We have already seen abstract groups as permutation groups, or, more generally, as groups of bijections, in a number of contexts:

- acting on itself by (left) multiplication, when we associated with each a bijection by
- acting on itself by
*conjugation*, when we associated with each a bijection by - acting on the set of (left) cosets of a subgroup when we associated with each a bijection by

We have also seen matrix groups at work as permutation groups (e.g. on the set of subspaces), permutation groups becoming matrix groups (e.g. represented by permutation matrices), etc.

What unites all these examples is the concept of a group action.

Let be a group and be a set, with a group of bijections An

actionof on is a homomorphism

There are many different types of actions, e.g. *permutation actions* when is a finite set and the symmetric group (of all permutations of

*linear actions,* when is a vector space, and a group of its linear transformations (i.e. matrix groups), etc.

By the homomorphism theorem, An action satisfying is called *faithful* (or *effective*).

For instance, the action of on itself by (left) multiplication is always faithful, but the action on cosets of an need not be faithful. E.g. it will not be faithful if is commutative and

**Orbits and point stabilisers**

An action of on defines an equivalence relation on as follows: iff there exists such that The equivalence classes of are called *orbits* (more precisely, orbits of on ).

Let and Then is called the *stabiliser* of in and denoted

Let for some What can be said about

Indeed, the coset consists of the elements of that map to Thus any fixes and any that fixes belongs to as

When we consider the action of on itself by conjugation, the orbits are called *conjugacy classes*.

For instance, the conjugacy classes of the symmetric group on letters, can be shown to be in 1-to-1 correspondence with the unordered partitions of into parts where as follows:

Let a permutation be written in the cyclic notation. Then for any one has

and so we can transform to any permutation with cycles of the same lengths

**Equivalent actions**

When do two actions and of a group on, respectively, sets and can be viewed as “the same”? Obviously we need a bijection to exist, that also “respects” in the sense that for all and Such an is called a –*equivariant* bijection, and is said to define an *isomorphism of actions*.

An action that has just one orbit is called *transitive*.

Let be a transitive action of on Then is isomorphic to the action on for any where we denoted by the preimage of in

This isomorphism of actions is given by such that where Indeed,

- is obviously a bijection
- iff iff iff
- is -equivariant. Indeed, let Then for any one has as claimed.

Finally, for an action of on we can characterise as

for an

**Actions of matrix groups** (a.k.a. *linear actions*)

Subgroups of the group, with respect to matrix multiplication, of invertible matrices with entries in a field have a number of natural actions, that all come one way or another from the vectorspace Indeed, each provides a bijection by for any Moreover, such a bijection is even an automorphism of the vectorspace (Automorphisms of vectorspaces are defined just as these for groups and rings: these are bijections preserving the corresponding algebraic structure (i.e. operations, etc)).

Apart from the action on the vectors of (called the * natural action * (by left multiplication)) there is another action on the vectors, namely, by (which one can equivalently writen as ). It is called *contragradient action*.

(Note that the map given by is **not always** an action, as and so is not a homomorphism, unless we have a commutative group.)

Natural and contragradient actions give a simple example of two non-equivalent faithful (i.e. with trivial kernel) actions of one group, say, , with on the same set Indeed, if were an equivalence of these actions, then for all and

Let and $f(v)=u$ so that and are linearly independent. In the basis , take Then We see that a contradiction.

If, on the other hand, and are always linearly dependent,

we take the same in the standard basis and see that for some again a contradiction. The argument showing non-equivalence in the case is similar. For these two actions are isomorphic, as is commutative.

Another important class of actions is the action on on subspaces of For simplicity, let us consider the action on the set of subspaces of dimension Given a -dimensional subspace and we define Then is a -dimensional subspace, as can be seen by choosing a basis of and observing that is a basis of Thus induces a bijection on

When fixes a we can choose a basis in such a way, that becomes a group of block matrices, discussed earlier here.In this case has a normal subgroup acting trivially on that is the kernel of the action of on the quotient space

The centre of acts trivially on as it preserves every 1-dimensional subspace (indeed, any equals with ).

**Projective line.**

The simplest nontrivial is when This set is called *projective line* and denoted and can be viewed as the set of pairs as follows: every 1-dimensional subspace of can be described by equation or by equation

To understand the action of an element of on consider and just as if we multiply a vector by and then normalize either to or to depending upon whether or not

30 May, 2010 at 3:51 |

In regard to the section on the action of a subset G of the general linear group over F^n on a subspace of F^n, if g is an nxn matrix and u is vector in F^k where k<n, then how is gu defined? The dimensions aren't right.

30 May, 2010 at 4:19 |

The dimensions ARE right. Indeed, a k-subspace of is just a subset of

In order to comput how the action works, take a basis of and apply the transformation (the action of which we are after) to each element of the basis. The result will be again a basis of a k-subspace of To see this as a linear action, read on on Plucker coordinates.

30 May, 2010 at 9:06 |

I see now. I had the wrong thing in mind. I was thinking of a subspace of V as consisting of a set of vectors with fewer coordinates, which is not correct (though it is isomorphic to such a vector space).

Sorry for the pointless reply, but at least you know someone is reading. 🙂