TITLE: Why are ring actions much harder to find than group actions?
QUESTION [19 upvotes]: I admit freely that the following question is a bit of a fishing expedition inspired by this lovely "definition" of a module as found on Wikipedia:

A module is a ring action on an abelian group.

It takes a while for the novice to unzip this definition into the usual list of identities, but for someone with a basic understanding of (group) actions who needs a quick shorthand to remember what being a module entails, it seems hard to beat this definition.
Now, volumes have been written about group actions. This makes sense, because set theory underlies modern mathematics and for any set $X$, the automorphisms $\operatorname{Aut}(X)$ naturally have a group structure. And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \operatorname{Aut}X$. 
On the other hand — aside from this definition of a module — it is hard to come across a general theory of ring actions. In order to find interesting actions of a ring $R$, one analogously needs the set of endomorphisms $\operatorname{End}X$ to have the structure of a ring so that one may search for ring morphisms $R \to \operatorname{End}X$.

What are the most general objects $O$ for which $\operatorname{End}O$ canonically has the structure of a ring?

Certainly, $O$ at least contains abelian groups, but is there anything else?

REPLY [7 votes]: $\newcommand{\Hom}{\operatorname{Hom}}$An example for a class of rings where the ring actions are very close to group actions are Hopf algebras. As a consequence many theorems from representation theory of groups generalize to the representation theory of Hopf algebras. 
In order to give a feeling on how definitions from group actions generalize to Hopf algebras consider a group $G$ and $kG$-Modules $M,N$ ($k$ a commutative ring). Then we have the invariants $M^G := \lbrace m \in M\mid \forall g \in G:\;gm=m\rbrace$ and $G$ also acts on $\Hom_k(M,N)$ via
$(g\cdot f)(m) := gf(g^{-1}m)$. Then the $G$-linear maps can be recovered as invariants by 
$\Hom_{kG}(M,N) = \Hom_k(M,N)^G$.
Now let $H$ be a Hopf algebra over $k$ (corresponds to $kG$) with antipode $S: H \to H$ (corresponds to $g \mapsto g^{-1}$), augmentation $\epsilon: H \to k$ (corresponds to $g \mapsto 1$) and coproduct $\Delta: H \to H \otimes_k H$ (corresponds to $g \mapsto g \otimes g$). Then one defines for $H$-modules $M,N$: $$M^H := \lbrace m \in M\mid \forall h \in H: \;hm=\epsilon(h)m\rbrace$$
and if $\Delta(h)=\sum_i h_i^{(1)} \otimes h_i^{(2)}$ then $H$ acts on $\Hom_k(M,N)$ by 
$$(h\cdot f)(m):= \sum_i h_i^{(1)}f\big(S(h_i^{(2)})m\big).$$
Now, as expected, the identity $\Hom_H(M,N)=\Hom_k(M,N)^H$ holds again.