## Group actions III — what’s the point of them?

Somebody told me recently that a few years ago they had a supervision with a colleague of mine (who shall remain nameless, but he or she is an applied mathematician) and asked what the point of group actions was. “I have absolutely no idea,” was the response, and the implication that one might draw from it was apparently intended.

No pure mathematician could hold such a view. I’ve stated a few times that group actions tell you a lot about groups. In this post I want to try to explain why that is, though there is far more to say than I am capable of explaining, let alone fitting into one blog post.

Several proofs that use group actions seem to depend on almost magically coming up with an action that just happens, when you analyse it the right way, to tell you what you wanted to know. I am not an algebraist and do not have a good all-purpose method for finding actions to prove given statements. I don’t rule out that such a method might exist, at least for reasonably simple statements, and would be interested to hear from anybody who thinks they can usefully add to what I have to say.

In the absence of a systematic method for coming up with group actions for specific purposes, the next best thing is to have a good supply of actions that one can simply look through in the hope of finding something useful. Very often, features of the problem will enable you to rule out many actions as unhelpful, which reduces the brute-force search to something manageable. Although this isn’t as systematic as the general method I described when discussing how one might come up with the notion of a quotient group (which was, roughly speaking, to pretend you’ve found the object you’re looking for and say so much about it that you end up determining it uniquely), it comes fairly close.

A supply of group actions.

I’ll divide these into intrinsic and extrinsic actions. By an intrinsic group action I mean one that is defined in a general abstract way in terms of the group itself, and by an extrinsic group action I mean one that is defined in terms of other objects that come into a particular description of the group.

Rather than attempt to make these definitions more precise, let me illustrate them with all the examples I can think of. There are bound to be many more, so again I’d be grateful if people want to suggest further examples.

Extrinsic group actions.

I’ll start with some extrinsic examples. A common way for these to arise is if the group is given to you as a group of symmetries, or more generally a group of transformations of some set $X$. Of course, that already gives us an action before we even start: the group acts on the set $X$. However, it potentially gives us many more actions.

As an example, let’s consider the symmetry group of a cube. As defined, this group acts on the entire cube, but we soon spot that certain subsets of the cube are invariant under the action — this means that if you take a point in the subset and apply one of the symmetries you will get another point in the set. For instance, the set of centres of faces is invariant, since every symmetry takes the centre of a face to the centre of some other (or possibly the same) face. So, writing $G$ for the symmetry group of the cube, we find that $G$ acts on the set of centres of faces.

Does this tell us anything interesting about $G$? Well, it tells us that $G$ is a subgroup of $S_6$, but that’s only mildly interesting. One of the reasons for the limited interest is that this action is faithful — that is, different symmetries permute the centres of the faces in different ways — so regarding $G$ as acting on the centres of the faces rather than on the cube feels a bit too close to the original description of $G$.

However, things get a bit more interesting if we spot that opposite points go to opposite points. This means that we can partition the six centres of faces into three sets, each consisting of two opposite points, and think of $G$ as acting on the set that consists of these three pairs of points. Here are two other ways of saying what I’ve just said. We could regard two face centres as equivalent if they are opposite each other and say that $G$ acts on the equivalence classes. Or if we wanted to be more geometrical, we could join the opposite centres by line segments and say that $G$ acts on the three line segments.

This time the action isn’t faithful, since a half turn takes each line segment (to use the geometrical viewpoint) to itself, as does a reflection in the plane through the centre that’s parallel to one (and hence two) of the faces. Earlier, I implied that that would be more interesting, so let’s try to think what it could tell us.

Another point that I’ve been trying to put across in these posts is that

• kernels of homomorphisms are normal subgroups are kernels of homomorphisms
• and yet another is that

• a group action is a homomorphism from a group to a group of transformations of a set.
• Now a group action is faithful if and only if the homomorphism to the group of transformations is an injection, which (as is easy to show and I’m sure you’ve seen) is the case if and only if the kernel of the homomorphism consists of just the identity element. So if we put the two points together, then we get a third point, and it’s a useful one:

• a non-trivial action that is not faithful gives rise to a non-trivial normal subgroup.
• Why is this? Well, the action is a homomorphism $\phi$. The kernel of $\phi$ is a normal subgroup. If the action isn’t faithful, then its kernes isn’t the identity, and if the action doesn’t just send every element of the group to the identity (that’s what I mean by the action being non-trivial) then its kernel isn’t the whole group.

Now let’s go back to the example we had of an unfaithful action. It was the action of the symmetry group of the cube on the set of three line segments that join opposite centres of faces. This has a non-trivial kernel, which is generated by the three reflections that keep two of the lines fixed and flip the other one round. This group is $C_2\times C_2\times C_2$, so we find that the symmetry group of the cube has $C_2\times C_2\times C_2$ as a normal subgroup. Also, the isomorphism theorem tells us that the quotient group is equal to the image of the homomorphism of which we have just calculated the kernel. Since it is possible to permute the three segments however you like, this image is $S_3$, so the quotient is $S_3$. (Incidentally, it may be worth my mentioning a mistake that you could conceivably make here, which is that you can’t multiply both sides of $G/H=K$ by $H$. In our case, it is not true that the symmetry group of the cube is isomorphic to $C_2\times C_2\times C_2\times S_3$. Something of the kind is true, but the “product” is more complicated than the straightforward group product.)

We’ve therefore used the action of the symmetry group on the three line segments to obtain information about the group itself. It’s not that hard to show that the subgroup generated by those three reflections is closed under conjugation, but if you want to know that it’s a normal subgroup, then it’s much easier just to observe that the group acts on those three line segments and that the subgroup is precisely the kernel of that action (when the action is considered as a homomorphism).

What other actions of this group might be interesting? Well, we quite like actions on small sets, since those are more likely not to be faithful. So one thing we could try to do is look for small orbits (of the action on the entire cube). The reason I said that is that if we put a point $x$ into the set on which we’d like $G$ to act, then obviously we have to put all the other points in the orbit of $x$. And if you think about it, that’s exactly what we did with the centres of the faces: once one of them was in there, we had to put in the others too.

That suggests that we can’t do much better, if we’re going for small sets, than what we’ve done already. (Well, we could make $G$ act on the centre of the cube, but that’s a trivial action, so uninteresting.) But let’s remember the second idea we had, which was to observe that certain relationships between points are preserved by the transformations in $G$. In the case of the previous action, we noted that opposite points go to opposite points, which led us to split them up into three pairs.

We can do exactly the same with a different set of segments, this time the four segments that join a vertex of the cube to the opposite vertex. It’s clear that $G$ acts on these four main diagonals, so what does that tell us about $G$?

Let’s think first of all about whether the action is faithful. Is there any symmetry of the cube that is not the identity but that nevertheless preserves all four diagonals? Yes there is: the function that takes each point to the opposite point, which is sometimes called reflection through the centre. (If your cube is sitting in $\mathbb{R}^3$ with its centre at the origin, then the symmetry is the map that takes each vector $x$ to $-x$.) A bit of further thought reveals that this is the only map, apart from the identity, that preserves the four diagonals. To see why, note that each vertex must go either to itself or to the opposite vertex. Since neighbouring vertices must also go to neighbouring vertices, once you’ve decided where one vertex goes, you’ve decided where all its neighbours go, and all their neighbours, and all their neighbours — which has ended up including all vertices.

So the kernel of the action of $G$ on the four diagonals is the cyclic group $C_2$. What about its image? That is, which permutations of the four diagonals can arise if you perform a symmetry of the cube?

A preliminary question we might ask ourselves is whether there are any restrictions at all. The way I like to approach a question like this is to ask myself whether there are any relationships that sometimes hold and sometimes don’t. That’s a bit of a vague statement (deliberately, because the idea works quite generally), but we get a good example of it if we go back to the set of eight vertices. Why can’t we permute them arbitrarily? Because some pairs of vertices are next to each other and others aren’t. Some pairs of vertices are opposite each other and others aren’t. These relationships are preserved by the symmetries of a cube, so we can’t, for instance, take two neighbouring vertices to two vertices that are not next to each other.

This doesn’t seem to work with diagonals, since for every pair of diagonals you can find two opposite edges such that the two diagonals join their two end points (making, with the edges themselves, a kind of bow-tie shape). That doesn’t prove that we can get all permutations of the diagonals, but it starts to suggest it, so let’s look in the other direction.

If you want to show that you can get an arbitrary permutation, then a good way of doing it is to show that you can get all transpositions, since these generate the symmetric group. So we find ourselves asking this: given two main diagonals of the cube, is there a symmetry that exchanges those diagonals and sends the other two diagonals to themselves? You might like to draw a picture here, but let me try to explain in words why the answer is yes. You take the plane that contains the other two diagonals and reflect in it. That obviously keeps the other two diagonals where they were, and it must interchange the two diagonals you want to exchange, since otherwise the map would have to be the identity by the argument we used earlier to identify the kernel of the action of $G$ on the diagonals. (Alternatively, just check that this reflection really does swap the diagonals it’s supposed to swap.)

A reflection isn’t the only way of swapping just two diagonals. Indeed, it can’t be because the kernel of the action is non-trivial. Since we know what the kernel is, we know that minus the reflection we’ve just defined must also swap the two diagonals. A more geometrical way of thinking of it is as a half turn about an axis perpendicular to the plane that contains the two diagonals you want to send to themselves.

We have now shown that the image of the action is the permutation group of the four diagonals, so it is isomorphic to $S_4$. By the isomorphism theorem, we find that $G/C_2=S_4$.

The fact that this particular copy of $C_2$ (consisting of the identity and the map that sends every point to the opposite point) is a normal subgroup of $G$ is easy to see directly, since it is given by the matrix $-I$ (where $I$ is the $3\times 3$ identity matrix), which commutes with all other $3\times 3$ matrices. Because of that, it follows that in this case we can “multiply through” and see that $G$ is isomorphic to $C_2\times S_4$. [Edit: that’s not quite enough of a reason — see Greg Martin’s comment below and my response to it.] Moreover, every symmetry of the cube is plus or minus a rotation, so the rotation group of the cube is isomorphic to $S_4$, a fact we can see another way by noting that the rotation group also acts on the four main diagonals, this time faithfully, and the image of the action is $S_4$.

I invite you to think how you might see that the group of rotations of the cube is isomorphic to $S_4$ without using the concept of group actions. If you see an easy argument, then let me know. But it’s sort of obvious that if you want to see it, then the best way is to find a set of four objects that the rotations permute.

Have we exhausted the interesting actions that we can build out of the action of $G$ on the cube? I’m afraid not. Recall that our techniques so far are these.

(i) Consider the action of $G$ on the orbit of a point (ideally chosen in a nice “symmetrical place” so that its orbit isn’t too big).

(ii) Find an equivalence relation on an orbit with the property that the transformations in $G$ take equivalent points to equivalent points, and look at the resulting action of $G$ on the equivalence classes.

Let’s continue to think about what happens when $G$ is the symmetry group of a cube and the point is a vertex. The orbit of that point is the set of all eight vertices of the cube. What equivalence relations are there on this set with the property that equivalent vertices go to equivalent vertices? There are two trivial ones (one where there is just one equivalence class and one where the equivalence classes are singletons), and one that we have already considered (where two vertices are considered equivalent when they are opposite each other). Are there any more?

Let’s answer this question using the pretend-you’ve-got-it-already technique. So let’s suppose that $\sim$ is an equivalence relation on the set of vertices and that if $v\sim w$ and $T$ is a symmetry of the cube then $Tv\sim Tw$. Given a pair of vertices $(v,w)$, what other pairs of vertices can arise as $(Tv,Tw)$? The answer is that the distance between vertices never changes as a result of a symmetry, and that if the distance between $t$ and $u$ is the same as the distance between $v$ and $w$, then there is a symmetry $T$ such that $Tv=t$ and $Tw=u$.

From this it follows that $\sim$ can depend only on the distance between $v$ and $w$. Incidentally, there are at least two sensible notions of distance here. One is just the physical distance in $\mathbb{R}^3$, but an easier one to take is the number of edges you have to walk along to get from one vertex to the other, so for instance the distance from a vertex to the opposite vertex is 3. It’s this notion of distance that I’ll be talking about in the next paragraph.

Since $\sim$ is transitive, if $v\sim w$ for any two vertices at distance $d$, then we also know that $t\sim u$ whenever you can get from $t$ to $u$ by a series of steps, each of which takes you a distance $d$. For example, if two neighbouring vertices are equivalent, then since you can get from any vertex to any other vertex by walking along edges, it follows that all vertices are equivalent. That isn’t very interesting. But if two vertices that are distance 2 apart (so they are opposite vertices of some face), then that tells you that two vertices are equivalent if they can be reached by a series of diagonal jumps across faces. And if you think about it for a bit, you’ll see that if you start at a vertex $v$, then you can’t reach all other vertices if you take steps of size 2. The set of vertices you can reach forms (the vertices of) a regular tetrahedron, as does the set of vertices you can’t reach. So we’ve found another equivalence relation: two vertices are equivalent if they are an even distance apart (and therefore either 0 or 2 apart), and the equivalence classes can be thought of as two tetrahedra.

If we look at distance 3 then we get back to the partition into pairs of opposite vertices, so that won’t give us anything new. But the tetrahedra should interest us: they’ve given us another action (of $G$ on the set that consists of these two tetrahedra).

Everything I’ve just said applies just as well if I consider the group of rotations of the cube rather than the full symmetry group. Since the full symmetry group is just the product of the rotation group by $C_2$, I’ll concentrate on the rotation group instead, and I’ll use this action to say something about it. Let’s call the rotation group $R$. Since $R$ acts non-trivially on the group of permutations of the two tetrahedra, which is just $S_2=C_2$, we know that we’ve got a surjective homomorphism from $R$ to $C_2$.

Writing that makes me realize that I have some more principles to set out, about using group actions to find interesting quotients.

• Quotient groups are images of homomorphisms are quotient groups.
• Group actions are homomorphisms to groups of transformations of a set.
• Putting those two principles together gives us this.

• Non-trivial unfaithful group actions give rise to non-trivial quotients.
• You might object that this is implicit in what I said before. As I pointed out then, non-trivial unfaithful group actions give rise to non-trivial normal subgroups, so the above principle follows if you simply quotient by the normal subgroups you obtain. But I’m actually saying slightly more than that: if you can identify the image of $G$ under the action, as you often can, then you can actually say what the quotient group is, rather than merely saying that it’s what you get when you quotient by the normal subgroup.

This is actually a general remark about the isomorphism theorem. Why is it useful? One reason is that it gives you a way of identifying quotient groups. If you’ve got some normal subgroup $H$ of a group $G$ and want to describe $G/H$, then a good strategy is to try to come up with a homomorphism defined on $G$ that has kernel $H$. Then the image of that homomorphism is isomorphic to the quotient group. A trivial way of carrying out this programme is to take the quotient map from $G$ to $G/H$, but that’s not very helpful. What one wants to do is to come up with a less abstractly defined homomorphism with kernel $H$. That does two nice things at once: it gives you an insight into why $H$ is normal, and it gives you a description of the quotient.

Anyhow, we’ve got a surjective homomorphism from $R$ to $C_2$, so $R$ has a normal subgroup of index 2 and the quotient is isomorphic to $C_2$ (as it must be since it has order 2). What is the normal subgroup? If you remember that $R$ is isomorphic to $S_4$ it will come as no surprise to learn that the normal subgroup is $A_4$.

Can we see this geometrically? Yes we can. The kernel of this particular action is the set of rotations that send the two tetrahedra to themselves, which, since there are only two tetrahedra, is the set of rotations that sends one of the tetrahedra to itself. In other words, it is isomorphic to the rotation group of a regular tetrahedron. It is fairly easy to show that using rotations you can achieve any even permutation of the vertices of a regular tetrahedron, but you can’t swap just two vertices. So that group is $A_4$.

The general point I’m trying to demonstrate by saying all this is that we can really get inside the group of rotations (or the full symmetry group) of the cube by considering the various different actions that we can define in terms of the original action of the group on the cube. In a systematic way we can come up with several actions, and these actions tell us that the group is isomorphic to $S_4$ (or $S_4\times C_2$) as well us giving us descriptions of various normal subgroups and quotients.

This post is getting long enough that I think I’d better split it in two. But before I finish this part, let’s have a quick look at another group, namely $A_4$. We could treat $A_4$ as the group of rotational symmetries of a regular tetrahedron, but instead let us think of it in the way that it is usually defined — as the group of even permutations of the set $\{1,2,3,4\}$.

Once again we have a group that is defined in terms of an action on a set. So can we use that action to define another action? The orbit of any point is the whole set, so that doesn’t help much. What’s more, given any two (unordered) pairs of elements, there is an even permutation that takes one to the other, so any equivalence relation that is preserved by $G$ would have to be trivial. So the equivalence-relations trick doesn’t work either.

Here’s a set on which $A_4$ acts interestingly: its elements are the three partitions of the set $\{1,2,3,4\}$ into two sets of equal size. (These partitions could be represented in a shorthand notation as {12,34}, {13,24} and {14,23}.) If you do a permutation, then it sends the partition to the corresponding permuted elements. For example, if you do the permutation $(123)$, then it changes {12,34} to {23,14}, which is the same as the partition {14,23}.

It doesn’t take long to see that every cyclic permutation of the three partitions is possible and that if one partition is fixed, then so are the other two. Therefore, the image of this action is the cyclic group $C_3$. Therefore, $A_4$ has $C_3$ as a quotient. What is the kernel of the action? We’ve just seen that a 3-cycle has a non-trivial effect, but if we try a transposition pair, we find that it sends each partition to itself. For instance, if we take the pair $(12)(34)$, then it sends {12,34} to {21,43}, {13,24} to {24,13} and {14,23} to {23,14}. In each case, I’ve just given a different notation for the partition we started with. It follows that the kernel is the identity together with the three transposition pairs. Therefore, this group, which is isomorphic to $C_2\times C_2$, is a normal subgroup of $A_4$ (a fact that you have almost certainly been shown, but quite possibly in a different way).

The idea of considering partitions might seem as though it sprang from nowhere. That is a common illusion in mathematics, which arises when an author neglects to mention the thoughts that provoke an idea. I confess that I’ve just done that, so let me give two other ways of coming up with essentially the same action, the first of which is the way that I used without saying so.

If we recall that $A_4$ is isomorphic to the rotation group of a regular tetrahedron, then we can apply the same kinds of tricks that we used for the cube. And we find that the group acts on the three lines that connect midpoints of opposite edges. (I’ve mentioned this action before, I think.) Again, if one line is fixed, then the other two must be as well. So now we have a geometrical way of seeing that $A_4$ has $C_3$ as a quotient. The kernel is given by the identity and three half turns about the lines just specified. We could even think of the tetrahedron as one of the ones made out of alternate vertices of a cube, in which case the three lines are the coordinate axes. I then wanted to forget the geometry, so I needed some kind of object made out of the elements of the set $\{1,2,3,4\}$ that could stand in for the lines. Since each line joins the midpoints of two opposite edges, and two opposite edges have disjoint sets of vertices at the end of them, it was natural to associate lines with partitions of the set $\{1,2,3,4\}$ into two sets of size 2. Thus, for instance, if we think of the numbers 1 to 4 as labels of vertices, we label the edges by pairs of such numbers, and the line joining the centre of the edge 13 to the centre of the edge 24 will be represented by the partition {13,24}.

A second way of thinking about it is one that I want to touch on only briefly because it is better thought of as an intrinsic action. However, it seems wrong not to mention it. The group $A_4$, like all groups, acts on itself by conjugation. You may remember that I discussed towards the end of a post on permutations how to conjugate permutations. The quick thing to remember is that if you want to calculate the cycle representation of $\sigma\rho\sigma^{-1}$ then you take the cycle representation of $\rho$ and apply $\sigma$ to everything inside it. (This is because if $\rho(x)=y$ then $\sigma\rho\sigma^{-1}(\sigma(x))=\sigma(y)$.)

A corollary of this observation is that conjugation preserves cycle type. But as I’ve just said, every group $G$ acts on itself by conjugation: with each element $g$ of $G$ we consider the function $\phi(g):G\to G$ that takes $h$ to $ghg^{-1}$ (which gives us a homomorphism $\phi$ from $G$ to the group of automorphisms of $G$). Since conjugation preserves cycle type, if $G$ is a group of permutations, then any two permutations in the same orbit of the conjugation action must be of the same cycle type.

Now this observation is particularly nice when it comes to $A_4$, because there is one cycle type, namely that of $(12)(34)$, that is shared by only three permutations. So that gives us a strong chance of finding a non-trivial homomorphism, as indeed we do. The conjugation action of $A_4$ permutes the three transposition pairs (which, you cannot fail to have noticed, correspond in an obvious way to partitions of $\{1,2,3,4\}$ into two sets of size 2), and the rest of the discussion is basically the same as the one I gave just now.

As a final thought, let’s consider what happens if we look at the action of $A_4$ on the set of 3-cycles. There are eight 3-cycles, which turn out to split into two conjugacy classes. (Why two? Well, if you want to conjugate $(123)$ so that it becomes $(132)$, then the obvious way of doing it is to use a permutation $\sigma$ that takes 1 to 1, 2 to 3 and 3 to 2. But the only such permutation is $(23)$, which is odd. That’s not quite your only option because $(132)=(321)=(213)$, but the other two possibilities fail for similar reasons — as they must, because they are derived from the first by composing with an even permutation.) So if we take one of these conjugacy classes, say the one that contains $(123), (142), (134)$ and $(243)$, we find that $A_4$ acts on it. If you look in detail at what that action does, you find that it is faithful: you can get any even permutation of those four 3-cycles. In fact, you’ll find a rather nice pattern: if you associate each 3-cycle with the element that it doesn’t do anything to — so, for example, you associate $(134)$ with 2 — then the effect of conjugating by an even permutation is to permute the 3-cycles according to that correspondence. For example, the permutation $(123)$ fixes $(123)$ when you conjugate, and sends $(243)$ to $(134)$ to $(142)$ to $(243)$, which in terms of the points missed out is taking 1 to 2 to 3 to 1. However, even if this pattern is nice, it doesn’t help us (as far as I can see) work anything further out about the group structure of $A_4$.

As you might suspect the pattern has a geometrical explanation. If we label the vertices, then we can think of a 3-cycle as a face with an orientation. Roughly speaking, an orientation is a little circle with an arrow on it that tells you which way of going round you regard as clockwise (which isn’t obvious in the abstract, since you can look at a plane from two different sides). So the 3-cycle $(123)$ represents the face with vertices 1, 2 and 3, together with the instruction that if you want to spin round clockwise then go the same way round as the path from 1 to 2 to 3 to 1. Each face has two possible orientations, so there are eight oriented faces. Moreover, the rotations of a regular tetrahedron act in an obvious way on oriented faces, and can never take an oriented face to the same face with the opposite orientation. So we have a geometrical interpretation of the fact that the eight 3-cycles split up into two conjugacy classes.

Summary.

Not for the first time in this series of posts, I have been surprised by how much there is to say, even when I thought I was restricting the scope considerably. This post will have a second part in which I discuss intrinsically defined actions, and I’ll try to provide some kind of summary too — since otherwise you may feel that all I’ve really done is discuss a few examples in detail. Of course, I have done that, but there are several general messages embedded in the discussion, of which perhaps the most important is the amount that you can learn about a group from the possible concrete ways in which it can be defined, and in particular from the actions that can be derived from those definitions.

### 19 Responses to “Group actions III — what’s the point of them?”

1. plm Says:

Finding group actions reduces often to finding subsets (we take permutations of the ambient set leaving the subset invariant), which is done using the axiom of separation. Then what subsets are interesting is a matter of taste (regular polygons in R^2/polyhedra in R^3 are common choices).

Other axioms of set theory yield other straightforward methods for constructing group actions. Taking cartesian product yield an action of Sym(X) on XxY, and you may have an action of Sym(X) on Y already. Actions on X pass to the power set P(X) (passing from the notation f(x) to f(A), x an element, A a subset), therefore quotients by equivalence relations also yield actions (taking the partition induced by a relation, a subset of P(X)).
These constructions are functors.

For instance Galois groups act on k, therefore on k^n, then on projective spaces, then on projective varieties, and on their quotients, and further functors like cohomology (which themselves are built in smaller (set-theoretic) steps) yield actions.

• plm Says:

I should have said k a number field, and we take the Galois group of k/Q.

• plm Says:

Thinking about it, it appears that most existence results (in particular ZF axioms) yield group actions, and then we naturally study their properties: trivial, faithful, transitive, etc.? And we have a supply of tools which we patiently apply, the orbit-stabilizer theorem, the kernel is a normal subgroup, etc.

This leads to a process of research bit by bit: constructing objects and studying their properties in steps, which can be combined and as a whole provide all possible approaches to a problem.

We form ever larger strategies, like having a family of group actions each extracting some information from a group.
The Langlands program and other large projects come to mind of course.

• gowers Says:

I haven’t fully thought through what I am going to say about intrinsically defined group actions, but these remarks will be very helpful when I do (though they make me realize that I don’t have a hope of doing justice to the subject).

• plm Says:

A further thought is that the whole idea of “geometry” is subsumed by group actions in a sense. This was remarked by Klein in one form. Most notions of space have homogeneous models (in a broad nonrigorous or rigorous sense), manifolds, Cartan geometries, etc., just because it is easy to put useful group structures on most sets, there are so many.

And then local models are glued along groups also, for instance the fundamental group of a space glues open subsets of its universal cover and combined with the local group it gives informative and universal (and natural and canonical) ways of going from one point to another.
This idea taken broadly is homotopy theory, in a sense.

Computationally this is very useful, there is a whole industry built around torus actions in symplectic/algebraic geometry (in mechanics).

And homotopy also works in algebraic geometry (with the help of heavy machinery).

Also, results like Hilbert’s 5th problem show that there are deep insights that can be neatly summarized (though many have said that this particular result is not very useful -but this seems to be changing).

2. Qiaochu Yuan Says:

A simple application of group actions is the following. If you present a group by generators and relations, it’s straightforward to use those relations to prove that two given words in the generators are equal. But how do you prove that two words in the generators are not equal?

A simple general method to do this is to find an action of the group such that you can prove using other means that the two words are not equal. For example, if you can find a group action with respect to which the two words have different numbers of fixed points, then they’re clearly not equal. Similarly if you can find a linear representation with respect to which the two words have different traces or determinants. I use essentially this idea in this math.SE answer to show that the product of two elements of fixed finite orders in a group need not have finite order in general.

(The relationship between a generators-and-relations presentation of a group and its actions is closely analogous to the relationship between syntax and semantics of a logic. There the analogous discussion is this: it is straightforward to show that one statement implies another syntactically, just by writing down a derivation of one from the other, but hard to show that one statement does not imply another. A simple general method to do the latter is to find a model in which one statement is true and the other is false.)

• Terence Tao Says:

The duality between “intrinsic” presentations and “extrinsic” representations can also be thought of category theoretically: “intrinsic” descriptions of an object X tend to focus on morphisms into X (e.g. a presentation of a group G is basically a morphism from the free group to G) while “extrinsic” descriptions of X tend to focus on morphisms out of X (e.g. a representation of a group G is basically a morphism from G to something else, such as a permutation group or linear group). When viewed that way, it is not so surprising that intrinsic and extrinsic descriptions of an object play complementary roles.

3. Greg Martin Says:

You say “Because of -that-, it follows that in this case we can ‘multiply through’ and see that G is isomorphic to C_2 x S_4.” As far as I can tell, “that” refers to the fact that C_2 is a normal subgroup of G. Surely that normality itself is not enough to conclude that G is the direct product?

• gowers Says:

“That” actually referred to the fat that everything in the copy of $C_2$ commutes with everything in $G$. Unfortunately, I was still wrong (I had one of those moments of anxiety that I stupidly suppressed instead of properly dealing with), since it is not true, for example, that $O(2n)$ is a direct product of $SO(2n)$ with $\{1,-1\}$. So I should have said explicitly that multiplication by $-I$ multiplies the determinant by -1 in odd dimensions, and that it is that that allows us to say that we’ve got a direct product, since it gives us a way of choosing an element from each coset.

4. Gabriel Verret Says:

The action of A_4 on the three partitions of the set {1,2,3,4} into two sets of equal size has kernel the Klein group and the quotient is cyclic of order 3, not S_3.

• gowers Says:

Thanks very much. Where I said, “It doesn’t take very long to see that every permutation is possible,” I would have done better to say, “It doesn’t take very long to see that not every permutation is possible.” I have corrected it now (which, amongst other benefits, means that I don’t present a statement that flatly contradicts the orbit-stabilizer theorem).

5. Window Actions « Log24 Says:

[…] A post by Gowers today on group actions suggests a review. […]

6. schlafly Says:

• gowers Says:

I don’t have strong views on that, but it’s undeniable that group actions tell you a lot about groups.

• schlafly Says:

To me, groups are interesting in all the other branches of mathematics because objects have symmetries in those branches, and group theory helps us understand those symmetries better. Yes, group actions tell you a lot about groups, but that applied mathematician may not care about groups for their own sake. But analysis, algebra, geometry, and even applied math have objects with symmetries, and that is the broad appeal of groups.

7. Kannappan Sampath Says:

(Finite) Group actions are useful not just in unraveling the group structure but also they offer very useful counting schemes, cf. the Polya and Szego’s Theory of Enumeration. For other deeper applications, one could consider reading Applied Finite Group Actions by Adalbert Kerber.

8. Dr Shirleen Stibbe Says:

And here’s another justification for teaching Group Actions – you can win loads of pints of lager from your students!

This is a game I played with students at a revision session on Group Actions at an Open University Pure Mathematics summer school.

9. Martin Says:

Hello Mr Gowers. First of all, thank you very much for your great “blogging”! Your posts about group-theory related topics are particularly good.

When you write
“Now a group action is faithful if and only if the homomorphism to the group of transformations is an injection, which (as is easy to show and I’m sure you’ve seen) is the case if and only if the kernel of the homomorphism consists of more than just the identity element. ”
I guess it should be: “if and only if the kernel of the homomorphism consists of \emph{no} more than just the identity element”?

Many thanks — corrected now.

10. macharia cyprian Says:

IF U DON MIND I REALLY WONNA KNOW MORE OF PRIMITIVITY OF THE DIHEDRAL GROUPS WHEN THEY ARE ACTING ON THE VERTICES OF A REGULAR N-GON