Why require the “identity” condition in the axioms for a group action?

Question

Reading the Wikipedia article on group action, I am wondering, why are the axioms stipulating that a group action obey both “compatibility” and “identity”?

If a group action is merely a group homomorphism from a group $G$ into the group of bijections on a set, that is, $\operatorname{Sym}(X)$, and a group homomorphism automatically sends the identity in the domain group to the identity in the range group, wouldn’t it suffice to merely require the “compatibility” condition and drop the “identity” condition?

What am I missing?

Answer 1

The "homomorphism from a group $G$ into $\operatorname{Sym}(X)$" interpretation is equivalent to these two axioms; you need both axioms to imply the homomorphism interpretation and vice versa. The Wikipedia page that you linked states this pretty clearly:

From these two axioms, it follows that for every $g$ in $G$, the function which maps $x$ in $X$ to $g.x$ is a bijective map from $X$ to $X$ (its inverse being the function which maps $x$ to $g^{−1}.x$). Therefore, one may alternatively define a group action of $G$ on $X$ as a group homomorphism from $G$ into the symmetric group $\operatorname{Sym}(X)$ of all bijections from $X$ to $X$.

Answer 2

Consider the trivial group $E = \{ e \}$. Forgetting about the "identity" axiom, we can let $E$ act on $\mathbb{R}$ by $e \cdot x = 0$ for all $x \in \mathbb{R}$.

This satisfies the "compatibility" axiom, as clearly $ee \cdot x = 0$ always. But it is not an actual group action, and it is not a homomorphism to the group of bijections of $\mathbb{R}$.

Answer 3

A group action is indeed nothing else than a homomorphism into the group of bijection on the set $X$. However, the first axiom does not guarantee that the image of g is actually a bijection. You could fix an element $x_0\in X$ and define $g.x:=x_0$ for every $g\in G$. This satisfies the first axiom, but does not give a bijection of $X$ (except for the trivial case $X=\{x_0\}$).

Together with the second axiom you can prove that $x=e.x=g^{-1}g.x=g^{-1}.(g.x)$, so the map $x\mapsto g.x$ must be a bijection. Its inverse is nothing else than $x\mapsto g^{-1}.x$.