Exercise 1.7.14 ($g\cdot a = ag$ does not define a group action)

Question

Let $G$ be a group and let $A = G$. Show that if $G$ is non-abelian then the maps defined by $g\cdot a = ag$ for all $g,a \in G$ do {\bf not} satisfy the axioms of a (left) group action of $G$ on itself.

Richard Ganaye · Accepted Answer

\begin{proof} 
Assume for the sake of contradiction that this action satisfies the two axioms of a left group action of $G$ on itself.

Since $G$ is non-abelian, there are two elements $g, h \in G$ such that $gh 
e hg$. Put $e = 1_G$. Then
$$hg = (hg)\cdot e = h\cdot(g\cdot e)= h \cdot (eg) = (eg)h = gh,$$
so $hg = gh$. This is a contradiction.

Therefore, if $G$ is non-abelian, then the maps defined by $g\cdot a = ag$ for all $g,a \in G$ do not satisfy the axioms of a (left) group action of $G$ on itself.

\end{proof}

Exercise 1.7.14 ($g\cdot a = ag$ does not define a group action)

Answers

Comments

Exercise 1.7.14 ($g\cdot a = ag$ does not define a group action)

Answers

Comments

Add answer