This question may be easy to solve with an example, but I am not able to think of any.
Let's say that you have a group G, and a subgroup H, it is show that a the cosets $(aH), a \in G$, form an equivalence relation, so they partition G, but the interesting part in this case is that every part of the partition has the same amount of elements.
But let's say you have a set X, which is not a group, but you have a group G, that acts on the group X.
Formally this means that:
$gx\in X, \forall g \in G, \forall x \in X$
$ex=x, \forall x \in X$
$(g_1g_2)x=g_1(g_2x)$.
It can be show that the relation $x_1$~$x_2$, if there is a g such that $gx_1=x_2$, exists. Hence since we have an equivalence relation we get a partition of X, and we call these partitions for orbits in X under G.
My question is: Are these orbits/partitions of the same size like in the coset case?
Best Answer
No, in general they aren't.
Easy example: Let $\mathbb R^\ast$ (the group of nonzero real numbers under multiplication) act on the set $\mathbb R$ via multiplication. There are exactly two orbits, $\{0\}$ and $\mathbb R \setminus \{0\}$.
One thing to point out though it that while cosets of $H$ can be acted on by $G$, this action is transitive and has only one orbit, so the cosets themselves are not orbits under this action.
The cosets $G/H = \{gH \ | \ g \in G\}$ are actually the orbits of the right action of $H$ on $G$ by multiplication. $G$ also has a left action on itself by multiplication and these actions are compatable, i.e., $g\cdot(x\cdot h) = (g\cdot x)\cdot h$ for all $g, x \in G$ and $h \in H$. Whenever you have a set $X$ with a left $G$ action and a right $H$ action which are compatible in this way, then $G$ acts on the $H$-orbits $X/H$ and if two $H$-orbits in $X/H$ lie in the same $G$-orbit (of the action of $G$ on $X/H$) then indeed they will be the same size.
So you could say that the reason the partitions of $G$ into cosets of $H$ are all the same is precisely because the action of $G$ on these cosets is transitive.