So first of all the group algebra $K[G]$ has, in addition to its algebra structure, a coproduct,
$$
\Delta: k[G] \rightarrow k[G] \otimes k[G] \\
g \mapsto g \otimes g
$$
an antipode,
$$
S: k[G] \rightarrow k[G]\\
g \mapsto g^{-1}
$$
and a counit $\epsilon:k[G] \rightarrow k$ sending $g \mapsto 1$ for all $g \in G$. It is these maps that give $k[G]$ the structure of a Hopf algebgra.
As far as structure theorems go, there are several. These are particularly striking in the case of the cohomology of a H-space $X$ over a field $k$ of characteristic 0, when Hopf proved that the Hopf algebra $H^\bullet(X; k)$ is:
- An exterior algebra generated by homogeneous elements of odd degree if $H^\bullet(X;k)$ is finite dimensional
- A free graded-commutative algebra if each $H^n(x;k)$ is finite dimensional.
I include these results as they provide both a motivation for what a structure theorem for Hopf algebras looks like, and some historical context, as these objects were the motivation for the definition of a Hopf algebra. As far as more general structure theorems for Hopf algebras go there are some nice results of Cartier, Gabriel and Milnor-Moore. Here are two theorems:
For the first theorem note that $U(\mathfrak{g})$ denotes the universal enveloping algebra. I remark here that $k[G]$ is an example of a cocommutative Hopf algebra.
Theorem (Cartier-Gabriel) Assume that $k$ is algebraically closed, that $A$ is a cocommutative Hopf algebra. Let $\mathfrak{g}$ be the space of primitive elements, and $\Gamma$ the group of group like elements in $A$. Then there is an isomorphism of $\Gamma \ltimes U(\mathfrak{g})$ onto $A$ as Hopf algebras, inducing the identity on $\Gamma$ and on $\mathfrak{g}$.
Theorem (Milnor-Moore) Let $A = \bigoplus_{n \geq0}A_n$ be a graded Hopf algebra over $k$. Assume
- $A_0=k$ (we say $A$ is connected in this case [HINT: think about motivating example above!])
- The product in $A$ is commutative
Then $A$ is a free commutative algebra (a polynomial algebra) generated by homogeneous elements.
Reference: Pierre Cartier has a survey paper called "A primer of Hopf algebras" which has all of this stuff and more. There are further structure theorems for Hopf algebras but they are too hard to state without the addition of more definitions (e.g. conilpotent), which would just take too long here.
Very short answer: Gradings and filtrations are used everywhere for associative algebras.
Not very detailed exposition (I expect you don't know most of the terms, but I am just listing things I know that form examples to the answer of your questions)
(Q1) Finding a good grading on (usually finite dimensional) associative algebra has become one of the main trend in recent years in representation theory. The gradings people are looking for are usually the Koszul grading, i.e. a grading so that the algebra is Koszul. People who are more DG(differential graded)-oriented may need more than just a positive Z-grading. This comes under the name of Adams grading (still a group-graded theory anyway)
Other use of grading I know of: grade the preprojective algebra of finite type and take a smash product with it, one would arrive to a mesh algebra.
In a nut shell, as you have said it yourself, it really depends on what you want at the end before you actually start looking for a grading.
(Q2) Again I can only give examples that I know of. If $A$ is quasi-hereditary algebra, then there are certain objects called the standard modules, which exhibit certain nice (homological) properties. The category of the objects which can be filtered by standard modules have very interesting homological properties (this actually happens not only in quasi-hereditary algebras), such as, they are contravariantly finite.
If $A$ is Koszul algebra, then the Koszul dual $A^! := Ext_A^\bullet(A_0,A_0)$ in fact "determines" the category of modules which are filtered by the 0-th graded piece $A_0$. When people try to generalise Koszul theory, one thing they would study is that if the algebra $Ext_A^\bullet(M,M)$ can "determine" the category of objects filtered by $M$, etc. etc. A good exposition on this direction is Keller's Introduction to $A_\infty$ algebras and modules.
Filtration, sometimes, comes very close to determining some sort of useful/interesting resolutions (for Koszul algebras, the filtration gives you the projective resolution). Whether you want a filtration by two-sided ideal, or one-sided ideal, or submodule, is entirely up to you. As you have expected, it really depends on what you want to do.
EDIT: In response to OP comments. I think you need to understand filtrations and gradings are two different concepts and should not to be confused the two. I myself fell into this trap before. This is understandable, because, most of the time people talk about positive $\mathbb{Z}$-grading, so then you can get a filtration $\bigoplus_{i\geq 0} A_i \supset \bigoplus_{i\geq 1} A_i \supset \cdots$. However, as the comments to the question already tell you, in order to have 'filtration', you need a concept of inclusion for the objects in your category (warning: this is not really true, but I should not confuse you further), and you usualy need "quotient" for objects so that you can talk about things like "filtered by a family of objects". The first definition you wrote down in the comment looks more like a grading thing rather than a filtration thing (sorry for bad wording). The subscript appearing in a grading has significance in the study; but the subscript appearing in a filtration, is merely used to label the position of an object in the chain (at least I would choose to think about them this way).
TBH, I don't understand why you need to be so precise on whether you should choose one-sided or two-sided ideal for a ring (for me, I look at algebras). You should start by thinking what you want at the end; or if you don't know what you want at the end, why don't you try fiddle with some examples first and see which kind of filtration would give something interesting? May be you want to define a type of filtration on something you are studying, I guess you can, for example, write down the definition you had in your mind first, and go around asking people and see what they think?
Again, take quasi-hereditary algebras as example; such algebra is required to possess a chain of inclusions of some special two-sided ideal (a filtration of the algebra).
On the other hand, you can successively take the (Jacobson) radical for an algebra, and this gives also gives you a filtration of the algebra. $A \supset Rad A \supset Rad^2 A \supset \cdots$. But these are two entirely different filtrations, and they have different uses.
Best Answer
No: you have to be careful not to confuse the "-" in the quaternion group with the "-" in $K$.
Explicitly, $(1_K)(-g)$ and $(-1_K)(g)$ are distinct elements of $K[G]$, and furthermore $(1_K)(g)$ and $(-1_K)(-g)$ are distinct from each other and from the last two.
$K[G]$ is necessarily a free $K$-module of rank $|G|$, one basis element per group element.
The same would be said of $A=\bigoplus_{g\in G}A_g$: the fact that $g$ and $-g$ "look" like scalar multiples of each other is just a red herring. They are distinct, and produce separate coordinates in the sum. This will be of rank $|G|$ too. The "-" in front of $g$ is just a notation from the group $G$, and not from the additive group of the module you're making.