How to understand the object in a category Am a real beginner in category theory... I read that a category can be defined as a $4$-tuple:
$$
\textbf{C}=(\text{Ob}(\textbf{C}),\text{Ar}(\textbf{C}),\mu(\textbf{C}),\text{id}(\textbf{C}))
$$
where $\text{Ob}(\textbf{C})$ is any collection of elements, $\text{Ar}(\textbf{C})$ is a family of sets $\textbf{C}(X,Y)$ indexed by pairs of elements of $\text{Ob}(\textbf{C})$, and $\text{id}(\textbf{C})$ are in $\textbf{C}(X,X)$ that satisfy the identity laws ...
But am still trying to understand $\text{Ob}(\textbf{C})$ ... I read that given a concrete group $G$, with the category of $G$ as $G_{cat}$, we have that $\text{Ob}(G_{cat})$ has only one member (which can be indexed as $\{0\}$) ... However, this member does not consist of $G$, but rather is a fictitious mathematical object on which $G$ acts...
The description above sounds very nebulous ... Does this mean that the single object ($\{0\}$) can simply be any set such that the $G$ can act on it ? In this case, can an infinite set $S$ isomorphic to $\mathbb{N}$ be considered also as "a specific case of" $(\{0\})$?
 A: What the single object is need not matter much. It is just some mathematical object... literally anything.
Where the group enters the picture is that it is encoded as the morphisms, and the group multiplication is encoded as the composition operation on the morphisms. So, for instance if we're trying to represent the group $(\mathbb Z, 0,+)$ as a category, the category will have a single object $*$ and will have morphisms $$\{f_n:*\to *\mid n\in \mathbb Z\}.$$ And then $f_0$ is the identity morphism, and for the composition operation, we have $f_n\circ f_m = f_{n+m}.$ There is nothing special about $\mathbb Z$ here. For a general group $(G,e,\cdot),$ we have morphisms $\{f_g: g\in G\},$ and $f_e=id_*$ and $f_{g^{-1}}\circ f_g=f_g \circ f_{g^{-1}}=id_*,$ and $f_{g_1}\circ f_{g_2}=f_{g_1\cdot g_2}.$
So as with many things in category theory, it's really all about the morphisms. The object $*$ is only there as a formality so there is a source and a target for these morphisms. It's not doing much of anything.
That said, there's nothing wrong about imagining that $*$ is some set that the group $G$ acts on (e.g. the group itself acted on by left-multiplication, or some set the group represents permuations of), but it's not strictly necessary to envision it that way. This is really just the (immaterial) distinction between groups in the abstract and groups as symmetry transformations, only in the perhaps unfamiliar territory where the group operation is being understood as composition of morphisms in a category with one object.
A: Forget categories for a moment, and think about some more familiar kind of algebraic structures — e.g. groups themselves.
Take, for instance, the cyclic group $C_5$ (just as a group, not as a category).  It has 5 elements.  What are those 5 elements?  Depending on what construction you use, they might be certain rotations of $\newcommand{\R}{\mathbb{R}}\newcommand{\C}{\mathbb{C}}\R^2$, or certain roots of unity in $\C$, or perhaps equivalence classes of integers modulo 5.  But it doesn’t really matter what the elements are individually — what matters is how many of them there are, and what the group operation does on them.  Precisely, I’ve described three different groups, but they’re all canonically isomorphic to each other — so we think of them as three different constructions of essentially the same group $C_5$, and when we say $C_5$, it doesn’t matter which specific construction we have in mind.  This is what’s abstract in abstract algebra: it doesn’t matter what the elements of the group are, what matters is the structure they form.
Now, back to categories.  The key point is: A category itself is an abstract algebraic structure.  It’s not important what the individual elements of $\mathrm{ob}(\mathbf{C})$ are; what matters is the structure they form, up to isomorphism.*
In particular, viewing a group as a one-object category: it doesn’t matter what the “one object” of the category is.  $\mathrm{ob}(G_{\mathrm{cat}})$ can be any one-element set.  If you choose $\mathrm{ob}(G_{\mathrm{cat}}) = \{G\}$, and I choose $\mathrm{ob}(G_{\mathrm{cat}}) = \{ \emptyset \}$, and someone else chooses $\mathrm{ob}(G_{\mathrm{cat}}) = \{ 5i + 8 \}$, it doesn’t matter: the categories we get will all be isomorphic!  (With categories, we are often even a bit looser than “isomorphism” — we consider categories essentially the same as long as they’re equivalent, and we expect/require all categorical constructions to respect equivalence of categories.  But isomorphism is certainly strong enough to justify considering two categories as the same.)
A: The key idea to view a group as a category with one object is to treat the elements in the set of morphisms as elements in the group. I think some examples will help you understand it better.
If you consider the symmetry group $S_n$, then you can take the object as the set $\{1,2,\dots,n\}$ and you can define the morphisms are the bijection from $\{1,2,\dots,n\}$ to itself.
If you consider the dihedral group, then the object can be a regular n-gon and the morphisms are rotations and reflections of the regular n-gon. I hope this helps you.
