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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.


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