2 Extra comment about coboundaries

I don't know if this is a "deep" or "shallow" explanation, but if anyone is still reading this thread, here is a different explanation. I'll start with the preliminary comment that the cohomology of a group $K$ is a special case of cohomology of topological spaces. In topology in general, you get the same phenomenon that $H^k(X,G)$ is well-defined either when $G$ is abelian or $k=1$.

Consider the definition of simplicial cohomology for locally finite simplicial complexes. Or, more generally, CW cohomology for locally finite, regular complexes — regular means mainly that each attaching map is embedded. You can define a $k$-cochain with coefficients in a group $G$ (or even in any set) as a function from the $k$-cells to $G$. In attempting to define the coboundary of a cochain $c$ on a $k+1$-cell $e$, you should multiply together the values of $c$ on the facets of $e$. The obvious problem is that if $G$ is non-abelian, the product is order-dependent. However, if $k=1$, geometry gives you a gift: The facets are cyclically ordered, and what you mainly wanted to know is whether the product is trivial. The criterion of whether a cyclic word is trivial is well-defined in any group, not just abelian groups. A similar but simpler phenomenon occurs for the notion of a coboundary: If $e$ is an oriented edge and $c$ is a 0-cochain, there is a non-abelian version of modifying a 1-cochain by $c$ because the vertices of $e$ are an ordered pair.

So far this is just a more geometric version of Eric Wofsey's answer. It is very close to the fact that $\pi_1$ is non-abelian while higher homotopy groups are abelian . and therefore non-commutative classifying spaces exist only for $K(G,1)$. However, in this version of the explanationmeans that , something extra appears when $X=M$ is a 3-manifold.

If $M$ is a 3-manifold, then not only are the edges of a face cyclically ordered, the faces incident to an edge are also cyclically ordered. It turns out that, at least at the level of computing the cardinality of $H^1(M,G)$, you can let $G$ be both non-commutative and non-cocommutative. In other words, $G$ can be replaced by a finite-dimensional Hopf algebra $H$ which does not need to be commutative or cocommutative. Finiteness is necessary because it is a counting invariant. The resulting invariant $\#(M,H)$ was a topic of my PhD thesis and is explained here and here. Although the motivation is original, the invariant is a special case of more standard quantum invariants defined by other people. (The same construction was also later found by three physicists, but I can't remember their names at all right now.)

Many 3-manifolds are also classifying spaces of groups, so for these groups there is the same notion of noncommutative, non-cocommutative group cohomology.

1

I don't know if this is a "deep" or "shallow" explanation, but if anyone is still reading this thread, here is a different explanation. I'll start with the preliminary comment that the cohomology of a group $K$ is a special case of cohomology of topological spaces. In topology in general, you get the same phenomenon that $H^k(X,G)$ is well-defined either when $G$ is abelian or $k=1$.

Consider the definition of simplicial cohomology for locally finite simplicial complexes. Or, more generally, CW cohomology for locally finite, regular complexes — regular means mainly that each attaching map is embedded. You can define a $k$-cochain with coefficients in a group $G$ (or even in any set) as a function from the $k$-cells to $G$. In attempting to define the coboundary of a cochain $c$ on a $k+1$-cell $e$, you should multiply together the values of $c$ on the facets of $e$. The obvious problem is that if $G$ is non-abelian, the product is order-dependent. However, if $k=1$, geometry gives you a gift: The facets are cyclically ordered, and what you mainly wanted to know is whether the product is trivial. The criterion of whether a cyclic word is trivial is well-defined in any group, not just abelian groups.

So far this is just a more geometric version of Eric Wofsey's answer. It is very close to the fact that $\pi_1$ is non-abelian while higher homotopy groups are abelian. However, this version of the explanation means that something extra appears when $X=M$ is a 3-manifold.

If $M$ is a 3-manifold, then not only are the edges of a face cyclically ordered, the faces incident to an edge are also cyclically ordered. It turns out that, at least at the level of computing the cardinality of $H^1(M,G)$, you can let $G$ be both non-commutative and non-cocommutative. In other words, $G$ can be replaced by a finite-dimensional Hopf algebra $H$ which does not need to be commutative or cocommutative. Finiteness is necessary because it is a counting invariant. The resulting invariant $\#(M,H)$ was a topic of my PhD thesis and is explained here and here. Although the motivation is original, the invariant is a special case of more standard quantum invariants defined by other people. (The same construction was also later found by three physicists, but I can't remember their names at all right now.)

Many 3-manifolds are also classifying spaces of groups, so for these groups there is the same notion of noncommutative, non-cocommutative group cohomology.