**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.16**

Let $G$ be a group. Show that the mapping defined by $g \cdot a = gag^{-1}$ does satisfy the axioms of a left group action of $G$ on itself. (Called *conjugation*.)

Solution: We have $1 \cdot a = 1a1^{-1} = a$. If $g_1, g_2 \in G$, then $$g_1 \cdot (g_2 \cdot a) = g_1 \cdot g_2ag_2^{-1} = g_1g_2ag_2^{-1}g_1^{-1} = (g_1g_2)a(g_1g_2)^{-1} = (g_1g_2) \cdot a.$$ Thus this mapping is a group action.