Exercise 1.7.15 (Action defined by $g\cdot a = ag^{-1}$)

Question

Let $G$ be any group and let $A = G$. Show that the maps defined by $g\cdot a = ag^{-1}$ for all $g,a \in G\ $  {\bf do} satisfy the axioms of a (left) group action of $G$ on itself.

Richard Ganaye · Accepted Answer

\begin{proof} If $g,h,a \in G$, then
\begin{enumerate}
\item[(i)] $g\cdot (h \cdot a) = g\cdot (ah^{-1}) = (ah^{-1}) g^{-1} = a (h^{-1} g^{-1}) = a (gh)^{-1} = (gh) \cdot a$.
\item[(ii)] $e \cdot g = g e^{-1} = g$.
\end{enumerate}
The maps defined by $g\cdot a = ag^{-1}$ for all $g,a \in G$ do satisfy the axioms of a (left) group action of $G$ on itself.

\end{proof}

Exercise 1.7.15 (Action defined by $g\cdot a = ag^{-1}$)

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Exercise 1.7.15 (Action defined by $g\cdot a = ag^{-1}$)

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