Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 5 / a (The symmetric group)

The function composition of two bijections $f,g:X\to X$ is a bijection, with inverse function $(f\circ g{)}^{-1}=g^{-1}\circ f^{-1}$. Thus function composition is a binary operation on $\mathrm{Sym}(X)$. It is associative, since both $(f\circ g)\circ h$ and $f\circ (g\circ h)$ send $x$ to $f(g(h(x)))$.

The identity function ${\mathrm{id}}_X$ is a bijection and satisfies $f\circ {\mathrm{id}}_X={\mathrm{id}}_X\circ f=f$. Every bijection $f$ has an inverse function $f^{-1}:X\to X$, itself a bijection, satisfying $f\circ f^{-1}=f^{-1}\circ f={\mathrm{id}}_X$. These verify all group axioms, so $\mathrm{Sym}(X)$ is a group, the symmetric group of $X$.

For a singleton $X$, the symmetric group contains only the identity function. For a two-element set, it contains the identity function and the transposition exchanging the two elements; the latter squares to the identity function. In both cases all elements commute, so the group is Abelian.

If $X$ contains three distinct elements $a,b,c$, define the transpositions $\sigma =(ab)$ and $\tau =(bc)$, fixing every other element of $X$. They are bijections, even when $X$ is infinite. With function composition read from right to left,Thus they do not commute, so the symmetric group is not Abelian. Since $X$ is nonempty, this proves

For each $g\in G$, left translation in a group $L_g:x\mapsto gx$ is a bijection, with inverse function $L_{g^{-1}}$. Thus $\theta (g)=L_g$ belongs to $\mathrm{Sym}(G)$. For $g,h,x\in G$, associativity of the group operation givesHence $\theta$ is a group homomorphism. If $\theta (g)=\theta (h)$, evaluation at the identity element $e$ gives $g=ge=he=h$. ThereforeThis is the left regular representation used in Cayley's theorem.