Is my understanding of the orbit-stabilizer theorem correct? So I just made a very short sketch proof of the orbit-stabilizer theorem, and I wanted to see if it was flawed because I'm fairly certain it is:
Start with $\phi(g)=g*x$, $g,g'\in G, x\in X$, G is a group action (composition) and X is a set. Check that $\phi$ is a homomorphism: $\phi(g*g')=(g*g')*x=g*(g'*x)=g*\phi(g')=\phi(g)*\phi(g')$
This is where I'm fairly certain I'm wrong. Since $g'*x$ is still in X, I'm assuming that the function can still act on this instead of just $x$. But I think this is right because $Ker(\phi)$=$\left \{ g\in G \mid \right  (g*x=x) \}$=$Stab_G(x)$, so by the first isomorphism theorem $G/Stab_G(x)\cong Im(\phi)=\left \{ g*x \mid \right (g\in G) \}=Orb_G(x)$, then by Lagrange's theorm, $|G|=|Orb_G(x)||Stab_G(x)|$
I'm only a junior in high school and haven't reviewed much abstract algebra in a very long time, so my understanding in several places might be very wrong. But I'm assuming that proving this theorem is somewhat like this. Any help is appreciated.
 A: Technically you shouldn't write $\phi(g)=g\ast x$, maybe write $\phi(g)(x)$. Note $\phi$ is a function of $g$, but the output $\phi(g)$ is itself a function of $x$. To check the two functions $\phi(g\ast g')$ and $\phi(g)\circ\phi(g')$ are equal (where $\circ$ is function composition), you need to check they yield the same output for every input, i.e.
$$ \phi(g\ast g')(x)=(\phi(g)\circ\phi(g'))(x) $$
for all $x$, which is equivalent to
$$ (g\ast g')\ast x=g\ast (g' \ast x). $$
(I wouldn't always recommend using the same symbol $\ast$ for both the group operation $G\times G\to G$ and the group action $G\times X\to X$ in elementary introductions to group actions, but whatever.) Since you're declaring $g\ast x$ is a group action from the get-go, I assume that you're assuming $(g\ast g')\ast x=g\ast(g'\ast x)$ for all $x$ as a given.
Your definition of $\ker\phi$ is missing an ingredient:
$$ \ker\phi=\{g\mid \color{red}{\forall x},~gx-x\}. $$
Thus, $\ker\phi$ does not depend on any one $x$; the $g\in\ker\phi$ fix all $x\in X$. This is different from the stabilizer $\mathrm{Stab}(x)$ of a specific element $x\in X$, which are all the $g\in G$ which fix $x$ but not necessarily all the other elements of $X$. A stabilizer $\mathrm{Stab}(x)$ can properly contain the kernel $\ker\phi$.
For example, suppose the symmetric group $S_3$ acts on $\{1,2,3\}$ in the standard way. Then the kernel is trivial; only the identity permutation fixes every element. But $\mathrm{Stab}(3)$ also contains the transposition $(12)$.
While the orbit-stabilizer theorem can be considered an analogue of the first isomorphism theorem for groups, it is not an application or direct corollary of it. You can say $G/\ker\phi\cong{\rm img}\phi$ are isomorphic as groups, but: (1) $\mathrm{Stab}(x)$ is not the same as $\ker\phi$, so you can't replace the one with the other (2) in general, $\mathrm{Stab}(x)$ is not a normal subgroup, so $G/\mathrm{Stab}(x)$ is not a quotient group, (3) $\mathrm{img}(\phi)$ is a set of permutations of $X$ which doesn't depend on any particular element $x\in X$, whereas $\mathrm{Orb}(x)$ is a set of elements of $X$ and depends on the particular element $x\in X$.
Again let's use the example of $S_3$. The map $\phi:S_3\to S_3$ is the identity map. So $\mathrm{img}(\phi)=S_3$, whereas the orbit $\mathrm{Orb}(x)$ will be the set $\{1,2,3\}$. (In this case the orbit didn't depend on $x$, but in general it will.) Or consider $S_2=\{\mathrm{id},(12)\}$ acting on $\{1,2,3\}$. Again $\mathrm{img}(\psi)=S_2$ but $\mathrm{Orb}(1)=\{1,2\}$ and $\mathrm{Orb}(3)=\{3\}$.
