Is the axiom $g1 = g$ essential for a group action A group $G$ acts on a set $\Omega$ if 
(1) $\omega\cdot 1_G = \omega$
(2) $(\omega \cdot g)\cdot h = \omega \cdot (gh)$
for all $\omega \in \Omega$ and $g,h \in G$. But is (1) really essential, what if we drop (1) and just require (2), then we must have
$$
 \omega\cdot 1_G = \omega \cdot (gg^{-1}) = (\omega\cdot g)\cdot g^{-1}
$$
for all $g \in G$.

So do you know any example where (2) holds, but (1) fails?

 A: Let $\Omega = \{1,2\}$, take $G$ to be any group and set $x \cdot g = 1$ for all $x \in \Omega$ and $g \in G$. Then of course $(x \cdot g) \cdot h = 1 \cdot h = 1 = x \cdot (gh)$, but $2 \cdot 1_G = 1 \neq 2$ .
A: The point is that axiom (2) alone tells you that the map
$$
\varphi : g \mapsto (\omega \mapsto \omega \cdot g)
$$
is a homomorphism of $G$ into the semigroup $M(\Omega)$ of maps on $\Omega$, i.e. $\varphi(g h) = \varphi(g) \circ \varphi(h)$. (Thanks to Derek Holt for correcting my previous mistake, I had written monoid instead of semigroup, of course a morphism of monoids takes $1$ to $1$.)
To get that $\varphi$ maps $G$ into the group $S(\Omega)$ of the permutations (bijective maps, invertible maps) on $\Omega$, you only need to require a single $\varphi(g_{0})$ to be bijective. Since this implies $\varphi(1) = 1$ (the latter being the identity of $S(\Omega)$), you might as well assume the latter, and this is axiom (1).

Proof of the claim. Suppose there is $g_{0} \in G$ such that $\varphi(g_{0})$ is invertible. Then axiom (2) implies that for each $g \in G$ we have
$$
\varphi(g) \circ \varphi(g^{-1} g_{0}) = \varphi(g_{0}),
\qquad
\varphi( g_{0} g^{-1}) \circ \varphi(g) = \varphi(g_{0}),
$$
so that $\varphi(g)$ is also invertible.
