So, let's consider a set of all (real-valued) differentiable real-valued functions over an interval (a, b): $C^1_{(a,b)}$ with an operation of addition +
It is clear that it is indeed an abelian group:
Addition is commutative and assosiative
The sum of two differentiable functions is a differentiable function, so the set is closed under addition
$0$ is the identity
for every fucntion f(x) there exists function -f(x), such that f(x) + (-f(x)) = 0
Let's consider the set of all continious functions over an interval (a,b): $C^0_{(a,b)}$
Any differentiable function over an interval is also continous over that interval, which implies $C^1_{(a,b)}$ $\leq$ $C^0_{(a,b)}$ Moreover, as both of those groups are abelian, that means that $C^1_{(a,b)}$ is a normal subgroup. Now, if $C^1_{(a,b)}$ $\triangleleft$ $C^0_{(a,b)}$ , we can construct $C^0_{(a,b)}$/$C^1_{(a,b)}$, so there exists a group $G$ and a homomorphism $g \space $: $C^0_{(a,b)}$ $\rightarrow$ $G$ with a kernel $ker\space g = C^1_{(a,b)}$
Are there any "more well-known" examples of $G$? I could not come up with any of those.
Oh, and on that topic, let's denote a set of all n-times differentiable functions over an interval (a,b) as $C^n_{(a,b)}$ and denote the set of infinitely differentiable function over an interval (a,b) as $C^\infty_{(a,b)}$
We can construct descending normal series:
$C^0_{(a,b)} \space \triangleright \space C^1_{(a,b)} \triangleright \space C^2_{(a,b)} \space \triangleright \space ... \space \triangleright \space C^\infty_{(a,b)} \space \triangleright \space R \space \triangleright \{0\}$
So, is there a general way to construct $C^n_{(a,b)}/C^{n+1}_{(a,b)}$ $\space$ , $\space$ $C^n_{(a,b)}/C^m_{(a,b)}$ $\space$ as well as $\space$ $ C^n_{(a,b)}/R$ and find "more well-known" groups isomorphic to those quotient groups?