Given a group *G*, we wish to find out more about its properties. Questions include: what subgroups does it have? And normal subgroups? How many elements of order *m* does it have (where *m* must divide the order of *G* if the latter is finite)? It turns out the most fruitful way of looking at this problem is by concretely representing the group as the set of symmetries of some object, then looking at the structure of the object.

For finite groups, some possibilities include:

- set of permutations on a finite set;
- set of automorphisms of some group
*H*(i.e. isomorphisms*H*→*H*); - set of invertible linear maps on a vector space.

We’ll be looking at the first two cases for now. Though the third case (representation theory) is actually much more interesting, it requires significantly more background to fully comprehend.

## Group Actions

Let *G* be a group.

Definition. Anactionof G on a set X is a group homomorphism , where is the group of all bijections X → X. The set X, together with the underlying action of G, is called aG-set.

Symbolically, the following notation is convenient. If and , then write “*g**·x*” or “*gx*” for the element . Notice that the following conditions are satisfied.

- for any
*x*in*X*. - for any .

[ The above notation does get a bit hairy (and will get worse), so if you’re new to this, do make sure everything works out. Or better yet, see if you can derive these results on your own. Diagrammatically, we have the following. ]

It turns out the above two conditions suffice to determine the original homomorphism. To put it explicitly, suppose we have a map *G* × *X* → *X*, denoted (*g*, *x*) → *gx*, such that: (i) *ex* = *x* and (ii) *g*_{1}(*g*_{2}*x*) = (*g*_{1}*g*_{2})*x*. Then there is a group homomorphism such that . The proof is as follows.

- For each , the map is bijective:
*injectivity*follows from ;*surjectivity*follows from .

- It follows that we have a map , where . Condition (ii) then implies that this is a homomorphism. ♦

Symbolically, this means that associativity holds for a word like:

, where and ,

so the brackets do not matter as long as the last “character” in the word belongs to *X*.

## Examples of Group Actions

- The symmetric group
*S*acts on the set {1, 2, …,_{n}*n*} naturally. - Any group
*G*acts on the underlying set of*G*itself by left multiplication, i.e. underlying map*G*×*G*→*G*is exactly group product, or*g*is taken to the permutation taking . - Related to (2),
*G*also acts on itself by right multiplication of*g*^{-1}, i.e.*g*is taken to the permutation . - One can compose (2) and (3) since they commute: this gives the
**conjugancy**map . We’ll have ample chance to use this later since is in fact an automorphism of*G*. - Let
*H*≤*G*be a subgroup and take the set*G*/*H*of left cosets*gH*. Then*G*acts on this set*G*/*H*via . - Let
*X*be the set of all subgroups of*G*. Each and subgroup give a**conjugate**. This gives an action of*G*on*X*.

Here’s an example of an application: we shall prove that the symmetric group *A*_{5} of order 60 has no subgroup of order 20. To show this, use the classical result that *A*_{5} is a simple group (i.e. it has no normal subgroups except {e} and itself). Now if is a subgroup of index 3, then *A*_{5} acts on the set of left cosets *A*_{5}/*H*, which has 3 elements. This gives a homomorphism . The kernel is a normal subgroup of *A*_{5} (which is simple) and so must be the whole group *A*_{5}. So *A*_{5} acts trivially on the set of left cosets, which is impossible (why?).

## Properties of Group Actions

Let’s look at a *G*-set *X* (i.e. a set equipped with an action of *G*). One can look at either the group or the set.

Definition. Thestabiliser group(orisotropy group) is defined by .For , we write x ~ y if there is a such that gx = y. If this holds for all x, y, we say the group is

transitive.

Some important properties need to be proven.

*The stabiliser group is a group*(as implied by the name) :- by definition it contains the identity and is closed and multiplication;
- for inverse, we have .

*The relation x ~ y is an equivalence relation*.- reflexive:
*x*~*x*since*ex*=*x*; - symmetric: if
*x*~*y*then*gx*=*y*so ; - transitive: if
*x*~*y*and*y*~*z*then*gx*=*y*and*g’y*=*z*, so*z*= (*g’g*)*x*.

- reflexive:

By property 2, each *G*-set *X* can be partitioned into a disjoint union of equivalence classes (known as **orbits**), each of which is a transitive *G*-set. For example, in example 6 above, since the conjugation map is bijective, the resulting subgroups *H* and *gHg*^{-1} have the same cardinality. So we could consider the action of *G* on the set *X _{m}* of all subgroups of order

*m*(for various

*m*), although this action is usually not transitive.

In short, it suffices to look at transitive *G*-sets and it turns out they can be nicely classified!

Theorem. Every transitive G-set is isomorphic to G/H for some subgroup H of G.

First, we explain what is meant by “isomorphism” of *G*-sets.

In set theory, we have sets and functions *f* : *S* → *T* from one set to another. In group theory, we’ve already seen that we should be looking at group homomorphisms *f* : *G* → *H*, i.e. functions which preserve the underlying algebraic structure (symbolically, we write *f*(*x*)*f*(*y*) = *f*(*xy*)). What about *G*-sets?

Likewise, when we consider functions *f* : *X* → *Y* between *G*-sets, we should focus on those functions which respect the action of *G*. Thus, a *homomorphism* of *G*-sets is just a function *f* : *X* → *Y* such that

*f*(*gx*) = *g*·*f*(*x*) for any *g* in *G*, *x* in *X*.

An isomorphism is then a bijective homomorphism as before, and intuitively it just means the two *G*-sets have identical structures up to relabelling.

**Proof of above theorem**. Pick any element x in X and let be its stabiliser group. We’ll map via . Let’s prove it’s an isomorphism of G-sets:

- well-defined and injective: ;
- surjective: obvious since action of
*G*on*X*is transitive; - homomorphism: . ♦

Thus, any *G*-set *X* can be pictorially represented as a union of orbits: