I am currently reading Abstract Algebra by Dummit and Foote Chapter 4. I was trying to grasp the idea behind proposition 2, I could prove it but could not construct a geometric understanding of the proposition. The proposition is as follows
Let $G$ be a group acting on a non-empty set $A$. The relation defined by $$ a \sim b \text{ if and only if for some $g \in G$, } b = g \cdot a $$ The relation is an equivalence relation and furthermore, for each $a \in A$, there is a one-to-one correspondence between $C_a$ and cosets of $G_a$ where $C_a$ and $G_a$ are respectively the equivalence class corresponding to $a$ and stabilizer of $a$ $$ C_a = \{ g \cdot a: g \in G \} $$ $$ G_a = \{ g \in G: g \cdot a = a \} $$
The one-to-one correspondence $\phi: C_a \to \text{cosets of } G_a$ is defined as follows: for all $b \in C_a$, there exists a $g \in G$ such that $b = g \cdot a$, the corresponding coset is $g G_a$ ($\phi$ is well-defined)
My current understanding is
Let $A$ be a finite set, then $G$ acting on $A$ can be considered as a permutation group (a subgroup of symmetric group) acting on elements of $A$. The equivalence relation part is easy as due to the structure of $G$, action on $A$ can be considered as an undirected graph and the equivalence class of $a$ can be considered as all the elements reachable from $a$. However, the one-to-one correspondence is a bit difficult for me, I was imagining the cosets of $G_a$ corresponding to $b$ is a sum of two parts: stabilizer of $a$ that maps $a$ into itself and $g$ the addon that translates $a$ into $b$ (i.e. $b = g \cdot a$) and that also translates stabilizer of $a$ into its coset corresponding to $g$.
Is there any representation that better explains the intuition behind this proposition?
My answer:
Let $G$ act on $A$, $a \in A$, and let the orbit of $a$ be $C_a$. We can impose a group structure on $C_a$ so that the Orbit-Stabilizer Theorem will be a consequence of Lagrange Theorem. Let $\star: C_a \times C_a \to C_a$ as follows $$ x \star y = (g_x g_y) \cdot a $$ where $g_x, g_y \in G$ such that $x = g_x \cdot a$ and $y = g_y \cdot a$. This map is well-defined and makes $C_a$ a group. We next define a homomorphism $\phi: G \to C_a$ as follows $$ \phi(g) = g \cdot a $$
The quotient group of this homomorphism is isomorphic to $C_a$ where the coset of $\ker \phi$ corresponding to $x \in C_a$ is exactly the set that sends $a$ to $x$. By Lagrange Theorem,$|G / \ker \phi| = |C_a|$
