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## Right multiplication by the inverse is a left group action of a group on itself

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.15

Let $G$ be any group. Show that the mapping defined by $g \cdot a = ag^{-1}$ does satisfy the axioms of a left group action of $G$ on itself.

Solution: We have $1 \cdot a = a1^{-1} = a$, and if $g_1, g_2 \in G$, we have $$g_1 \cdot (g_2 \cdot a) = g_1 \cdot ag_2^{-1} = ag_2^{-1}g_1^{-1}=a(g_1g_2)^{-1} = (g_1g_2) \cdot a.$$ Thus the mapping is a group action.