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GILA II: Orbits, stabilizers, and classifying group actions

Now that we’ve discussed group actions a bit, it’s time to characterize them. In this post I’d like to take a leaf from Tim Gowers’ book and try to make each step taken in the post “obvious.” While the content of the proofs is not too difficult, its motivation is rarely discussed.

First, it’s important to note that there is a way to take direct sums or disjoint unions (category theorists would say coproducts) of group actions: given a group $G$ acting on two sets $S, T$, one defines an action on their disjoint union $S \sqcup T$ in the obvious way: pick one action or the other. (Disjoint unions differ from unions in the usual sense because we relabel the elements of the sets so that they cannot intersect.) There’s a great reason to do this, and cycle decomposition showed us a special case: every group action is a direct sum or disjoint union of the action on its orbits.

This is the first step toward a structure theorem. Since the group action cannot “mix” between two orbits, it acts “independently” on orbits, and any question we might want to ask about the group action can be answered by looking at each orbit separately. A group action with a single orbit is called transitive, which means that for every $x, y$ in the underlying set there exists $G$ such that $gx = y$. So to classify group actions it suffices to classify transitive group actions.

First, given an element $x$ in an orbit, the orbit itself consists of the $|G|$ not necessarily distinct elements $\{ gx : g \in G \}$; it follows that a set with a transitive group action has size at most $|G|$. When does equality occur? Precisely for torsors!

If John Baez’s discussion (which you can find in the above link) doesn’t help, an example you want to keep in mind is that a regular $n$-gon is a torsor for the cyclic group $C_n$ acting by rotation. This is because if we fix any particular vertex of the $n$-gon, we can identify each vertex with the unique rotation going to it. Torsors are always free group actions, which means that the only group element with a fixed point is the identity. So torsors are very easy to understand. In fact, try to show that every torsor is (non-canonically) isomorphic to the action of $G$ on itself by left multiplication, so every group is a torsor for itself. (This action is called the regular representation of $G$.)

If an orbit isn’t a torsor, then the number of elements in the orbit must be less than the number of elements in the group, so some of the permutations will “collide”; in other words, for a particular $x$ there may be two different group elements $g, h$ with $gx = hx$, hence with $h^{-1} gx = x$. So collisions correspond to the existence of non-identity permutations with fixed points. This suggests that smaller orbits correspond to more fixed points. So what we want to do is find some way to measure how far an orbit comes from being a torsor, and it should be related to fixed points.

Let’s try the next simplest example after cyclic groups. The dihedral group $D_n$ is the group of symmetries of the regular $n$-gon, and it has $2n$ elements but acts faithfully on a set of size $n$, so the “natural” group action isn’t a torsor. Although the rotations have no fixed points, the reflections do. If $n$ is odd, every reflection in the group fixes one vertex, and if $n$ is even, half of the reflections fix two opposite vertices and half of the reflections don’t. Every stabilizer subgroup of a vertex is a subgroup of order $2$: it consists of the identity and the reflection about the line passing through that point and the origin.

This is important. One way to characterize a torsor is that for every pair of elements $x, y$ there exactly one element $G$ such that $y = gx$; in other words, it is possible to “subtract” elements of the set and get a group element back. The stabilizer subgroup $\text{Stab}(x)$ appears when we try to “subtract” $x$ from itself, so here’s an idea: instead of only requiring that we get a single element, for an arbitrary group action let’s define a “subtraction” map that, given $x, y$, returns every $g$ such that $y = gx$; we’ll call this $\text{Hom}(x, y)$ as a nod to the abstract perspective, so that $\text{Hom}(x, x) = \text{Stab}(x)$. In the dihedral group, there are always two ways to get from $x$ to $y$: one can either rotate to $y$ or apply the reflection fixing $x$ and then rotate to $y$. That second choice is equivalent to the other obvious thing to do, which is to rotate to $y$ and then apply the reflection fixing $y$. But there’s no reason this shouldn’t hold generally. In other words, $\text{Hom}(x, y) = g \, \text{Stab}(x) = \text{Stab}(y) \, g$

for every group action and any choice of $G$ such that $y = gx$. The inclusion of the right two sides into the left side is obvious. For the reverse inclusion, let $y = g' x$ for some $g' \in \text{Hom}(x, y)$; then $gx = g'x \Leftrightarrow (g')^{-1} gx = x \Leftrightarrow (g')^{-1} g \in \text{Stab}(x)$, and similarly for $\text{Stab}(y)$. In other words, $\text{Hom}(x, y)$ is always a (left) coset of $\text{Stab}(x)$. This defines an action of $G$ isomorphic to the original one where $G$ acts on the cosets of one of its subgroups $\text{Stab}(x)$ by left multiplication, and every transitive (left) group action must arise in this fashion; this is the classification we wanted. So we’ve proven the following.

Structure theorem for group actions: Every group action of a group $G$ is isomorphic to a direct sum (or disjoint union) of actions on the cosets of some sequence of subgroups $H_1, ... H_n$ of $G$. Such a group action has one orbit for every subgroup, and the stabilizer subgroup of any element in orbit $i$ is isomorphic to $H_i$.

If the subgroup $H_i$ has the additional property of being normal, then the action of $G$ on its cosets is the definition of a quotient group, but it’s important to recognize that this isn’t always the case. Nevertheless, this is a nice generalization of the special case of torsors: in any transitive group action we can “subtract” $x$ from $y$ to get $\text{Hom}(x, y)$, or we can “add” $x$ and $\text{Hom}(x, y)$ to get $y$. I might go so far as to call a transitive group action a “generalized torsor.”

Now, it’s always true that the cosets of a given subgroup define an equivalence relation, so they partition a group into disjoint sets. Moreover, cosets are all the same size. This gives the following important counting result.

Orbit-stabilizer theorem: Let $S$ be a set on which a group $G$ acts transitively, i.e. an orbit. Then $\displaystyle |S| = \frac{|G|}{\text{Stab}(x)}$, where $x \in S$.

This is because every element of $S$ is in bijection with a coset of $\text{Stab}(x)$. So the size of the stabilizer subgroup is our desired measure of how far an orbit is from being a torsor, which is defined in terms of fixed points as we expected. In the next post, we’ll use this observation to show how to count orbits in terms of fixed points and prove an easier version of the PET.

Exercises

If you enjoy this kind of thing, explicitly write down all the isomorphisms I claimed exist. Try to keep track of which ones are canonical and which ones aren’t.

Show that every group action of a group $G$ is isomorphic to a faithful group action of one of its quotient groups. (The first isomorphism theorem is relevant.)

If $G, H$ are two groups acting on sets $S, T$, the direct product of their group actions can be defined as the action of the direct product $G \times H$ on the Cartesian product $S \times T$ as follows: $(g, h)(s, t) = (gs, ht)$. Do we get every group action of $G \times H$ in this way?

4 Responses

1. on May 16, 2011 at 1:30 pm | Reply Artem Pelenitsyn

> let’s define a “subtraction” map that, given x, y, returns every G such that y= gx;

G should be g.

2. […] Comments J. H. S. on GILA II: Orbits, stabilizers,…GILA II: Orbits, sta… on GILA I: Group actions and equ…GILA II: Orbits, sta… on […]

3. on June 15, 2009 at 12:28 pm | Reply J. H. S.

A cool exercise on group actions.

Best of luck…