Past exam of the topology and groups course of the University of Oxford / 2010 / 2 / Solution

The normal closure $⟨⟨B⟩{⟩}_G$ is the smallest normal subgroup of $G$ containing $B$, equivalently the intersection of all normal subgroups containing $B$. Any such normal subgroup contains every $g{b}^{\pm 1}{g}^{-1}$ and hence every finite product of these elements. Thus $N\subseteq ⟨⟨B⟩{⟩}_G$.

Include the empty product so that $1\in N$, in particular when $B=\varnothing$. Products of elements of $N$ are obtained by concatenating their factor lists. Inversion reverses the list and replaces ${ϵ}_i$ by $-{ϵ}_i$, so $N$ is closed under inverses. For any $k\in G$,Therefore $N$ is a normal subgroup and it contains $B$. Minimality of the normal closure gives the reverse inclusion, provingThe notation $n\in \mathbb{N}$ in a finite-product description is understood to allow the empty product; otherwise the empty generating subset would require a separate identity convention.

In a group presentation, $X$ is a set of formal generators and $R$ is a subset of the free group $F(X)$, consisting of relators which are required to become the identity. The presented group isThe universal property of a group presentation says that a function $X\to H$ extends to a group homomorphism from the presented group precisely when every relator evaluates to $1$ in $H$: first extend freely to $F(X)$, then factor through the quotient because its kernel contains the normal closure of $R$.

A rotation $\alpha$ by $2\pi /n$ and a reflection $\beta$ of a regular $n$-gon satisfyThe last identity holds because reflection reverses the sense of rotation, and it implies $\alpha \beta \alpha \beta =1$. Hence all defining relators vanish, giving a homomorphism $\theta :D_{2n}\to M_n$ with $\theta (\sigma )=\alpha$ and $\theta (\tau )=\beta$.

To prove it is an isomorphism without presupposing the order of the presented group, use $\sigma \tau =\tau {\sigma }^{-1}$. Move every occurrence of $\tau$ to the left, cancel pairs by ${\tau }^2=1$, and reduce exponents modulo $n$. Every element has one of the formsThus $|D_{2n}|\le 2n$. For the usual regular polygon with $n\ge 3$, the $n$ rotations ${\alpha }^j$ are distinct and the $n$ reflections $\beta {\alpha }^j$ are distinct; no rotation is a reflection, since their planar orientation behavior differs. The supplied description says these exhaust $M_n$, so $|M_n|=2n$, and every one lies in the image of $\theta$. Surjectivity and the order bound force $|D_{2n}|=2n$ and trivial kernel. Thus $\theta$ is an isomorphism.

For the requested Cayley graph, use right multiplication by the generators. The $\sigma$ edges are the two directed 3-cyclesThe $\tau$ edges are the undirected pairsThese pairs use ${\sigma }^j\tau =\tau {\sigma }^{-j}$. Thus the underlying graph is a triangular prism, with the two rotation cycles oriented oppositely when reflection pairs are drawn as radial edges.