Need some help with understanding symmetry group of some $X \subset \mathbb{R}^n$ and its normal subgroup's quotient group. I'm a student who is currently studying Finite Groups of Isometries in Euclidean Geometry.
Symmetry Group is the first topic of this section, but I am having trouble understanding the concept clearly. So, I would like to get some help since it looks like the most fundamental thing in this section.
Definition of Symmetry Group in my textbook:

If $X \subset \mathbb{R}^n$ contains $(n+1)$ independent points, its symmetry group $S(X)$ is the subgroup of $I(\mathbb{R}^n)$ (the isometry group of $\mathbb{R}^n$).
If $X \subset \mathbb{R}^n$ doesn't contain $(n+1)$ independent points, the first is to consider isometries of $\mathbb{R}^n$ that sends $X$ to itself. (But, there may be many).
The second is to regard $X$ as a subset of the affine space $A$ spanned by it and consider isometries of $A$ that send $X$ to itself.

Especially, the second method is not really easy to understand.
And there is a simple example which I think the method 1 is applied.

If $l$ is a line in $\mathbb{R}^2$, it doesn't contain $3$ independent points and the subgroup of $I(\mathbb{R}^2)$ that preserves $l$ contains a normal subgroup with two elements whose quotient is $I(\mathbb{R}^1)$.

As any isometry in $I(\mathbb{R}^2)$ is either Translation, Reflection, Rotation, or Glide, we can see that any Translation (parallel to $l$) and a Reflection (by $l$) will send $l$ to $l$. Let the set $S(l)$ be $\{Id,T_{a_1},T_{a_2},...,R_l\}$ for any translation $T_{a_i}$ s.t $a_i$ is parallel to $l$. Then, $S(l)$ has a normal subgroup $\{Id,R_l\}$.

But, I don't quite understand why $S(l)/ \{Id,R_l\}= \{ sb | s \in S(l) \text{ and } b \in \{Id,R_l\} \} =I (\mathbb{R}^1)$


Also, if we want to use the second method for this example, what should I do? I think $l$ is a affine line that only contains $2$ independent points. But, how do I define the affine space $A$ that is spanned by $l$? If $l$'s independent points are $\{a_1,a_2\}$, then does $A$ have basis $\{a_0,a_1,a_2\}$ s.t $\lambda_1 (a_1-a_0) + \lambda_2 (a_2-a_0)=0$ iff $\lambda_1=\lambda_2=0$?

Please enlighten me. Thank you for your time!
 A: John.
In the first case, if $X\subset \mathbb{R}^n$ contains $(n+1)$ independent points, then every isometry $T\in S(X)$ can be uniquely extended to an isometry $\overline{T}\in I(\mathbb{R}^n)$.
This gives an injective morphism of groups $S(X)\rightarrow I(\mathbb{R}^n)$ which associates to each $T$ its unique extension $\overline{T}$. That's the way we can consider $S(X)$ as a subgroup of $I(\mathbb{R}^n)$.
In the other case, if $X$ doesn't contain $(n+1)$ independent points, it's contained in a hyperplane. In any case, we want to define the symmetry group of $X$ in relation to $I(\mathbb{R}^n)$, so that matrix methods can be applied.
One way of defining them is a bit "extrinsic". That is, define $S(X)$ to be the group of bijections of $X$ onto $X$ which are obtained by restricting to $X$ isometries in $I(\mathbb{R}^n)$. Not every isometry $T$ in $I(\mathbb{R}^n)$ defines a bijection of $X$ by restriction. It is necessary that $T$ maps $X$ into itself (otherwise the image of a point in $X$ could be outside $X$, and then $T$ wouldn't define a symmetry of $X$).
Denote by $Bij(X)$ the group of all bijective maps $X\rightarrow X$, and by $I_X(\mathbb{R}^n)$ the set of all isometries in $I(\mathbb{R}^n)$ which map $X$ onto itself (it's a subgroup of $I(\mathbb{R}^n)$). The "restriction to $X$" gives a morphism of groups $I_X(\mathbb{R}^n) \rightarrow Bij(X)$. By this method, $S(X)$ is defined to be the image of this morphism (by the isomorphy theorem, it's isomorphic to the quotient of $I_X(\mathbb{R}^n)$ by the normal subgroup $N_X$ leaving all points in $X$ fixed, that is, maps whose restriction is the identity).
Now, it seems too much to require that symmetries of $X$ come from isometries of the whole space, so it makes more sense to consider the "smallest" affine space $A$ containing $X$. We can now substitute $\mathbb{R}^n$ for this $A$. If this space $A$ is $k$-dimensional, then $X$ will contain $(k+1)$ independent points (otherwise $A$ wouldn't be the "smallest"), so we can define $S(X)$ by the same "first" method (applied to $A$) as a subgroup of $I(A)$.
Now, if $l$ is a line, it's equal to the affine set it spans. In this case, one can see that every isometry of the line can be obtained by restriction to $l$ of an isometry in $I(\mathbb{R}^2)$. So, the two methods give the same, and show that $I(\mathbb{R}^1)$ is a quotient of $I_l(\mathbb{R}^2)$ (this last group is what you denote $S(l)$).
Now, as I said before, the normal subgroup consists of the isometries in $I(\mathbb{R}^2)$ which leave $l$ fixed. If you take two points $p,q$ in $l$ such that $\vec{pq}$ has length $1$ and a third point $r$ such that $\vec{pr}$ has length $1$ and is orthogonal to $\vec{pq}$, then any isometry that leaves $l$ fixed must also leave $p, q$ fixed, and so must map $r$ to itself or to $p-\vec{pr}$. Since $p, q, r$ are independent, there are only two isometries satisfying this condition: the identity map and reflection with respect to $l$.
By the way, which book are you following?
