Actions and Sylow theorems · Definition

Orbit and stabilizer

modern algebragroups

Before you start you need to know

To understand how a group action splits a set we ask two questions: where can a point be taken, and which group elements leave it still.

Definition

If a group \(G\) acts on a set \(X\), the orbit of \(x\in X\) is the set of all its images \(g\cdot x\), and its stabilizer is the subgroup of elements of \(G\) that leave it fixed.

\[\mathrm{Orb}(x)=\{g\cdot x : g\in G\},\qquad \mathrm{Stab}(x)=\{g\in G : g\cdot x=x\}\]

The orbit of \(x\) answers the first question: it collects the points reachable from \(x\) by some \(g\in G\). Two orbits are equal or disjoint, since \(y\in\mathrm{Orb}(x)\) is equivalent to \(\mathrm{Orb}(y)=\mathrm{Orb}(x)\). Hence the orbits partition \(X\), just as cosets partition a group. If there is only one orbit, the action is called transitive.

The stabilizer answers the second. It is a subgroup of \(G\): the identity fixes everything, and if \(g\) and \(h\) fix \(x\), so do the product \(gh\) and the inverse \(g^{-1}\). When \(\mathrm{Stab}(x)=G\), the point \(x\) is fixed; when \(\mathrm{Stab}(x)=\{e\}\) the action is free at \(x\). In the action of \(G\) on itself by left multiplication, every stabilizer is trivial and the only orbit is \(G\).

Guiding example: \(S_3\) acts on \(\{1,2,3\}\). The orbit of \(1\) is the whole set and its stabilizer is \(\{e,(2\ 3)\}\). The stabilizers of \(2\) and \(3\) are \(\{e,(1\ 3)\}\) and \(\{e,(1\ 2)\}\): different, but conjugate to each other, because if \(y=g\cdot x\) then \(\mathrm{Stab}(y)=g\,\mathrm{Stab}(x)\,g^{-1}\).

A common mistake is to think the stabilizer is a subset of \(X\) or the orbit a subgroup; it is the other way around: the orbit lives in \(X\) and the stabilizer in \(G\). These two objects are linked by an equality of sizes, which is the subject of the orbit-stabilizer theorem.

Think about it

Can a point of an action have a different stabilizer from another point in its own orbit?

Yes. They are conjugate, \(\mathrm{Stab}(g\cdot x)=g\,\mathrm{Stab}(x)\,g^{-1}\), but may be different subgroups, as in \(S_3\) acting on \(\{1,2,3\}\). They always coincide only when the stabilizer is normal.

Exercises 10

  1. The group \(S_3\) acts on \(\{1,2,3\}\) by \(g\cdot x=g(x)\). How many elements does the orbit of \(1\) have?
  2. Let \(G=\langle(1\ 2\ 3)(4\ 5)\rangle\le S_5\) act by \(g\cdot x=g(x)\). What is the orbit of \(4\)?
  3. With the same group \(G=\langle(1\ 2\ 3)(4\ 5)\rangle\), what is the stabilizer of \(1\)?
  4. The symmetry group of the square (order \(8\)) acts on its four vertices. How many elements does the stabilizer of a vertex have?
  5. For an action of \(G\) on \(X\) and \(x\in X\), which statement is always true?
  6. Let \(G=\langle(1\ 2)(3\ 5)\rangle\le S_5\) act on \(\{1,2,3,4,5\}\). How many distinct orbits are there?
  7. A group \(G\) acts on itself by \(g\cdot x=gx\). What is the stabilizer of an element \(x\)?
  8. If \(y=g\cdot x\) in an action of \(G\), what is the stabilizer of \(y\) in terms of that of \(x\)?
  9. \(S_4\) acts on the two-element subsets of \(\{1,2,3,4\}\) by \(g\cdot\{a,b\}=\{g(a),g(b)\}\). How many elements does the stabilizer of \(\{1,2\}\) have?
  10. In an action of \(G\) on \(X\), the stabilizer of \(x\) is all of \(G\). What is the orbit of \(x\)?

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §1.7 y §4.1
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §16
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 29