How does this definition capture the intuitive notion of an algebra? On page 15 of this document, the author writes:

Definition 1.1.1. Let $\mathcal{E}$ be any category. Given an endofunctor $\Gamma : \mathcal{E} \rightarrow \mathcal{E}$, a $\Gamma$-algebra consists of a pair $\langle A, \alpha \rangle$ , where $A$ is an object of $E$ and $\alpha : \Gamma A \rightarrow A$ an arrow in $\cal E$. We call $A$ the carrier and $\alpha$ the structure map of the algebra.

I don't get it; how does this definition capture the intuitive notion of an algebra? For a more concrete question, how would an abelian group or a ring be represented under this an approach?
 A: If $T$ is a monad on a category $\mathcal{E}$, then one can define $T$-algebras (I prefer the terminology $T$-modules) and this generalizes all the usual notions of algebraic structures. They are objects $A$ equipped with a morphism $T(A) \to A$ such that the two obvious compatibility conditions are satisfied. For example, if $\mathcal{E}=\mathsf{Set}$ and $G$ is a monoid, then it induces a monad $G \times -$. An algebra for this monad is a set $A$ equipped with a map $G \times A \to A$ written $(g,a) \mapsto ga$ such that $1a=a$ and $g(ha)=(gh)a$. So this is the same as a $G$-set. In general, $T$ contains the "operators" which should "act" on the underlying object of your $T$-algebra. The "action" is precisely the morphism $T(A) \to A$.
Now if $T$ is a plain endofunctor of $\mathcal{E}$, without any monad structure, you can still define $T$-algebras as above but without any compatibility condition. The category of $T$-algebras is just the comma category $T \downarrow \mathrm{id}_\mathcal{E}$. Still $T$ contains the operators which act on the underlying object of a $T$-algebra, but in this setting these operators don't have to satisfy any compatibility condition (in fact we haven't got a chance to formulate any such conditions).
A very basic but nice example is the functor $1 + \mathrm{id} : \mathsf{Set} \to \mathsf{Set}$ which sends a set $X$ to $1 + X$, i.e. $X$ with one additional element adjoined. An algebra for this endofunctor is a set $X$ together with a map $1+X \to X$, i.e. an element $x_0 \in X$ and a map $S : X \to X$. This is nothing else than a Peano structure. The recursion theorem tells us precisely that $(\mathbb{N},0,\mathrm{successor})$ is an initial algebra. In fact, this is how one defines abstract natural number objects. Initial algebras turn out to be very interesting, even for very simple functors. For example you can construct the Banach space $L^1[0,1]$ as the initial algebra for the endofunctor $X \mapsto X \times X$ on the category of pointed Banach spaces.
Finally an exercise which may help to understand this concept: Describe the category of algebras for the endofunctor $\mathrm{Sym}^2 : \mathsf{Set} \to \mathsf{Set}$ which maps $X$ to $\mathrm{Sym}^2(X) = X^2 / ((x,y) = (y,x))_{x,y \in X}$.
