be a Jordan block. Let be the standard basis vectors, i.e. the -th component of is . Note the action of on the basis vectors: for each (where we take ).

Let’s look at in the disjoint cycle notation. One of the cycles is , where is the order of modulo . What about the others? In fact, every cycle is of the form for some . So every cycle has length . And the number of cycles is .

Let’s look at the sign of . There are cycles each of length , and each cycle of length can be written as a product of transpositions, so

where the last equality is obtained by looking at the cases when is even/odd (and using the fact that is even).

Remark. is saying that the number of transpositions in has the same parity as the number of disjoint cycles in it.

Hence , being a composition of homomorphisms, is a homomorphism from to . It might be interesting to know for which we get all of as , and for which we just get the trivial group .

Fact. if and only if is odd for some .

Corollary. If there exists such that , then there are exactly such ‘s. ( is the largest integer such that .)

Proposition. If is a power of two other than , then .

Proof. We want to show that is even for each . If then the integers coprime to are precisely the odd integers, so . So for some . If there is nothing to prove. Otherwise, suppose that . This means is a primitive root modulo . A well-known theorem says that primitive roots exist only for for an odd prime. Hence must be or .

On the other hand:

Proposition. If there is a primitive root modulo , then .

Proof. Let , where is odd. Let be a primitive root modulo . Then , so is odd, and so .

So both cases in fact occur infinitely often.

I don’t know whether it might be possible to completely classify the integers based on , nor do I know what any of this actually means, but perhaps it is something worth pondering.

where is the extended complex plane and the ‘point at infinity’ is defined so that

if then and ;

if then .

The following video gives a very illuminating illustration. The sphere in the video is called the Riemann sphere, which in a sense ‘wraps up’ the extended complex plane into a sphere. Each point on the sphere corresponds to a unique point on the plane (i.e. there is a bijection between points on the extended plane and points on the sphere), with the ‘light source’ being the point at infinity. This bijective correspondence is the main reason for including the point at infinity.

According to the video any Möbius transformation can be generated by the four basic ones: translations, dilations, rotations and inversions:

Translation: ,

Dilation: ,

Rotation: ,

Inversion:

Exercise. Show that any Möbius transformation is a composition of these four operations.

The Möbius transformations in fact form a group under composition which acts on . Moreover, we have a surjective homomorphism

.

Möbius transformations exhibit very interesting properties, some of which are:

Proposition 1. Given distinct , there is a unique Möbius map such that

Proof. It is not difficult to work out that the unique is given by

.

Proposition 2. The action of on is sharply triply transitive: if are distinct and are distinct, then there eixsts a unique such that for .

Proof. By proposition 1, there is a unique such that

and a unique such that

.

Then is the unique map satisfying the required property.

The cross-ratio of four distinct points is defined to be the unique such that if is the unique map satisfying

then , i.e.

.

One nice thing about cross-ratios is that they are preserved by Möbius transformations.

Proposition 3. If , then .

Proof. Let such that

.

Then . Likewise, if satisfies

then by proposition 1, so .

From we observe that some permutations of leave the value of the cross-ratio unaltered, e.g. is one such. What about the others?

Let act on the indices of the cross-ratio . The permutations that fix form the stabiliser subgroup of of this action. Using transitivity and invariance (propositions 2 and 3), the orbit of is just the different assignments of the values to ; i.e. the distinct cross-ratios that we get by permuting the indices are just

, , ,

, , .

Let such that

.

Then writing shows that the values of the above cross-ratios are (not necessarily in this order—too lazy to work out the precise order)

and they (as functions of ) form the subgroup of that fixes the set . This group is isomorphic to .

So there are in fact four permutations such that

and they form a subgroup of that is isomorphic to the Klein four-group