## Homomorphisms

[ This post roughly corresponds to Chapter VI of the old blog. ]

For sets, one considers functions *f* : *S* → *T* between them. For groups, one would like to consider only actions which respect the group operation.

Definition. Let G and H be groups. A function f : G → H is ahomomorphismif f(xy) = f(x)f(y) for all x, y in G. Note that the product on LHS is in group G, while the product on RHS is in group H.A homomorphism is called a

monomorphism(resp.epimorphism,isomorphism) if it is injective (resp. surjective, bijective).

Clearly, if *f* : *G* → *H* and *g* : *H* → *K* are both homomorphisms of groups, then so is *gf* : *G* → *K*. Furthermore, *f* satisfies the following.

- : we have , so cancel on both sides;
- : we have , so multiply both sides by .

These are the least we should expect out of homomorphisms. It also turns out a group homomorphism preserves subgroups, in the following sense:

Theorem. Let f : G → H be a group homomorphism.

- If K < G is a subgroup then f(K) < H is a subgroup.
- If L < H is a subgroup, then f
^{-1}(L) < G is a subgroup.- If L is a normal subgroup of H, then f
^{-1}(L) is a normal subgroup of G.

Definition. In particular, theimageim(f) := f(G) is a subgroup of H, and thekernelker(f) := f^{-1}({e}) is a normal subgroup of G.

The proof is easy so we’ll leave it to the reader. Notice one missing case: i.e. if *K* is a normal subgroup of *G*, is *f*(*K*) normal in *H*? A moment’s thought tells us this is obviously not true: pick a subgroup *H* of *G* which is not normal; then the inclusion *H* → *G* has image precisely *H*, which is not a normal subgroup of *G* by construction.

*This will be a recurring theme throughout algebra: pullbacks are generally more well-behaved than images f(X).*

## Examples

- If
*N*is a normal subgroup of*G*, then the**projection**map*f*:*G*→*G*/*N*which takes*g*to*gN*is an epimorphism. The kernel is precisely*N*. In fact, this characterises all epimorphisms from*G*as we shall see from the isomorphism theorems. - If
*G*=*S*(_{n}*n*>1), the set of*n*-degree permutations, then the sign map which takes even (resp. odd) permutations to +1 (resp. -1) is an epimorphism. The kernel is*A*._{n} - If
*G*= GL_{n}(**R**) denotes the group of*n*×*n*invertible matrices with real entries, then the determinant map det :*G*→**R*** is a group homomorphism. The kernel is SL_{n}(**R**), the set of matrices with determinant = 1.

## First Isomorphism Theorem

Although there’re three, it’s really the first that’s most important. The remaining two can be derived from this first one.

First Theorem of Isomorphism. Let f : G → H be a group homomorphism with kernel N = ker(f). This gives an isomorphism φ : G/N → im(f), where φ(gN) = f(g).

**Proof**. There’re several things to check for the map φ.

- Well-defined & injective: .
- Surjective: follows from definition of im(
*f*). - Homomorphism: .

Proof complete. ♦

Pictorially, we get:

Here’s one application of the theorem. Suppose *H* < *G* is a subgroup and *N* is a normal subgroup of *G*. This gives homomorphisms *H* → *G* → *G*/*N* so the composed map is also a homomorphism. The kernel of the resulting map is . So by the first isomorphism theorem, we get an injective map

.

Here’s another example: let’s prove that

- is isomorphic to the group
**R***; - if
*G*denotes the set of complex numbers*z*with |*z*|=1, prove that*G*is isomorphic to**R**/**Z**, where**Z**=group of integers,**R**=group of real numbers, both under addition.

The method in both cases is identical. Construct a map from one group to the other; find the kernel, then apply the above first isomorphism theorem.

- Take the determinant map . The map is surjective (e.g. the diagonal matrix with one entry
*x*and remaining entries 1 has determinant*x*). The kernel is . So the result follows from the first theorem of isomorphism. - Take the exponential map
**R**→**C***, where*x*goes to exp(2π*ix*). The kernel is exactly the set of integers while the image is the set of complex numbers*z*with |*z*|=1. The result follows from the theorem again.

**“Factoring Through”**

Let *f* : *G* → *H* be a homomorphism. Suppose we know of elements *x*, *y* for which *f*(*x*) = *f*(*y*) = *e*. Then since the kernel of *f* contains both *x* and *y*, it must contain *N*(*x*, *y*), the normal subgroup generated by {*x*, *y*}. So we get a series of group homomorphisms:

,

where the first two maps are projections (since *N*(*x*, *y*) ≤ ker(*f*)), the third is due to the first isomorphism theorem and the fourth is just good old inclusion. We thus conclude that *f* gives rise to *G*/*N*(*x*, *y*) → *H*.

This often happens when one doesn’t quite know the kernel, but can obtain a few explicit elements in it. The question is whether these elements generate the full kernel. One can, at the very least, say that the map *f* **factors through** *G*/*N*(*x*, *y*) → *H*.

## Second and Third Theorems of Isomorphism

Let’s look at the remaining two theorems of isomorphisms.

Second Theorem of Isomorphism.Let H, K ≤ G be subgroups with H normal in G. Then we already know KH is a group. We get the isomorphism.

**Proof**.

Map *K* → *KH* by inclusion and *KH* → *KH*/*H* by projection (this is well-defined since *H* is normal in *KH*). The composition is surjective because every element of *KH*/*H* is of the form (*kh*)*H* = *kH*. The kernel is the set of *k* in *K* such that *kH* = *H*, i.e. kernel = *K* ∩ *H*. Thus the result follows from the first theorem of isomorphism. ♦

Third Theorem of Isomorphism. Let N ≤ H ≤ G be subgroups such that N and H are normal in G (this also means N is normal in H). Then we have the isomorphism.

**Proof.**

Construct a map *f* : *G*/*N* → *G*/*H* via *f*(*gN*) = *gH*. This is well-defined since

.

It is clearly surjective. The kernel is the set of *xN* such that *xH* = *H*, i.e. the kernel is exactly *H*/*N*. Now apply the first theorem of isomorphism. ♦

## Correspondence between *G* and *G/N*

Let . The key result here is the interplay between subgroups of *G* and subgroups of *G*/*N*.

Theorem. There is a one-to-one correspondence :which preserves:

- inclusion : ;
- normality : ;
- indices : whenever .
Here, and are subgroups on the LHS corresponding to subgroups and on the RHS.

**Sketch of Proof.**

Let be the projection map. Let’s construct maps between the two sides.

- RHS → LHS : for subgroup
*K*≤*G*/*N*, take which is a subgroup of*G*containing*N*= ker(π); - LHS → RHS : for
*N*≤*H*≤*G*, take .

We’ll leave it to the reader to show that the composition of the two maps – in either order – is the identity map (on either the LHS or RHS). This shows the bijection.

- Inclusion follows immediately from the definition of the two maps.
- For normality, prove that
*K*is normal in*G*/*N*iff is normal in*G*. - For indices, prove that a complete set of coset representatives of is exactly a complete set of coset representatives of . ♦

## Example of a “Universal” Proof

Of course there’re more properties we haven’t mentioned: e.g.

If is a collection of correspondences, then the intersection also corresponds: .

This can be proven in a straightforward way, but it’s “better” to use the existing properties. Now if , then *H* satisfies the property:

- for each
*i*; - if there is an
*H’*such that for each*i*, then .

The corresponding *K* must then satisfy the same properties, albeit with *K _{i}*‘s:

- for each
*i*; - if there is a
*K’*such that for each*i*, then .

Thus the only possibility for *K* must be the intersection of the *K _{i}*‘s.

[ Note: this proof is “better” in the sense that the approach is more high-level (technical term: it is more **universal**). While the universal method doesn’t seem any shorter now, it lends itself to replication for many different types of algebraic structures, some of whose definitions are extremely intricate and involved. ]