This is, I'm sure, an incredibly naive question, but: is there a simple explanation for why one should be interested in 1-cocycles?
Let me explain a bit. Given an action of a group $G$ on another group $A$ (the group structure of $A$ is respected by the action, in the sense that $\tau(ab)=(\tau a)(\tau b)$ for $a, b\in A$ and $\tau \in G$), a (1-)cocycle is a map $a: G\rightarrow A$ satisfying $a(\sigma\tau)=a(\sigma)\sigma(a(\tau))$. Now these objects are closely connected with group cohomology. My question is the following. Suppose that I didn't know the language of cohomology or homological algebra (which isn't much of a supposition). Is there a simple explanation for why these objects are interesting?
Note: I don't mean to seem dismissive of the larger edifice of group cohomology; even knowing nothing about it, it's clear to me that it is both interesting and incredibly powerful. But the cocycle condition is so simply stated, that it seems there should be a clear picture for what these objects are doing without delving into the more abstract machinery of group cohomology.
Here's on simple interpretation of $H^1(G, A)$ that you can check directly.
If $G$ acts on $A$, then we may construct a semidirect product $G \ltimes A$. Then $H^1(G,A)$ corresponds to the conjugacy classes of complements to $A$ in the semidirect product. (In general, $H^1(G,A)$ only a pointed set - a set with a distinguished element (the identically 1 function); it is not a group unless $A$ is abelian.)
Proof: Any complement must be of the form $\{(g, a(g)) : g \in G\}$ for some $a : G \to A$. Check that the condition for such a function $a$ to be a complement is precisely the cocycle condition, and that if you change a cocycle by a coboundary, that corresponds to conjugation. The "special" cocycle that is identically 1 corresponds to the original complement $G$.