Easy# Cayley's Theorem: Group Acting on Itself and Permutation Groups

ABSALG-QCHJUC

Let $G$ be a **finite** group of order $|G|=n$, with neutral element $e$ and composition law $\ast$.

Let $G^{\rm \,op}$, called the opposite to $(G, \ast)$, be the group whose set of elements equals the set $G$, but whose group composition law $\bar{\ast}$ is $h\bar{\ast}g:= g\ast h$.

We identify $S_n$, the symmetric group on $n$ symbols, with the group of permutations of the **set** $G$, and of the **set** $G^{\rm\,op}$, since $G$ and $G^{\rm\,op}$ agree as sets.

If $G$ is abelian, then $(G,\ast)=(G^{\rm \,op},\bar{\ast})$, so we assume in this question that $G$ is **not necessarily** abelian.

For $g\in G$, let $L_g:G\rightarrow G$ be the map $L_g(h)=g\ast h$ and let $R_g:G\rightarrow G$ be the map $R_g(h)=h\ast g$.

Which of the following are always true?

Select **ALL** that apply.