Orbits, Stabilizers, and Conjugacy Classes

December 1, 2010, 23:45
Filed under: Algebra, Math | Tags: algebra, combinatorial, group theory, Math

We’ve seen a couple of ways to cut a group into pieces. First, we can look at its subgroups, which I visualize as irregular blobs all containing the identity. Under inclusion, these subgroups form a lattice, a partially ordered set in which every two elements have a greatest lower bound (here their intersection) and a least upper bound (here the group generated by their union). The structure of this lattice reveals a lot about the structure of the group and the things attached to it, the fundamental theorem of Galois theory being one powerful example. Second, given one subgroup, we can look at its cosets, which I visualize as parallel slices, and the quotient groups they form.

But cosets are tied to a specific subgroup and aren’t groups themselves, and the lattice of subgroups is in a sense too much information. One of the common problems of math is to find invariants — simpler objects that encode a lot of the data in a given structure and are easier to find. The only real way to get simpler than a group is with numbers, and one sequence of numbers is the class equation, which describes the conjugacy classes of the group. I visualize these as radial slices, like the layers of an onion.

A conjugacy class of a group is a set of elements of the form $C_h=\{ghg^{-1}:g\in G\}$, where $h$ is fixed. Like cosets, these aren’t subgroups, other than the conjugacy class of the identity, which is just $\{e\}$ . But conjugacy does give an equivalence relation, so the conjugacy classes partition the group. The class equation is just the equation $|C_{g_1}|+|C_{g_2}|+\dotsb+|C_{g_k}|=|G|$ , expressing the order of the group as a sum of orders of conjugacy classes partitioning it.

The main result on conjugacy classes actually comes from a more general result on group actions. Let $G$ act on a set $X$ . Recall that the orbit of $x\in X$ was $Gx=\{gx:g\in G\}$ . The dual notion is the stabilizer, which is the set of group elements that fix $x$ : $G_x=\{g:gx=x\}$ . (god, this notation is atrocious.) Note that $Gx\subset X,$ but $G_x\subset G$ , and it is in fact a subgroup (if $g$ and $h$ fix $x$ , then so does $gh$ , etc.), though not usually normal.

Even if it’s not normal, we can still look at its left cosets, and in particular the index $[G:G_x]$ (which is equal to the number of left cosets). Let’s define a map $\phi:g\mapsto gx$ . This is surjective onto $Gx$ . If $g$ and $h$ have the same image, then $\phi(g^{-1}h)=g^{-1}(hx)=g^{-1}(gx)=x$ , so $g^{-1}h\in G_x$ , and $g$ and $h$ are in the same coset. Thus, each $\phi$ gives a bijection between cosets and elements of $Gx$. We have proved:

Theorem (orbit-stabilizer). $|Gx|=[G:G_x]=|G|/|G_x|.$

There’s a similar statement worth mentioning about things in $X$ . It’s called Burnside’s Lemma, even though he cited it as being proved by Frobenius. Let $X/G$ be the set of orbits of $X$ under the $G$ -action. (If $X$ has a topology, then this can be the quotient space.) Let $Fix(g)$ be the set of elements in $x$ that $g$ stabilizes. The following really only makes sense when $X$ and $G$ are finite.

Theorem (not-Burnside’s Lemma). $|X/G|=\left(\sum_{g\in G}|Fix(g)|\right)/|G|$ .

Proof. We have $\sum_g|Fix(g)|=|A|$ , where $A=\{(g,x)\in G\times X:gx=x\}$ . Count these ordered pairs in the other direction, and you get $|A|=\sum_x|G_x|$ . We can break this down further into orbits: $|A|=\sum_{O\in X/G}\sum_{x\in O}|G_x|$ .

But if $x,y\in O$ where $O$ is an orbit, then there must be some $g$ such that $y=gx$ . As you can check, $h\mapsto ghg^{-1}$ is an isomorphism from $G_x$ to $G_y$ . In particular, all the stabilizers of elements of an orbit have the same cardinality,. Then $\sum_{O\in X/G}\sum_{x\in O}|G_x|=\sum_{O\in X/G}|O|\cdot |G_x|$ . By the orbit-stabilizer theorem, $|O|\cdot|G_x|=|G|$ . So this is $|G|\cdot |X/G|$ . Put everything together, and you get $|X/G|=\left(\sum_{g\in G}|Fix(g)|\right)/|G|$ . $\square$

One of the effects of the orbit-stabilizer theorem is that $|G_x|$ and $|Gx|$ must divide $|G|$ for any $x$ . In particular, $G$ acts on itself by conjugation, with the orbits being exactly the conjugacy classes. It follows that the order of each conjugacy class divides the order of the group!

So the class equation is a set of numbers summing to $|G|$ , and each of whom divides $|G|$ . Every normal subgroup has as its class equation a subset of these numbers, also adding to a factor of $G$ . This “subequation” will also have to contain the $1$ corresponding to the identity, so it looks like $1+(\text{factors of }G)=\text{factor of }G$ . This suggests that the number and order of normal subgropus is severely controlled by the order of $G$ . To make things a bit easier, there’s also a $1$ in the equation for every element of the center $Z(G)$ . (Remember that the center of a group is the set — and subgroup — of elements that commute with everything.)

Let’s do an example. A $p$ -group is one of order $p^n$ , for $p$ a prime. Because these guys have so few divisors, tools like the class equation work well on them. In this case, every number in the class equation that isn’t $1$ will be of the form $p^k$ for $1\le k$ , so a $p$ -group has to have a nontrivial center.

Since the center is a subgroup, it must be of order $p^k$ for some $1\le k\le n$ . If $n=1$ , then the center is the whole group! This gives an important corollary — every group of prime order is abelian. And the same reasoning tells us that no group of prime order has any nontrivial subgroup, which means that every element generates the whole group. Choose one element $a$ and write the elements as $e=a^0,a,a^2,\dotsc,a^{p-1}$ . The map $a^k\mapsto k$ is an isomorphism from the group to $\mathbb{Z}/p\mathbb{Z}$ . Every group of prime order is cyclic.

Here’s a second example. A group is simple if its only normal subgroups are itself and $\{e\}$ . You can think of these like prime numbers, and in fact the theory of solvability is based on a certain kind of “prime factorization.” From the above reasoning, it’s clear that $\mathbb{Z}/p\mathbb{Z}$ is simple if $p$ is prime; in the case of $\mathbb{Z}/n\mathbb{Z}$ with $n$ composite (not prime), meanwhile, then each of its divisors $k$ generates a subgroup $\{0,k,2k,\dotsc\}$ , which is normal because the group is abelian. The question is then: are there other simple groups?

The answer is yes, and the smallest one is the alternating group $A_5$ . Alternating groups are index-two subgroups of symmetric groups that preserve the “parity” of the symbols they act on; I haven’t said much about the internal structure of symmetric groups, so don’t worry if this doesn’t make much sense. Fortunately, $A_5$ is also the group of orientation-preserving symmetries of the dodecahedron. If you do not have a (real or imagined) model dodecahedron, please build one now.

Okay. With model dodecahedron in hand or mind, try to find some symmetries. If you visualize the dodecahedron as centered at the origin in $\mathbb{R}^3$ , then it’s clear that $A_5$ embeds into $SO(3)$ . In particular, they’re all rotations about some axis. There are three different kinds of axes that work: an axis that goes through a center of a face gives an order-5 rotation; one that goes through a vertex gives an order-3 rotation; and one that goes through the center of an edge gives an order-2 rotation.

The dodecahedron has 12 faces, 30 edges, and 20 vertices. Each of the above rotations has its axis through two opposite faces/edges/vertices, and we don’t count the identity, so the number of each kind of rotation is $12*(5-1)/2=24$ rotations of order 5, $20*(3-1)/2=20$ rotations of order 3, and $30*(2-1)/2=15$ rotations of order 2. Add in the identity and you get $1+15+20+24=60$ .

Since conjugate elements have the same order, the conjugacy classes have to be subsets of these three sets. In most cases, given two elements of the same order, one can be shown to be conjugate of the other via a rotation sending the axis of the first to that of the second. The exception is the rotations of order 5: we get one conjugacy class with all the rotations of angle $2\pi/5$ or $-2\pi/5$ , and one with those of angle $4\pi/5$ or $-4\pi/5$ . It’s much easier to prove this using the theory of the symmetric group than geometry, so I’ll leave it for later. In any case, this gives the class equation $60=1+15+20+12+12$ .

At this point, all we have to do is check that no subset of terms of the equation divides $60$ ! This proves that $A_5$ is simple.

I haven’t been doing much of this combinatorial group-classification work, but it’s sort of fun and I’ll probably do a bit more to lead up to solvability. We do need to talk about symmetric groups in detail. The other natural point of departure is the Sylow theorems, which are pretty much the Hebrew Hammer of combinatorial group theory, giving very strict rules about which subgroups of a given group can be $p$ -groups. Some non-combinatorial algebra is on the way, too: I want to introduce other algebraic structures, like rings and fields, and use them to construct the sets of numbers we know and love.

Gracious Living

Categories

Archives

Links

Gracious Living

Categories

Archives

Links

Share this:

Related

Leave a comment Cancel reply