Categorical description of algebraic structures

Well, every ring is isomorphic to a subring of the endomorphism ring of some abelian group. See also this MO question.

Besides that, a $R$-module can be understood as a representation of $R$ in the category of abelian groups, in the sense that an $R$-module $M$ is the same as a ring homomorphism $R \to \text{End}(M)$. Similarly, a $K$-vector space is the same as a ring homomorphism $K \to \text{End}(V)$.

I'm not entirely certain that the Yoneda lemma implies Cayley's theorem for groups. Certainly there are analogies to be drawn between the two, but I think at some stage, in order to translate what the Yoneda lemma implies into the language of group theory, you'll end up proving Cayley's theorem in some form.

Let $G$ be a category with one object $*$ and suppose all arrows $* \to *$ are isomorphisms. We note that a contravariant functor $\rho : G^{\text{op}} \to \mathbf{Set}$ is the same thing as a right (not left!) action of $G$ on the set $\rho(*)$. The Yoneda embedding gives us an isomorphic copy of $G$ in the functor category $\mathbf{Set}^{G^{\text{op}}}$. In particular, $*$ is mapped to the functor $\rho = \text{Hom}(-, *) : G^{\text{op}} \to \mathbf{Set}$. Consider the set $X = \rho(*) = \text{Hom}(*, *)$. Then, $X$ is the set of all arrows $* \to *$ in $G$, or in traditional language, the set of all elements of $G$. Each arrow $g : * \to *$ is an isomorphism, so $\rho(g) : X \to X$ must be a bijection, and $\rho(g)(h) = h g$ by definition. What we want to show is that $G$ embeds faithfully into the automorphism group $\text{Sym}(X)$ of $X$.

The Yoneda lemma also tells us there is a bijection between $X = \text{Hom}(*, *)$ and the set of all natural transformations $\rho \to \rho$. However, a natural transformation $\rho \to \rho$ is just a map $f : X \to X$ such that for every arrow $g : * \to *$, $f \circ \rho(g) = \rho(g) \circ f$. This tells us that $G^{\text{op}}$ embeds faithfully into the monoid $\text{End}(X)$, but we need something stronger. Let's see what this does to the identity arrow $e : * \to *$. $(\rho(g) \circ f)(e) = f(e) g$, and $(f \circ \rho(g))(e) = f(g)$, so $f(g) = f(e) g$. So every such natural transformation is determined by $f(e)$, which could be any element of $X$, i.e. arrow of $G$. But every arrow of $G$ is an isomorphism, and this implies $f$ must be a bijection. So now we know $G^{\text{op}}$ embeds faithfully into the group $\text{Sym}(X)$. But $G$ and $G^{\text{op}}$ are isomorphic, so we are done.

Let $C$ be a category, $C'$ the opposite category, and $S$ the category of sets.

Recall that the natural embedding $e$ of $C$ in the category $S^{C'}$ of functors from $C'$ to $S$ is given by the following formulas.

$\bullet$ If $c$ is an object of $C$, then $e(c)$ is the functor $C(\bullet,c)$ which attaches to each object $d$ of $C$ the set $C(d,c)$ of $C$-morphisms from $d$ to $c$.

$\bullet$ If $x:c_1\to c_2$ is in $C(c_1,c_2)$, then $e(x)$ is the map from $C(c_2,d)$ to $C(c_1,d)$ defined by $$ e(x)(y)=yx. $$

In particular, if $C$ has exactly one object $c$, then $e$ is the Cayley isomorphism of the monoid $M:=C(c,c)$ onto the monoid opposite to the monoid of endomorphisms of $M$.

One can also view $e$ as an anti-isomorphism of $M$ onto its monoid of endomorphisms.