In group cohomology, the 2-cocycle condition emerges from associativity (see e.g. here). From the answer to a previous question we see (at least for normalized cocycles) that the cocycle condition ensures invertibility too. But how do we know a priori, that associativity will ensure invertibility too?
Edit
We are talking about the algebraic structure $\Gamma = (A\times G,\bullet)$ where $A$ is an Abelian group, $G$ is a group, and $\bullet$ is the following binary operation:
$$(a_1,g_1)\bullet(a_2,g_2) = (a_1+\varphi_{g_1}(a_2)+f(g_1,g_2),\,g_1g_2)\tag{1}$$
where
- $\varphi:(A\times G)\to A: (a,g)\mapsto \varphi_g(a)$ is a group action of $G$ on $A$
- $f: G\times G\to A$
We can derive the
$$ f(g_1,g_2g_3)+\varphi_{g_1}(f(g_2,g_3)) = f(g_1g_2,g_3) + f(g_1,g_2)\tag{8}$$
cocycle condition purely from the requirement of associativity of $\bullet$. What is interesting, that this condition ensures automatically also
- The existence of an identity element in $\Gamma$, that is, if the cocycle condition holds, then $\Gamma$ will be not only a semigroup, but also a monoid.
- The existence of a left inverse and a right inverse for every $\gamma\in \Gamma$
(better to say, 2. depends on the cocycle condition only through 1. If an identity element is already given, then we don't need the cocycle condition any more for 2.)
Associativity, 1. and 2. together means that $\Gamma$ is a group, since in every monoid, left inverses are equal to the right inverses:
$$x\bullet \gamma = e = \gamma\bullet y \implies x=y$$ due to the associativity:
$$y=e\bullet y=(x\bullet\gamma)\bullet y=x\bullet(\gamma\bullet y) = x\bullet e = x$$
