If you find any mistakes, please make a comment! Thank you.

Demonstrate that right multiplication is not a left group action of a group on itself


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.14

Let $G$ be a nonabelian group. Show that the action of $G$ on itself by $g \cdot a = ag$ does not satisfy the axioms of a left group action.


Solution: Since $G$ is nonabelian, there exist $g_1, g_2 \in G$ such that $g_1 g_2 \neq g_2 g_1$. Then we have $$g_1 \cdot (g_2 \cdot 1) = g_1 \cdot 1g_2 = (1g_2)g_1 = 1(g_2g_1) \neq 1(g_1g_2) = (g_1g_2) \cdot 1.$$ Thus this mapping fails to be a group action.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu