If $G$ is a crystal group, $\overline{G}$ acts on $G_t$. This is the lemma from my Geometry textbook (Notes on Geometry by E.Rees), and I would like to get some help on interpreting the proof of this lemma.
Before we go on, I would like to specify what $G$, $\overline{G}$, $G_t$ stands for.

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*$G$ is a crystal group, and by the definition of my textbook, this is a subgroup of isometry group of $\mathbb{R}^n$, which contains $n$ independent translations, but no arbitrarily small translations. My textbook really doesn't say anything more, but I think this is a symmetry group of the lattice of $\mathbb{R}^n$.


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*Am I understanding the concept of the crystal group correctly?



*$G_t = G \cap \mathbb{R}^n$ is called a point group, and according to my textbook, this is a normal subgroup (of $G$) of translations.


*$\overline{G}=G/G_t$. And, my textbook says (without a proof) it is a subgroup of $O(n)$, the orthogonal group of $\mathbb{R}^n$. And I think this just means that $\overline{G}$ is a set of orthogonal matrices that preserves the lattice.



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*I am kinda lost here already because if we let $f \in G$ be a translation, then I think the left coset $fG_t$ is a set of translations, but why $fG_t$ could become a set of orthogonal matrices? This question might sound silly due to my lacking of knowledge in Group Theory (especially, quotient group). But, please help.

Proof of the lemma:

Given $A \in \overline{G}$, there is an $x \in \mathbb{R}^n$ such that $(A,x) \in G$. If $v \in G_t$, then $(I,v) \in G$. We show that $Av \in G_t$, that is $(I,Av) \in G$:
$$(A,x)(I,v)(A,x)^{-1}=(I,Av).$$
As this is independent of $x$, the action is well defined. (End of Proof)


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*First of all, I don't know why my book used notations such as $(A,x),(I,v)$. What do these notations really mean? If I could know what this means, then I would be able to find out what $(A,x)^{-1}$ means.


*Assuming I know what $\overline{G}$ is, $A$ must be some rotation about the origin. And, we just want to show that $Av \in G_t$. But, why do we have to show $Av \in G_t$ like the given proof? If $A$ preserves the lattice, then $Av$ is in the lattice, so considering it as a translation, we just get what we desire.
 A: $G$ contains a subgroup $G_t$ of translations, which may be viewed as a lattice in $\mathbb{R}^n$.
(1) $G$ may not be the full symmetry group of the lattice. In general, if $S$ is the full symmetry group, then $G$ is somewhere between $G_t$ and $S$ (i.e. $G_t$ is a subgroup of $G$ and $G$ is a subgroup of $S$).
(2) No, $G/G_t$ is the point-group, not $G_t$. You can think of it as the stabilizer of a point.
Normally you don't think of a quotient group as a subgroup, because it's not, but there are exceptions to this rule. The loosest exception is when $G$ is a knit product of subgroups. Suppose a group $G$ acts transitively on a set $\Omega$ and one of the points $x$ (element of $\Omega$) has stabilizer subgroup $K\le G$. The orbit-stabilizer theorem says that $\Omega$ is equivalent to the coset space $G/K$ (that is, there is a one-to-one correspondence which intertwines the action, given by $gK\leftrightarrow gx$). If $H\le G$ is a subgroup, then: (a) $H$ is transitive iff $G=HK$ (i.e. all group elements $g\in G$ are expressible as $hk$ with $h\in H,k\in K$), (b) $H$ is free iff $H\cap K$ is trivial, and therefore (c) $H$ is regular iff $G$ is a knit product of $H$ and $K$ (i.e. all elements $g\in G$ are expressible as $g=hk$ for some unique pair $h\in H,k\in K$). Subgroups of a knit product are not necessarily normal. If $K$ is normal, then the knit product is in fact a semidirect product $G=K\rtimes H$. In this case, the coset space $G/K$ is in fact a quotient group which is isomorphic to $H$. Thus, the quotient group $G/K$ may be identified with an actual subgroup $H$ of $G$.
In this context, $G_t$ is a regular subgroup of $G$, so $G$ is a knit product of $G_t$ and a stabilizer. In fact, $G_t$ is normal, so $G$ is a semidirect product of $G_t$ and any point-stabilizer.
(3) >>>
$\bullet$ If $f\in G$ is a translation, then $f\in G_t$, so the coset $fG_t$ is just the subgroup $G_t$ itself, which in the quotient group corresponds to the identity matrix of $O(n)$. In general, if $f\in G$ is arbitrary, then $f=(A,x)$ for some $A\in O(n),x\in\mathbb{R}^n$, and the coset $fG_t$ corresponds to $\bar{f}=A$. In general, $f(v)$ will not be linear, but $f(v)-f(0)$ will be linear represented by a matrix in $O(n)$. That matrix is what $fG_t$ corresponds to.
$\bullet$ The book explains that all isometries of $\mathbb{R}^n$ are of the form $(A,x)v=Av+x$ where $A\in O(n)$ and $x\in\mathbb{R}^n$. You should also be able to find lots of proofs of this if you google.
Suppose $(A,x)^{-1}=(B,y)$. Composing the two gives
$$ (A,x)(B,y)v=(A,x)(Bv+y)=ABv+Ay+x. $$
For $(A,x)(B,y)$ to be $(I,0)$, we need $AB=I$ and $Ay+x=0$. Thus $B=A^{-1}$ and $y=-A^{-1}x$.
$\bullet$ The lemma proof is arguably worded poorly. We're not trying to show $Av\in G_t$, we're using the fact $Av\in G_t$ (which follows by definition) to show $(A,x)(I,v)(A,x)^{-1}\in G_t$, which means $G_t$ is normal in $G$.
