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?

  • 2
    $\begingroup$ How do you define $Aut(X)$? As the set of permutations maybe? Then you better write $S(X)$. Condition 2. on its own does not guarantee that the $\phi(g)$ are bijective, i.e. elements of $S(X)$. $\endgroup$ – drhab Aug 10 '18 at 11:16
  • $\begingroup$ You are correct in your argument, if you assume that your group actions are going to be automorphisms of $X$. However, the definition of group action is used in more general cases, mapping elements of the group to functions of $G$ that may not necessarily cancel. $\endgroup$ – user583012 Aug 10 '18 at 11:16
  • $\begingroup$ @drhab indeed, I think I understand now that that is what I was getting wrong. I implicitly assumed the image of $\phi$ to be the set of bijections on $X$ (this is what I meant with $\operatorname{Aut}(X)$, but without the identity axiom this is not necessarily true $\endgroup$ – glS Aug 10 '18 at 11:34

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.

  • $\begingroup$ ah, I think I'm getting it now. To be clear: with $\operatorname{End}(X)$ you mean the set of all maps from $X$ to $X$, right? Am I correct in saying that the gist of the problem is that, without the identity axiom, it is not ensured that $g.x\neq g.y$ for $x\neq y$? $\endgroup$ – glS Aug 10 '18 at 11:19
  • 1
    $\begingroup$ Yes, $\operatorname{End}(X)$ is the set of all maps from $X$ to $X$. And yes : if you add the assumption that $g\cdot x=g\cdot y\Rightarrow x=y$, that is, every map $\phi(g)$ is injective, then the definition become equivalent. $\endgroup$ – Arnaud D. Aug 10 '18 at 11:43

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 $e.x = e.x$.

  • $\begingroup$ this is true (I actually realised that argument was flawed but didn't have time to fix the question). However, I still don't get what is the fault in the rest of the argument. In particuar, why isn't the second axiom enough for $\phi$ to be a group homomorphism? $\endgroup$ – glS Aug 10 '18 at 11:09
  • $\begingroup$ Because you haven't proved that the image is in Aut(X). $\endgroup$ – Kenny Lau Aug 10 '18 at 11:10

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).


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


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