## Conjugation is a group action

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.)

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.