# Rings are abelian

In this post, we prove the following well-known lemma:

Lemma. Let satisfy all axioms of a ring, except possibly the commutativity . Then is abelian.

That is, additive commutativity of a ring is implied by the other axioms.

Proof. By distributivity, we have , so multiplication by is a homomorphism. By our previous post, this implies is abelian.

# Criteria for groups to be abelian

This is a review of some elementary criteria for a group to be abelian.

Lemma. Let be a group. Then the following are equivalent:

1. is abelian,
2. The map given by is a group homomorphism;
3. The map given by is a group homomorphism;
4. The diagonal is normal.

Proof. We prove that each criterion is equivalent to (1).

For (2), note that , which equals if and only if .

For (3), note that , which equals if and only if .

For (4), clearly is normal if is abelian. Conversely, note that , which is in the diagonal if and only if .

# A fun example of a representable functor

This post is about representable functors:

Definition. Let be a functor. Then is representable if it is isomorphic to for some . In this case, we say that represents .

Exercise. If such exists, then it is unique up to unique isomorphism.

Really one should encode the isomorphism as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation is uniquely determined by the element of corresponding to the identity of .

When is a natural isomorphism, the corresponding element is called the universal object of . It has the property that for every and any , there exists a unique morphism such that .

Example. The forgetful functor is represented by . Indeed, the natural map

is an isomorphism. The universal element is .

Example. Similarly, the forgetful functor is represented by . The universal element is .

A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.

The main example for today is the following:

Lemma. The functor that associates to a topological space its topology is representable.

Proof. Consider the topological space with topology . Then there is a natural map

Conversely, given an open set , we can associate the characteristic function . This gives an inverse of the map above.

The space we constructed is called the Sierpiński space. The universal open set is .

Remark. The space represents the data of open sets for : for any continuous map , we have , where . If denotes the complementary open, then the form a cover of if and only if . This corresponds to the statement that lands in .

Thus, the open cover is the universal open cover, i.e. for every open covering there exists a unique continuous map such that .

# Cauchy’s theorem

This is a really classical result, but this proof is just so magical that I have to include it.

Lemma. Let be a finite group, and let be a prime dividing . Then has an element of order .

Proof. Consider the action of on

given by

Then , since the orbits of size all have size (and the fixed points are exactly the union of the orbits of size 1). The right hand side is as . Thus,

But there is a bijection

The former trivially contains the element 1, hence (since its size is divisible by ) it has to contain some other element . This element will then have (exact) order .

Remark. Of course, the result also follows from the much stronger Sylow theorems (of which we only need the existence statement).

# Separation properties for topological groups

Although this is quite a classical result, I really like it.

Lemma. Let be a topological group. Then is if and only if is Hausdorff.

Proof. One implication is clear. Conversely, suppose is . Then the identity element is closed. The map

is continuous. Hence, the inverse image of the identity is closed. But this is the diagonal, hence is Hausdorff.

Exercise. Prove that Hausdorff is in fact equivalent to .