Are subgroups kernels of some $1$-cocycles? If an $1$-cocycle $\sigma :G\to  K$ has image in a group on which $G$ acts, then the set of element of $G$ that is mapped to $\text{$e_K$}$ by $\sigma$ forms a subgroup of $G$. This could be proved by the cocycle relation.
Since, in the case of a normal subgroup, the cocycle in question can even be a homomorphism, we are thus led to ask:  

Let $H\le G$ be a subgroup of $G$. Does there exist an action of $G$ on a group $K$, such that $H$ is exactly the kernel of an $1$-cocycle with respect to that action?  

If so, then we can introduce cocycles under this setting, and say affirmatively that $1$-cocycles are generalisations of homomorphisms, and that subgroups are generalisations of normal subgroups.
I have tried to find a similar map to the natural map $G\to G/H$ in the case of a normal subgroup, but to no avail.
Any hint or solution is well-appreciated. Thanks in advance.  
 A: Short answer: Yes. Every subgroup $H$ is the kernel of a principal crossed homomorphism from $G$ to the permutation module $k[G/H] = k \otimes_{kH} kG$.
Basic definitions
Definition: Let $G$ act on the group $K$, that is, $f:G \to \operatorname{Aut}(K)$ abbreviated as $k^g := k^{f(g)}$ so that $(kl)^g = k^g l^g$ and $k^{gh} = (k^g)^h$ for $k,l \in K$ and $g,h \in G$. Let $\delta:G \to K$. Then $\delta$ is a crossed homomorphism if and only if $\delta(gh) = \delta(g)^h \delta(h)$.
Proposition: If the kernel $\ker(\delta)$ of a crossed homomorphism $\delta$ is defined as $\ker(\delta) = \{ g : \delta(g) = 1_K \}$, then the kernel is a subgroup of $G$ and $\delta(g) = \delta(h)$ if and only if $\ker(\delta)g = \ker(\delta) h$.
Proof: These are all just calculations. The kernel is non-empty:  $\delta(1_G) = \delta(1_G \cdot 1_G) = \delta(1_G)^{1_G} \delta(1_G)$ so $\delta(1_G)^{1_G} = 1_K$ and $\delta(1_G) = 1_K$. The kernel is closed under products: If $\delta(g) = \delta(h) = 1_K$ then $\delta(gh) = \delta(g)^h \delta(h) = 1_K^h \cdot 1_K = 1_K$. The kernel is closed under inverses: If $gh=1_G$ then $1_K = \delta(g)^h \delta(h)$ so $\delta(g^{-1})^g = \delta(g)^{-1}$. If $g = xh$ for some $x \in \ker(\delta)$, then $\delta(g) = \delta(x)^h \delta(h) = 1_K^h \delta(h) = 1_K \delta(h) = \delta(h)$. Conversely if $\delta(g) = \delta(h)$ then $\delta(gh^{-1}) = \delta(g)^{h^{-1}} \delta(h^{-1}) = \delta(g)^{h^{-1}} \delta(h)^{-h^{-1}} = 1_K$. $\square$
Proposition: If a principal crossed homomorphism $\delta_k$ corresponding to $k \in K$ is defined by $\delta_k(g) = (k^{-1})^g k$, then $\delta_k$ is in fact a crossed homomorphism. Its kernel is called the centralizer or stabilizer of $k$, $\ker(\delta_k) = C_G(k) = \{ g : k = k^g \}$.
Proof: Just a calculation: $\delta_k(gh) = (k^{-1})^{gh} k$ while $\delta_k(g)^h \delta_k(h) = (k^{-1})^{gh} k^h (k^{-1})^h k$ and the middle terms
cancel to give equality. The kernel is all $g$ such that $(k^{-1})^g k = 1_K$, that is, all $g$ such that $k = k^g$. $\square$
Every subgroup is a kernel
Let $k$ be a commutative, associative, unital ring like $\mathbb{Z}/2\mathbb{Z}$. Let $V$ be a free $k$-module with basis the set of cosets $G/H$, that is, $V = k \otimes_{kH} kG$. $V$ is a called a permutation module and $G$ acts on the basis via $x \cdot e_{yH} = e_{xyH}$ for $x,y \in G$ and $e_{yH}$ in the basis $\{ e_yH : y \in G\}$. Notice that $C_G( e_H ) = \{ g \in G: g \cdot e_H =e_H \} = \{ g : e_{gH} = e_H \} = \{ g : gH = H \} = \{ g \in G : g \in H\} = H$.
Define $\delta : G\to V$ as $\delta = \delta_{e_H}$ with kernel $\ker(\delta_{e_H}) = C_G(e_H) = H$.
Other standard examples
Proposition: If $K$ is a normal subgroup of $G$, then $\delta_k(g) = [g,k]$ is the standard commutator, and its kernel is the standard centralizer.
The standard 1-cocycle (crossed homomorphism) and 1-coboundary (principal crossed homomorphism) setup is from the following:
Proposition: If $G= H \ltimes K$, that is, $G$ has a normal subgroup $K$ and subgroup $H$ such that $G=HK$ and $H \cap K=1$, then $\delta(hk) = k$ defines a crossed homomorphism from $G$ to $K$ with kernel $H$. The conjugacy classes of complements to $K$ are in bijection with the set of crossed homomorphisms modulo multiplication by principal crossed homomorphisms.
So already, two important types of subgroups are kernels of crossed homomorphisms: centralizers and complements.
Relation to homomorphisms
Crossed homomorphisms are related to homomorphisms in a fairly simple way:
Proposition: Let $X = G \ltimes K$ be the semidirect product of $G$ and $K$ and set $\phi(g) = (g,\delta(g))$. Then $\phi:G \to G \ltimes K$ is a homomorphism iff $\delta:G \to K$ is a crossed homomorphism. The kernel of $\delta$ is the preimage of $G$ under $\phi$.
Proof: Indeed, $\phi(gh) = (gh,\delta(gh)) = (gh,\delta(g)^h \delta(h))$ and $\phi(g) \phi(h) = (g,\delta(g)) (h,\delta(h)) = (gh, \delta(g)^h \delta(h))$.
Indeed, there are 3 ingredients to an automorphism of an extension that fixes the kernel $K$: an automorphism of $G$, an automorphism of $K$, and a crossed homomorphism from $G$ to $K$.
