Is orbit a group? Let a group $G$ act on a set $S$, and let $s$ be an element in $S$. The identity of the orbit of $s$ is $s$ itself and if $a$, $b$ are in the orbit of $s$, then $a b = g_1 g_2 s$, where $g_1$, $g_2$ are from $G$ and $g_1 s = a$, $g_2 s = b$. Using this operation, I claim that the orbit of $s$ is a group. Is that right?
 A: Consider the example of $G = S_3$, the permutations of $\{1, 2, 3\}$ and let $S = \{1, 2, 3\}$.  Choose $s = 1$.  What is $2 \cdot 3$ under your operation.  Well, for $g_1$ we could use either the permutation $213$ or $231$.  (Here, I write one-line notation for permutations, e.g. $231$ stands for the permutation which is the mapping $1 \rightarrow 2, 2 \rightarrow 3, 3 \rightarrow 1$.)  And for $g_2$ we could use either $312$ or $321$.  Thus, we can compute $2 \cdot 3$ as $g_1 g_2 s$ for any of the four choices for the pair $(g_1, g_2)$.  If $g_1 = 213$ and $g_2 = 312$, then $g_1 g_2 = 321$, so $g_1 g_2 s = 3$.  If $g_1 = 231$ and $g_2 = 312$, then $g_1 g_2 = 123$, so $g_1 g_2 s = 1$.  But then we have defined $2 \cdot 3$ as both $3$ and $1$ -- ergo, the given operation is not well-defined!
A: No it's not correct. Suppose $gs=\bar{g}s$ and $hs=\bar{h}s$ for $g,\bar{g},h,\bar{h}\in G$. Your operation is defined so that we have $(gs)(hs)=ghs$ but also $=(\bar{g}s)(\bar{h}s)=\bar{g}\bar{h}s$. But are these both the same element, that is do we have $ghs=\bar{g}\bar{h}s$? One can easily provide examples where this doesn't work.
Indeed, we expect a canonical isomorphism $G/{\rm Stab}_G(s)\cong{\rm Orb}_G(s)$ not as groups but as $G$-sets, that is as sets equipped with left $G$-actions (where morphisms are intertwining/equivariant maps between $G$-sets). Notice that if $H={\rm Stab}_G(s)$ is normal in $G$, then multiplication in $G/H$ can be transported to a multiplication operation on ${\rm Orb}(s)$, and this operation is precisely the one that you've defined. Yet if $H$ is not normal, $G/H$ has no inherent meaning as a group (only a space of left cosets), so we can expect ahead of time for your operation on ${\rm Stab}(s)$ to fail exactly as it fails for the space $G/H$. If $H$ isn't normal, then any instance of $aHbH\ne abH$ for $a,b\in G$ will provide an instance of the failure of the group operation in ${\rm Orb}(s)$ to be well-defined, specifically by considering the example of $(\bar{a}s)(\bar{b}s)$ for choices of $\bar{a}\in aH$, $\bar{b}\in bH$.
Even if $H$ is normal, indeed even if $H=1$ is trivial there is still some "unnaturalness" in thinking of the orbit's elements as identified with $G$, since this identification depends on which orbit representative you use (i.e. which choice of $s$ one uses). Even if ${\rm Orb}(s)={\rm Orb}(r)$ whilst $r\ne s$, we will necessarily have $G\cong{\rm Orb}(r)$ and $G\cong{\rm Orb}(s)$ be different isomorphisms. A special case of this fact is he distinction between vector spaces and affine spaces. Given an affine space, one can select a point to be the origin, and this induces an isomorphism of the affine space with the vector space, but this depends on an arbitrary choice of origin.
