Left multiplication action

Proposition (Left multiplication action). Let $G$ be a group . Define a map $G\times G\to G$ by

Then this defines a group action of $G$ on the underlying set of $G$.

Moreover, this action is:

• transitive (there is one orbit), and
• free (only the identity fixes any element),

hence it is a regular action , often called the left regular action.

Context. This action is the input for Cayley's theorem : it produces an injective homomorphism from $G$ into a symmetric group by viewing elements as permutations of $G$.

Proof sketch. Check the two action axioms: $e\cdot x=ex=x$ and $(g_1g_2)\cdot x=g_1\cdot(g_2\cdot x)$ by associativity. Transitivity: given $x,y\in G$, take $g=yx^{-1}$ so that $g\cdot x=y$. Freeness: if $g\cdot x=x$ then $gx=x$, so multiplying by $x^{-1}$ gives $g=e$.