Is the identity axiom in the definition of group action redundant?

Question

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

1. $\forall x\in X,\, e.x=x$
2. $\forall( x\in X, \,g,h\in G),\,\, (gh).x=g.(h.x)$

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?

Answer 1

In your "more direct way", the problem is that you can't simplify $g$ without knowing that $e\cdot x=x$ in the first place.

In fact, if your set $X$ is non-empty, then you can always choose some $x_0\in X$ and define an "action" as $g \cdot x=x_0$ for all $x\in X$ and $g\in G$. This satisfies the second condition, but not the first one.

But you are right that any group homomorphism preserve the identity. The problem here is that in general, a map $G\times X\to X$ only gives you a function from $G$ to the monoid $\operatorname{End}(X)$ of maps $X\to X$, and the second condition only tells you that this map preserve the products of these monoids. But for monoids, it is not true that any map that preserve the product also preserve the identity, so you need to ask for that to hold as well, i.e. you need to add the first condition.

Answer 2

If you multiply $g.x = g.(e.x)$ on the left by $g^{-1}$, then you get $g^{-1}.g.x = g^{-1}.g.(e.x)$, which you can simplify to $e.x = e.(e.x)$, but it carries no more information than $e.x = e.x$.

Answer 3

Fix some $y\in X$ and define $g.x=y$ for every $g\in G$ and every $x\in X$.

Then $(gh).x=y=g.(h.x)$ for $g,h\in G$ and $x\in X$ but not $e.x=x$ for every $x\in X$ (unless $X$ is a singleton).

Answer 4

Without the identity axiom we would get that $e.x=0$ for every $x \in \mathbb{R}$ is an action of {$e$} on $\mathbb{R}$.