## Intermediate Group Theory (1)

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. An action of G on a set X is a group homomorphism $\phi : G \to S_X$, where $S_X$ is the group of all bijections X → X. The set X, together with the underlying action of G, is called a G-set.

Symbolically, the following notation is convenient. If $g\in G$ and $x\in X$, then write “g·x” or “gx” for the element $\phi(g)(x) \in X$. Notice that the following conditions are satisfied.

• $ex = \phi(e)(x) = id_X (x) = x$ for any x in X.
• $g_1(g_2 x) = g_1(\phi(g_2)(x)) = \phi(g_1)(\phi(g_2)(x)) = \phi(g_1 g_2)(x) = (g_1 g_2)x$ for any $g_1, g_2 \in G, x\in X$.

[ 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 (gx) → gx, such that: (i) exx and (ii) g1(g2x) = (g1g2)x. Then there is a group homomorphism $\phi : G \to S_X$ such that $gx = \phi(g)x$. The proof is as follows.

• For each $g \in G$, the map $X \to X, x \mapsto gx$ is bijective:
• injectivity follows from $gx = gy\implies g^{-1}(gx) = g^{-1}(gy) \implies (g^{-1}g)x = (g^{-1}g) y \implies ex = ey \implies x=y$;
• surjectivity follows from $g(g^{-1}x) = (gg^{-1})x = x$.
• It follows that we have a map $G \to S_X$, where $g\mapsto (x \mapsto gx)$. Condition (ii) then implies that this is a homomorphism. ♦

Symbolically, this means that associativity holds for a word like: $g_1 g_2 \dots g_n x$, where $g_1, g_2, \dots, g_n \in G$ and $x\in X$,

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

## Examples of Group Actions

1. The symmetric group Sn acts on the set {1, 2, …, n} naturally.
2. Any group G acts on the underlying set of G itself by left multiplication, i.e. underlying map × G → G is exactly group product, or g is taken to the permutation taking $x\in G \mapsto gx\in G$.
3. Related  to (2), G also acts on itself by right multiplication of g-1, i.e. g is taken to the permutation $x\in G\mapsto xg^{-1}\in G$.
4. One can compose (2) and (3) since they commute: this gives the conjugancy map $\phi_g : G \to G, \phi_g(x) = gxg^{-1}$. We’ll have ample chance to use this later since $\phi_g$ is in fact an automorphism of G.
5. 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 $(g, g'H) \mapsto gg'H$.
6. Let X be the set of all subgroups of G. Each $g\in G$ and subgroup $H\le G$ give a conjugate $gHg^{-1} \le G$. This gives an action of G on X.

Here’s an example of an application: we shall prove that the symmetric group A5 of order 60 has no subgroup of order 20. To show this, use the classical result that A5 is a simple group (i.e. it has no normal subgroups except {e} and itself). Now if $H\le A_5$ is a subgroup of index 3, then A5 acts on the set of left cosets A5/H, which has 3 elements. This gives a homomorphism $A_5 \to S_3$. The kernel is a normal subgroup of A5 (which is simple) and so must be the whole group A5. So A5 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. The stabiliser group (or isotropy group) is defined by $G_x = \{g\in G : gx = x\}$.

For $x, y\in X$, we write x ~ y if there is a $g\in G$ such that gx = y. If this holds for all x, y, we say the group is transitive.

Some important properties need to be proven.

1. 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 $x = ex = g^{-1}(gx) = g^{-1}x$.
2. The relation x ~ y is an equivalence relation.
• reflexive: xx since exx;
• symmetric: if xy then gxy so $g^{-1}y = g^{-1}gx = ex = x$;
• transitive: if xy and yz then gxy and g’yz, so z = (g’g)x.

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 $G\to G, x \mapsto gxg^{-1}$ is bijective, the resulting subgroups H and gHg-1 have the same cardinality. So we could consider the action of G on the set Xm 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 fG → 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 Gx 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 $H = G_x$ be its stabiliser group. We’ll map $f : G/H \to X$ via $gH \mapsto gx$. Let’s prove it’s an isomorphism of G-sets:

• well-defined and injective: $gH = g'H \iff g^{-1}g' \in H=G_x \iff g^{-1}g'x = x \iff gx = g'x$;
• surjective: obvious since action of G on X is transitive;
• homomorphism: $f(g'(gH)) = f((g'g)H) = g'gx = g'\cdot f(gH)$. ♦

Thus, any G-set X can be pictorially represented as a union of orbits: This entry was posted in Notes and tagged , , , , , , . Bookmark the permalink.