatives are not scalar multiples of one another, the two off-diagonal representatives are not scalar multiples of one another, and a diagonal representative cannot be a nonzero scalar multiple of an off-diagonal one.

Although $A$ and $B$ anticommute as matrices, their projective classes commute:Furthermore $[A{]}^2=[B{]}^2=[I_2]$ and $[C{]}^2=[-I_2]=[I_2]$. Thus every nonidentity element of the image has order two. An explicit group isomorphism to the Klein four-group isIndeed reordering products introduces only a sign, which disappears in the projective linear group, so this map turns multiplication into addition modulo two in each coordinate. Equivalently, $\ker \tilde{R}=G\cap N=\{\pm I_2\}$ and the first isomorphism theorem givesThis projective image of the square symmetry group illustrates how a noncommutative group can have an Abelian projective image: its scalar commutator becomes the identity class.