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Abstract Algebra

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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.

A

The map $g\mapsto L_g$ defines a group embedding of $G$ into $S_n$.

B

The map $g\mapsto L_g$ defines a group embedding of $G^{\rm\,op}$ into $S_n$.

C

The map $g\mapsto R_g$ defines a group embedding of $G$ into $S_n$.

D

The map $g\mapsto R_g$ defines a group embedding of $G^{\rm\,op}$ into $S_n$.

E

$\{L_g\mid g\in G\}$ equals the set $S_n$ of permutations of $G$.

F

$\{R_g\mid g\in G\}$ equals the set $S_n$ of permutations of $G$.