Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 9

A subgroup $N\le G$ is a normal subgroup when conjugation by every group element preserves it:Equivalently, $gN=Ng$ for every $g$; normality permits the cosets to inherit a quotient-group multiplication.

A group homomorphism is a map preserving multiplication:It consequently sends identity to identity and inverses to inverses, since $\varphi (e)=\varphi (e{)}^2$ and $\varphi (g)\varphi (g^{-1})=\varphi (e)$.

The image of a group homomorphism contains $e_{G_2}=\varphi (e_{G_1})$. If $u=\varphi (g)$ and $v=\varphi (h)$ lie in the image, then $u{v}^{-1}=\varphi (g{h}^{-1})$ also lies there. The subgroup test proves $\mathrm{im}\varphi \le G_2$.

The kernel of a group homomorphism is $K=\{g:\varphi (g)=e_{G_2}\}$. It is nonempty and closed under $k{h}^{-1}$ for $k,h\in K$, so it is a subgroup. For any $g\in G_1$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Replacing $g$ by $g^{-1}$ gives the reverse inclusion, provingFinally, if $K=\{e\}$ and $\varphi (g)=\varphi (h)$, then $\varphi (h^{-1}g)=e$. Thus $h^{-1}g\in K$ forces $g=h$, establishing injectivity. Conversely an injective homomorphism has trivial kernel, soNone of these arguments requires either group to be finite.

Label the triangle vertices $1,2,3$ counterclockwise. Let $R$ rotate by $2\pi /3$ counterclockwise and let $m$ reflect across the line through vertex 1 and the midpoint of the opposite side. The dihedral group has elements $I,R,R^2,m,Rm,R^{2}m$, with $R^3=m^2=I$ and $mRm=R^{-1}$.

The action on vertices identifies them with these permutations, with the rightmost operation applied first:The minimal number of generators is two, and $\{R,m\}$ is one minimal generating set. They produce every displayed element. A one-element generating set would make the group cyclic, but every nonidentity element has order two or three, never six.

The vertex-action map preserves composition, hence is a group homomorphism from $D_3$ to the symmetric group $S_3={\Sigma }_3$. The displayed actions are all six distinct permutations of three vertices, so the map is both injective and surjective. ThereforeThis identification uses the convention $|D_3|=6$, with the subscript counting polygon vertices.