Cayley’s theorem

• $G$,
  • group
• $\mathrm{Sym}(G)$,
  • symetric group

\[\mathrm{Sym}(G) := \{ f: G \rightarrow G \mid f: \text{bijection} \} .\] \[g \in G, \quad l_{g}: G \rightarrow G, \quad l_{g}(x) := g x \quad (x \in G) .\]

Obviously, $l_{g} \in \mathrm{Sym}(G)$.

\[K: G \rightarrow \mathrm{Sym}(G), \quad K(g) := l_{g} .\]

Then for each $G$, there is a subgroup $A \subseteq \mathrm{Sym}(G)$ and ismorhism $f: G \rightarrow A$.

proof

TBA

$\Box$

• $\mathrm{Grp}$,
  • group category
• $K: \mathrm{Grp} \rightarrow \mathrm{Set}$,
  • functor

\[\begin{eqnarray} KG & := & \mathrm{Sym}(G) \nonumber \\ f: G \rightarrow G^{\prime} \in \mathrm{Grp}, \quad Kf: \mathrm{Sym}(G) \rightarrow \mathrm{Sym}(G^{\prime}), \quad f^{\prime}: G \rightarrow G, \ (Kf)(f^{\prime}) & := & f \circ f^{\prime} \nonumber . \end{eqnarray}\]

Reference

• Cayley’s theorem - Wikipedia