Motivation for the relations defining $H^1(G,A)$ for non-commutative cohomology First let me review the definition of first non-commutative cohomology. Let $G$ be a group and $A$ a left $G$-group, i.e. for any $\sigma, \tau\in G$ and $a, b\in A$, one has $\sigma(\tau(a))=(\sigma\tau)(a), \sigma(ab)=\sigma(a)\sigma(b)$. 
A 1-cocycle from $G$ to $A$ is a function $f: G\to A$ such that $f(\sigma\tau)=f(\sigma)\sigma(f(\tau))$ for any $\sigma, \tau\in G$. Two 1-cocycles $f, g$ are equivalent if there exists $c\in A$ such that $f(\sigma)=c^{-1}g(\sigma)\sigma(c)$ for all $\sigma\in G$, this is easily verified to be an equivalence relation. Then the first non-commutative cohomology $H^1(A,G)$ is defined to be the set of 1-cocycle modulo equivalence relation. This is a pointed set with base point given by the trivial cocycle.
My question is: both relations $f(\sigma\tau)=f(\sigma)\sigma(f(\tau))$ and $f(\sigma)=c^{-1}g(\sigma)\sigma(c)$ seem to be unnatural. Does there exists a good motivation for these relations?
 A: Here is one application that might provide some motivation. Given a finite group $G,$ one might consider a representations of $G.$ These are finite dimensional $\mathbb{F}$-vector spaces $V$ on which $G$ acts as linear transformations. But what is a linear transformation? 
You may be thinking it's a matrix. But you'd be wrong. 
A technical point that you may have glossed over when first learning linear algebra is that matrices are representations of linear transformations with respect to a chosen basis. Said in another way, given a basis $\beta$ of $V$ one obtains a homomorphism from $[\cdot]_{\beta}: End(V) \rightarrow M_n(\mathbb{F})$ from the ring of linear transformations of $V$ to $n$ by $n$ matrices over $F.$ And given a different basis $\beta'$ one may obtain a different homomorphism $[\cdot]_{\beta'}.$ Luckily, these homomorphisms are related by conjugation by the change of basis matrix $M_{\beta, \beta'}.$ Now one might like to think of a representation of $G$ as a homomorphism from $G$ to $GL_n(\mathbb{F}),$ but in order to associate such a map to a representation one must first choose a basis and in general there is no best choice. To avoid such a choice one instead associates to the representation the set of all homomorphisms from $G$ to $GL_n(\mathbb{F})$ obtained upon choosing a basis for $V.$ This set of homomorphisms is exactly a $GL_n(\mathbb{F})$ conjugacy orbit in $Hom(G,GL_n(\mathbb{F}))$ i.e. an element of  $H^1(G,GL_n(\mathbb{F}))$  where $G$ acts trivially on $Gl_n(\mathbb{F}).$ On the other hand, to any element of $c \in H^1(G,GL_n(\mathbb{F}))$ we may associate the representation given by any lift of $c$ to $C^1(G,GL_n(\mathbb{F})).$ One checks that these maps are inverses and thus we see there is a bijection 
$$H^1(G,GL_n(\mathbb{F})) \cong \{\text{n-dimensional Linear Representations of G over } \mathbb{F}\}.$$ 
One can broaden the category of representations of $G$ by considering semilinear representations. These are finite dimensional vector spaces $V$ together with an action of $G$ on $V$ and $\mathbb{F},$ such that 


*

*$g(cv)=g(c)g(v)$ for all $c \in \mathbb{F}$ and $v \in V$

*$g(v_1 + v_2)=g(v_1) + g(v_2)$ for all $v_1,v_2 \in V.$

*$G$ acts on $\mathbb{F}$ as field automorphisms 


One can then tell a similar story as in the case of linear representations. Given a semilinear representation $V$ of $G$ one obtains for every basis $\beta$ of $V$ a map $f_{\beta}:G \rightarrow GL_n(\mathbb{F})$ which sends $g$ to $[g]_{\beta}.$ One checks that such a map satisfies $f_{\beta}(g_1g_2) = f_{\beta}(g_1)g_1(f_\beta(g_2))$ and for a second basis $\beta'$ of $V$ the two maps $f_{\beta}$ and $f_{\beta'}$ are related by $M_{\beta,\beta'}^{-1}f_{\beta}(g)g(M_{\beta,\beta'}) = f_{\beta'}.$ Associating to $V$ the set of all such maps we obtain an element of $H^1(G,GL_n(\mathbb{F}))$ where the action of $G$ on $GL_n(\mathbb{F})$ is induced from the action of $G$ on $\mathbb{F}.$ Calling the action of $G$ on $\mathbb{F}$ $\Sigma.$ It follows by an identical argument as above  
$$H^1(G,GL_n(\mathbb{F})) \cong \{\text{n-dimensional Semi-Linear Representations of G over } \mathbb{F} \text{ where G acts on } \mathbb{F} \text{ by } \Sigma\}.$$
A: It is a well-known fact in algebraic topology that for an abelian group $A$ (or more generally, any sheaf of abelian groups $A$) and any topological space $X$, the sheaf cohomology group $H^1 (X, A)$ is in bijection with the set of $A$-torsors (a.k.a. principal $A$-bundles) on $X$. Roughly speaking, a local trivialisation for an $A$-torsor gives a Čech $1$-cocycle for $A$, and the cohomology class of this cochain depends only on the isomorphism class of the $A$-torsor and not the choice of trivialisation. As I understand it, the definition of $H^1$ for non-abelian $A$ is designed so that this theorem remains true.
So how do we bring this idea to group cohomology? Well, for any discrete group $G$ and any $G$-module $A$, the group cohomology $H^* (G, A)$ is naturally isomorphic to topos  cohomology $H^* (\mathbf{B} G, A)$, where $\mathbf{B} G$ is the category of all left $G$-sets. The definition of $A$-torsor makes sense in any topos and for any internal  group $A$, and in the case of $\mathbf{B} G$ it comes down to this: an $A$-torsor is an inhabited left $G$-set $X$ equipped with a $G$-equivariant right $A$-action such that the map $X \times A \to X \times X$ sending $(x, a)$ to $(x, x \odot a)$ is a bijection.
In particular, for any choice of $x$, the map $a \to x \odot a$ is a bijection of sets $A \to X$. But this is not a $G$-equivariant bijection: after all, $g \cdot (x, a) = (g \cdot x, g \cdot a)$, and that gets mapped to $(g \cdot x, (g \cdot x) \odot (g \cdot a))$, which by $G$-equivariance is equal to $g \cdot (x, x \odot a) = (g \cdot x, g \cdot (x \odot a))$; so
$$(g \cdot x) \odot (g \cdot a) = g \cdot (x \odot a)$$
but it is not necessarily true that $x \odot (g \cdot a) \stackrel{?}{=} g \cdot (x \odot a)$. Regardless, we have a set-theoretic bijection, so for each $g$ in $G$ there exists a unique element $f(g)$ of $A$ such that
$$g \cdot x = x \odot f(g)$$
and given another $h$ in $G$, we have
\begin{align}
x \odot f (h g)
& = h g \cdot x \\
& = h \cdot (g \cdot x) \\
& = h \cdot (x \odot f(g)) \\
& = (h \cdot x) \odot (h \cdot f(g)) \\
& = (x \odot f(h)) \odot (h \cdot f(g)) \\
& = x \odot (f(h) (h \cdot f(g)))
\end{align}
which by uniqueness implies
$$f (h g) = f(h) (h \cdot f(g))$$
which lo and behold is the condition defining a $1$-cocycle for $A$. (You could do all this for a left $A$-action on $X$, but that gives a different equation.)
What if we choose a different element of $X$, say $x'$? Again, whatever $x'$ is, there is a unique element $c$ of $A$ such that 
$$x' = x \odot c$$
and so if $f'$ is the $1$-cocycle associated with $x'$, then 
$$g \cdot x' = x' \odot f'(g) = x \odot (c f'(g))$$
and
$$g \cdot x' = g \cdot (x \odot c) = (g \cdot x) \odot (g \cdot c) = x \odot (f(g) (g \cdot c))$$
and therefore
$$f'(g) = c^{-1} f(g) (g \cdot c)$$
which is precisely the definition of equivalence of $1$-cochains.
OK, you say, so every $A$-torsor gives rise to a cohomology class of $1$-cocycles, and obviously this construction depends only on the isomorphism class of the $A$-torsor. What about the other way around? Let $f : G \to A$ be a $1$-cocycle and let $X = A$ with the following $G$-action:
$$g \cdot x = x f(g)$$
This is a $G$-action, because $f(1_G) = 1_A$ and 
$$h g \cdot x = x f(h g) = x f(h) (h \cdot f(g)) = (h \cdot x) (h \cdot f(g)) = h \cdot (x f(g)) = h \cdot (g \cdot x)$$
and obviously $A$ acts on itself $G$-equivariantly because $A$ is a $G$-group. Observe also that $X$ has a canonical element, namely $1_A$, and it is easy to see that the $1$-cocycle that it defines is exactly $f$. On the other hand, if $f$ is the $1$-cocycle corresponding to an $A$-torsor $\tilde{X}$ and an element $\tilde{x}$, then there is a unique $(G, A)$-equivariant isomorphism $X \to \tilde{X}$ sending $1_A$ to $\tilde{x}$. Thus the two constructions are mutually inverse up to isomorphism, and $H^1(G, A)$ really does classify $A$-torsors up to isomorphism. 
