The definition of a group action, as given on wikipedia, is the following:
Let $G$ be a group and $X$ a set. A (left) group action of $G$ on $X$ is a function $$G\times X\ni(g,x)\mapsto g.x\in X$$ such that
- $\forall x\in X,\, e.x=x$
- $\forall( x\in X, \,g,h\in G),\,\, (gh).x=g.(h.x)$
What I do not understand is why do we have to assume $e.x=x$.
Indeed, given a group homomorphism $\varphi:X\to Y$, the identity is always preserved: $\varphi(e_X)=e_Y$.
But the second axiom above seems to be enough for the mapping $\phi:G\to\operatorname{Aut}(X)$, defined as $$\phi(g)(x)\equiv g.x,$$ to be a group homomorphism: $$\phi(gh)(x)=(gh).x=g.(h.x)=\phi(g)(\phi(h)(x))\equiv[\phi(g)\circ\phi(h)](x).$$ If $\phi$ is a group homomorphism, then it should be ensured that $\phi(e_X)=e_{\operatorname{Aut}(X)}$, so that $$\forall x\in X,\,\, \phi(e)(x)=e.x=x.$$
A more direct way to see this is to consider that $\forall g\in G,\,\,x\in X$, $$g.x=(ge).x=g.(e.x),$$ and multiplying on the left by $g^{-1}$ we should get $x=e.x$.
In conclusion, am I missing something in the reasoning above, or is the identity axiom in the definition redundant?