Exercise 1.7.16 (Action by conjugation)

Question

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

Richard Ganaye · Accepted Answer

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

Exercise 1.7.16 (Action by conjugation)

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Exercise 1.7.16 (Action by conjugation)

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