Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.1
Let $F$ be a field. Show that the multiplicative group of nonzero elements of $F$, denoted $F^\times$, acts on the set $F$ by $g \cdot a = ga$ – that is, by left multiplication.
Solution: Let $a \in F$. We have $1 \cdot a = 1a = a$ by the definition of identity. Now let $g_1, g_2 \in F^\times$. Then $$g_1 \cdot (g_2 \cdot a) = g_1 \cdot g_2 a = g_1(g_2a) = (g_1g_2)a = (g_1g_2) \cdot a,$$ by the associativity of multiplication in $F$. So left multiplication by nonzero elements is in fact a group action.