Proper symmetry group/Finite/Class/Group homomorphism/Fact/Proof

Proof

Due to the definition of a class of semiaxes, for ${}H\in K$ we also have ${}g(H)\in K$ for all ${}g\in G$. Therefore, the mapping ${}\sigma$ is well-defined. Let ${}f,g\in G$. Then we have

${}\sigma _{g\circ f}(H)=(g\circ f)(H)=g(f(H))=g(\sigma _{f}(H))=\sigma _{g}(\sigma _{f}(H))=(\sigma _{g}\circ \sigma _{f})(H)\,.$

To fact