This article assumes you know some basic group theory. The motivation here is to consider groups whose underlying operations are continuous with respect to its topology.

Definition. Atopological groupG is a group with an underlying topology such that:

- the product map m: G × G → G is continuous;
- the inverse map i: G → G is continuous.

**Examples**

- Any group can be a topological group by endowing it with the discrete topology.
- The complex plane
**C**, real line**R**, and their subspaces**Z**,**Q**are all topological groups under addition. **C*** =**C**– {0} and**R*** =**R**– {0} are also topological groups under multiplication.- More interesting topological groups include GL
_{n}(**R**) and SL_{n}(**R**), the group of non-singular*n*×*n*real matrices, and the group of*n*×*n*real matrices with determinant 1 (respectively). The topology of these groups are given by the subspace topology upon embedding them in

The first property we want to prove is:

Proposition. On a topological group G, the following maps are homeomorphisms.

- Left-multiplication: for g in G,
- Right-multiplication: for g in G,
- Inverse: above.
- Conjugation: for g in G,

**Proof**.

Since multiplication and inverse maps are continuous, the above maps are all continuous too. Plus, they’re bijective and their inverses are given by respectively, which are all continuous. So the maps are homeomorphisms. ♦

In particular, for any there’s a homeomorphism of *G* which maps *x* to *y*; indeed,

On an intuitive level, this means the topology of *G* is completely uniform: e.g. to check what happens near any point *g*, it suffices to check near the identity *e*. [ This should not be confused with *uniform topological spaces*, which is an entirely different thing. ]

Next, the following result is useful.

Proposition 2. Let A, B be subsets of a topological group G. If A is open in G, then so are AB and BA, where

**Proof**.

This follows from which is a union of open subsets of *G* since right-multiplication is a homeomorphism. Same goes for *BA*. ♦

The case for closed subsets is not so nice, but at least we have:

Proposition 3. Let C be a closed subset of G and K be a compact subspace of G. Then CK and KC are closed in G.

**Proof**.

We’ll show that *G*–*CK* is open in *G*; the case for *KC* is similar.

Fix *x* in *G-CK* and take the map which takes Clearly *f* is continuous, so is open. Now for any the ordered pair because if we would have

So for some open subsets and such that By compactness of *K*, it can be covered by finitely many

let and

Then Thus so indeed *G*–*CK* is open. ♦

## Subgroups

First, we look at the separation axioms on a topological group.

Proposition 4. The following are equivalent for a topological group G.

- G satisfies T3.
- G satisfies T2.
- G satisfies T1.
- The trivial subgroup {e} is closed.

**Proof**

The only non-trivial direction is (4→1). In fact, we’ll prove any topological group *G* is regular, from which (3→1) follows. The final case of (4→3) follows by left-multiplying *G* with *g*, thus every {*g*} is closed.

Consider the continuous map which takes If *x* is contained in an open subset *U* of *G*, then contains (*x*, *e*) so we have for some open subsets *V*, *W* of *G* containing *x*, *e* respectively.

Now . Furthermore, since any element in the intersection corresponds to such that which is a contradiction. By proposition 2, (*X*–*U*)*W* is an open subset containing *X*–*U*, so any topological group is regular. ♦

It’s ok if {*e*} is not closed, for we can take the closure which will turn out to be a subgroup, in fact, a normal subgroup of *G*. More generally:

Proposition 5.If H is a subgroup of G, then so is cl(H). If N is a normal subgroup of G, then so is cl(N).

**Note**.

The term “*normal*” unfortunately has a slight ambiguity here. It can refer to *normal subgroups*, or normal topological spaces (as in, any two disjoint closed subsets can be separated by open subsets). We’ll only refer to the former in this article.

**Proof**.

Take *x*, *y* in cl(*H*). Now the continuous map which takes gives us Since cl(*H*) is a closed subset of *G* containing *H*, is a closed subset of *G* × *G* containing *H* × *H*. Thus contains cl(*H* × *H*) = cl(*H*) × cl(*H*) and we have which, together with , proves that cl(*H*) is a subgroup of *G*.

For the second statement, each conjugancy map is a homeomorphism and maps *N* to *N*. Thus it must map cl(*N*) to cl(*N*) also. ♦

The closure can be nicely classified as follows:

Proposition 6. If H is a subgroup of G, then cl(H) is the intersection of all HU, where U is an open subset containing e.

**Proof**.

Now iff every open subset *V* containing *x* intersects *H*, which holds iff every open subset *U* containing *e* intersects And this holds iff Now recall that *U* is open iff is open. ♦

Proposition 7. The connected component Y containing e is a closed normal subgroup of G.

**Proof**.

*Y* is closed in *G* since every connected component is closed.

Now suppose Since left-multiplication by *x* is a homeomorphism of *G*, it must map the connected component of *e* to that of *x*. But *x* and *e* both have *Y* as their connected component, so *xY* = *Y* and thus

Next, inverse is a homeomorphism so it must take the connected component of *e* to the connected component of *i*(*e*)=*e*, i.e. *i*(*Y*) = *Y*. This proves *Y* is a subgroup. Normality follows from the fact that the conjugation map is a homeomorphism taking *e* to itself. ♦

**Note**.

Proposition 7 still holds if we replace “connected components” with “path-connected components“.

So far we’ve been looking at closed subgroups. What about open ones?

Proposition 8. An open subgroup H of G is also closed. Furthermore, a closed subgroup H of finite index in G is open.

**Proof**.

The complement *G*–*H* is a union of (left) cosets of *H*. For the first statement, each coset is open since left-multiplication is a homeomorphism. Thus *G*–*H* is open and *H* must be closed. For the second statement, each coset is closed and *G*–*H* is a union of finitely many closed subsets, so it’s closed too, and *H* is open. ♦

Corollary. If G is a connected group, then any neighbourhood U containing e generates the whole group (in the algebraic sense).

**Proof**.

Replace *U* by *U* ∩ *i*(*U*) and we may assume *i*(*U*) = *U*. Now consider where *U ^{n}* is the collection of products of

*n*elements from

*U*. It’s easy to see that

*V*is an open subgroup of

*G*. By the above proposition

*V*is a clopen subset so

*V*=

*G*. It’s clear that the group <

*U*> generated by

*U*must contain

*V*; thus <

*U*>=

*G*. ♦

## Continuous Homomorphisms

Between topological groups, naturally we’ll be looking at homomorphisms which preserve the topology, i.e. continuous homomorphisms. For example: the following is obvious.

Property. If is a continuous map of topological groups, then for any closed (resp. open) subgroup K of H, is a closed (resp. open) subgroup of G.If K is normal in H, then is normal in G.

In particular, if *H* is Hausdorff, then the kernel of *f* is a closed normal subgroup of *G*.

The corresponding result doesn’t hold for the im(*f*), though we do know it’s a subgroup of *H*. E.g. consider which takes (*a*, *b*) to *a*+*b*√2. The homomorphism is clearly continuous if **Z**^{2} is given the discrete topology. However, the image of *f* is not closed.

But since the continuous image of a connected (resp. compact) set is also connected (resp. compact), at least we have:

Property. If is a continuous homomorphism of topological groups and G is connected (resp. compact), then im(f) is a connected (resp. compact) group.

Next, we’ll talk about isomorphisms.

Definition. Anisomorphismof topological groups is a group isomorphism f:G→H which is also a homeomorphism.

Note that it’s not enough to say *f* is a bijective and continuous homomorphism of groups, since its inverse may not be continuous (even though it’s guaranteed to be a group homomorphism).

For example, take the above map which takes (*a*, *b*) to *a*+*b*√2. This induces a group isomorphism from **Z**^{2} to im(*f*) which is continuous but not a homeomorphism.