Should be: $u\in S_{xy}$. ]]>

I suppose one can get away with equivalence classes in this context but here is another example of the utility of the Orbit-Stabilizer theorem. Namely the Sylow theorems: let G be a finite group of order p^a*m, where p does not divide m and p prime.

Let S denote the collection of all subsets of G of size p^a. S has size p^a*m choose p^a. It is (if I may dare say this to you Tim) a relatively simple exercise in combinatorics to show that p does not divide |S| if p is prime. Now G acts on S by right multiplication. Since p does not divide |S|, there must be an orbit O = G(X) of the G-action of size not divisible by p. But by the Orbit-Stabilizer theorem |O| divides |G| and hence |O| divides m. Let G_X be the stabilizer of the subset X in O. Again by the Orbit-Stabilizer theorem, |G_X| = p^a*k, where k = m/|O|. On the other hand, for an element x in X, the set {xg: g in G_X} is a subset of X so |G_X| is at most |X| = p^a. Hence, G_X is a subgroup of G of size p^a, i.e., we have shown the existence of a Sylow p-subgroup.

Once can continue further: if C denotes the set of G conjugates of G_X (note now we considering a different action), then by considering the sub-action of G_X on C, one can show that G_X is the only fixed point (again using the Orbit-Stabilizer theorem) and hence all Sylow p-subgroups are conjugate. This also yields that the number of Sylow p-subgroups is congruent to 1 mod p, again by the O-S theorem, each orbit has size 1 or a power of p etc.

Of course there are other proofs of the above theorems but to my taste this seems quite elegant and one can see for instance why the number of Sylow p-subgroups is 1 mod p.

]]>Having just written that, I read what you wrote more carefully, and I now see that you do in fact use the orbit-stabilizer theorem to save a tiny bit of time, so I take back slightly what I’ve just written. In the argument I suggest above, you have to argue that if the sequence is equal to a cyclic permutation of itself, then we get some such that for every , and since every element of is a multiple of it follows that all the are equal. Your argument is that the stabilizer is a subgroup of and hence has size 1 or p, from which it follows that the orbit has size 1 or p.

That still leaves me with the feeling that Cauchy’s theorem isn’t really an application of the orbit-stabilizer theorem: it’s more that a small part of the proof can be made slightly quicker (at the cost of making the argument look a bit more mysterious and magical).

]]>I was going to write about that argument (which appears in one of the questions on the Cambridge examples sheet mentioned in the post) in a further post on group actions, but in the course of thinking about it, I came to the conclusion that it was a bogus application, in the sense that essentially the same argument can be written out more simply with no mention of the orbit-stabilizer theorem. It goes like this. (Most of it is the same as the argument as you’ve written it.)

1. The set S has size .

2. If and are not all equal, then all cyclic permutations of belong to and are distinct. (This depends on being prime.)

3. On the other hand if then all cyclic permutations are trivially equal.

4. Hence, if we define two elements of to be equivalent if one is a cyclic permutation of the other, then the equivalence classes all have size 1 or p.

5. Since belongs to , at least one equivalence class has size 1.

6. Therefore, there must be another equivalence class of size 1, which implies that there is some such that .

In this proof there is no need to mention the action, though I suppose implicitly it’s there. Similarly with orbits. But the real point is that there is absolutely no need to use the orbit-stabilizer theorem, since we can see directly what the sizes of the orbits are.

]]>To get the theorem from this by counting, we need for all . (Admittedly, once you have that it is hardly worth bothering with the argument we are in the middle of.) Then we proceed as follows:

]]>Another thought: You mention that the way to show that two sets are the same size is to find a bijection between them. I was trying to think how other proofs, such as proof 4, show that the two sets are the same size. The way it does it is by showing that the sets are equal to a pair of sets we know to be equal, a subgroup and one of its cosets. However, it is not much different, because we would show a bijection to prove that a subgroup is equal in size to one of its cosets, so all that is different here is that the use of the technique is buried in the proof of a result that we rely on.

A typo: Where you wrote “What we’d really like is to define an element of g”, I believe this should be “What we’d really like is to define an element of G”.

*Thanks — fixed. (And I like that bijection.)*

Consider the set S, a subset of GxGx…xG, the p-fold Cartesian product of G, where S = {(x_1, x_2, …, x_p): x_1*x_2*…*x_p = 1}. S is non-empty and in fact it is fairly easy to see that |S| = |G|^{p-1}.

Define an action of Z/p on S, where Z/p acts via cyclic permutations (as a subgroup of the symmetric group S_p). Then by the Orbit-Stabilizer theorem (and the fact that p is prime) every orbit of this action on S has size either 1 or p. An orbit of size 1 is a fixed point, and an element of S is fixed for all cyclic permutations if and only if it looks like (x,x,…,x). The collection of fixed points in non-empty; the element (1,1,…,1) is fixed. If there are ‘k’ fixed points, where k is at least 1, and ‘m’ orbits of size p, then we have

|S| = |G|^{p-1} = k + m*p

The left side is divisible by p so that implies that k>1. So there is a non-trivial element (x,x,…,x) in S (in fact at least p-1 such elements) such that x^p = 1 which is the required element of order p.

]]>The picture looks correct to me too.

]]>I am not sure what you mean. To get (iii) from (i). Rotate \pi/4 about the central axis (connecting the midpoints of the top face and the bottom face) and then \2pi/3 about the diagonal of the cube through . To get (ii) from (i) rotate \pi/4 about the central axis and reflect along the plane containing and normal parallel to (this is still a reflection… to see this, notice that the det of the operation is +1.(-1)=-1)

]]>*Many thanks — fixed now.*

I think you probably have to invoke orientations somehow, even if you disguise them. One way one might disguise them is to say something like this: imagine that you are walking from b to a in the first picture; then c is over to the right; the same must be true after a rotation; so once you’ve fixed a and b then you’ve fixed c and hence the whole linear map. (This is a bit like the right-hand rule applied to 0, a, b and c.)

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