Here are the operation tables for two groups of order 4:

There is an obvious sense in which these two groups are "the same": You can get the second table from the first by replacing 0 with 1, 1 with a, and 2 with .

When are two groups the same?

You might think of saying that two groups are the same if you can get one group's table from the other by substitution, as above. However, there are problems with this. In the first place, it might be very difficult to check --- imagine having to write down a multiplication table for a group of order 256! In the second place, it's not clear what a "multiplication table" is if a group is infinite.

*One way to implement a substitution is to use a function.* In
a sense, a function is a thing which "substitutes" its
output for its input. I'll define what it means for two groups to be
"the same" by using certain kinds of functions between
groups. These functions are called * group
homomorphisms*; a special kind of homomorphism, called an * isomorphism*, will be used to define
"sameness" for groups.

* Definition.* Let G and H be groups. A * homomorphism* from G to H is a function such that

Group homomorphisms are often referred to as * group
maps* for short.

* Remarks.* 1. In the definition above, I've
assumed multiplicative notation for the operations in both G and H.
If the operation in one or both is something else, you must adjust
the definition accordingly. For instance:

2. You have seen patterns like this before; for example, "The derivative of a sum is the sum of the derivatives".

* Lemma.* Let G be a group and let H be a
subgroup.

(a) The identity map defined by is a group map.

(b) The inclusion map defined by is a group map.

* Proof.* I'll prove (a); the proof of (b) is the
same. Let . Then

Hence, is a group map.

* Example.* (* Constant maps are
usually not group maps*) For the group under addition, define by

Show that f is not a group map.

* Example.* (* Logs and
exponentials*) (a) Prove that the exponential function given by is a group map.

(b) Prove that the natural log function is a group map.

(a) Let . Then by properties of exponentials,

(b) Let . Then by properties of logarithms,

* Example.* (* Checking whether a
function is a group map*)

(a) Define by

Prove or disprove: f is a group map.

(b) Define by

Prove or disprove: f is a group map.

(a) f is a group map: If , then

(b)

Since , g is not a homomorphism.

* Lemma.* Let V and W be vector spaces over a
field F, considered as groups under vector addition. Let be a linear transformation. Then T is a group map.

* Proof.* This follows immediately from one of
the axioms for a linear transformation: If , then

* Example.* and are groups under vector addition. Define by

Prove that T is a group map.

Write T as a matrix multiplication:

From linear algebra, this defines a linear transformation. Hence, T is a group map by the previous lemma.

* Example.* (* A group map on a
matrix group*) Let be the group of reals matrices under matrix addition. Let denote the * trace
map*:

Show that is a group homomorphism.

Now

Thus,

Therefore, is a homomorphism.

* Lemma.* Let be a
group homomorphism. Then:

(a) , where is the identity in G and is the identity in H.

(b) for all .

* Proof.* (a)

If I cancel off both sides, I obtain .

(b) Let .

This shows that is the inverse of , i.e. .

* Warning.* The properties in the last lemma are
not part of the *definition* of a homomorphism. To show that f
is a homomorphism, all you need to show is that for all a and b. The properties in the
lemma are automatically true of any homomorphism.

On the other hand, if you want to show a function is *not* a
homomorphism, do a quick check: Does it send the identity to the
identity? If not, then the lemma shows it's *not* a
homomorphism.

* Example.* (* Group maps must
take the identity to the identity*) Let denote the group of integers with addition. Define
by

Prove that f is not a group map.

Note that . Since the identity is not mapped to the identity , f cannot be a group homomorphism.

Warning: If a function takes the identity to the identity, it may or may not be a group map. Consider given by

, but this doesn't mean that g is a homomorphism. In fact,

The point is that simple-looking functions you may have seen in other math classes need not be homomorphisms. When in doubt, check the definition.

There are several important subsets associated to a group homomorphism .

* Definition.* Let be a
group homomorphism.

(a) The * kernel* of f is

(b) The * image* of f is (as usual)

(c) Let . The * inverse image* of
is (as usual)

* Warning.* The notation *does not imply* that the *
inverse* of f exists. is simply the
set of inputs which f maps into ; this *is* applied to the set if there is a (but there need not be).

* Lemma.* Let be a group map.

(a) is a subgroup of G.

(b) is a subgroup of H.

(c) If is a subgroup of H, then is a subgroup of G.

* Proof.* (a) First,

Suppose . Then

Hence, .

Finally, suppose . Then

Hence, . Therefore, is a subgroup of G.

(b) since .

Suppose . Then

Finally, suppose . Then

Therefore, is a subgroup of H.

(c) Let be a subgroup of H. I want to show that is a subgroup of G. Reminder: The criterion for membership in is that f takes the element into .

Since and , it follows that .

Suppose . This means that and are in . Since is a subgroup, is in as well. But

Therefore, is in , which means that .

Finally, suppose , so . Since is a subgroup, . But , so . This means that .

Hence, is a subgroup of G.

* Example.* (* Finding the kernel
and image*) (a) Let

Show that is a group under multiplication of complex numbers.

(b) Define by

Show that f is a group map, and find its kernel and image.

(a) Each element can be uniquely written in the form

Note that

This shows that multiplication is closed (hence a binary operation) on S.

Complex number multiplication is associative. The identity element is 1; the inverse of is .

(b) To see that f is a homomorphism, note that

From the representation of elements of S as , I have .

The kernel of f is

Using , you can see that .

* Example.* (* Kernel, image, and
inverse image*) is defined by

Take for granted that f is a group map. Find , , and , where H is the subgroup of .

The kernel consists of elements of which f takes to 0. Since 0 "is" 12 in , and since f multiplies inputs by 3, I'll get multiples of 12 out if I feed multiples of 4 in:

Hence, .

consists of the set of outputs of f. Since f multiplies its inputs by 3, the outputs are the multiples of 3:

Finally, consists of elements of which are mapped by f to either 0 or 6. So you need to find the elements in which give 0 or 6 when multiplied by 3. Obviously, an "odd" input will give an "odd" output, and I already know 0 and 4 are mapped by f to 0, so I just try 2 and 6:

Hence, .

* Definition.* Let G and H be groups. An * isomorphism* from G to H is a bijective
homomorphism . If there is an isomorphism , G and H are * isomorphic*;
notation: .

* Remarks.* 1. To say that two groups are
isomorphic is to say that they are the same *as groups*. The
elements of the two groups and the group operations may be different,
but the two groups have the same structure. This means that if one
has a certain group-theoretic property, the other will as well.

What is a *group-theoretic property*? Well, it's a bit
circular: a group-theoretic property is a property preserved by
isomorphism. For this to be a useful concept, I'll have to provide
specific examples of properties that you can check.

2. Some older books define an isomorphism from G to H to be an
injective homomorphism . That is, f need
not map G onto H. One then says G and H are isomorphic if there is an
isomorphism from G *onto* H. Unfortunately, one then has the
odd situation that there may be an isomorphism from G to H, yet G and
H may not be isomorphic! I'll always use the word
*isomorphism* to mean a bijective map.

Here is an easy way to tell that a group map is an isomorphism.

* Lemma.* A group map is an isomorphism if and only if it is invertible. In
this case, is also a homomorphism, hence an
isomorphism.

* Proof.* The first statement is trivial, since a
map of sets is bijective if and only if it has an inverse.

Now suppose that is an isomorphism. I must show the inverse is a homomorphism. Let . I need to show that

Since is onto, there exist such that and . Then

Therefore, is a homomorphism.

Since is invertible --- its inverse is f --- it is an isomorphism by the first part of the lemma.

* Example.* (* A group
isomorphism*) Show that the exponential map given by is a group isomorphism.

I showed earlier that and the natural log function are group maps. They're also inverses:

By the lemma, is an isomorphism (as is ). The groups and are isomorphic.

* Example.* (* A group
isomorphism on the integers mod 2*) Consider the set . Make G into a group using multiplication as the
group operation. Show that G is isomorphic to .

Define a map by

Clearly, f is invertible: Its inverse is

I'll show f is a homomorphism, hence an isomorphism, by simply checking cases:

The brute force approach above can be used to construct an
isomorphism from to any group of order 2. *There
is only one group of order 2, up to isomorphism.*

Here are some examples of "group-theoretic properties". Thus, if two groups are isomorphic and one of the groups has such a property, the other must as well. On the other hand, if one of two groups has one of these properties but the other group does not, then the two groups cannot be isomorphic.

* Proposition.* Suppose G and H are isomorphic
groups. If G is abelian, so is H.

* Proof.* Let . I must show
that . Since f is surjective, there exist such that and . Then

Therefore, H is abelian.

* Example.* (* Non-isomorphic
groups*) is the group of symmetries of an
equilateral triangle. and are both groups
of order 6. Why aren't they isomorphic?

is abelian, while is nonabelian. Therefore, and are not isomorphic.

* Proposition.* Suppose G and H are isomorphic
groups. If G is finite, so is H. If G is infinite, so is H.

In other words, isomorphic groups have the same cardinality.

* Proof.* Since G and H are isomorphic, there is
a bijective (group map) . Since f is bijective, (since that's what it means for two sets to have the
same cardinality).

* Example.* (* Groups of
different cardinalities aren't isomorphic*) Why can't and be isomorphic?

Both groups are infinite, but the integers are countable, while the reals are uncountable. Since they don't have the same cardinality, they can't be isomorphic.

* Proposition.* Suppose G and H are isomorphic
groups. If G has a subgroup K of order 42, so does H.

* Proof.* If and , then and (since f maps K bijectively
onto ) .

Obviously, there's nothing special about "42". If G has a subgroup of order 117, so does H. If G has a subgroup of order 91, so does H. And so on. This proposition is not very useful as is, and is just here to show you a property shared by isomorphic groups.

There are clearly infinitely many properties that will be shared by
isomorphic groups. However, the earlier examples show that some
properties are *not* shared by isomorphic groups. For example,
the elements of one group may be letters, while the elements of the
other are numbers. "Having the same kind of elements" is
*not* a group-theoretic property. Likewise, the operation in
one group may be addition of numbers, while the operation in the
other could be composition of functions. "Having the same kind
of binary operation" is *not* a group-theoretic property.

* Example.* (* Showing groups
aren't isomorphic by considering orders of elements*)

(a) Show that and are *not* isomorphic.

(b) Show that , , and are not isomorphic.

(a) Both groups have 4 elements; however, every element of has order 1 or 2. If , then

Therefore, the order of divides 2, and the only positive divisors of 2 are 1 and 2.

On the other hand, has two elements of order 4 (namely 1 and 3). Having different numbers of elements of a given order is a group property. Since these groups differ in this respect, they aren't isomorphic.

(b) , , and are all abelian groups of order 8. However, their elements have different orders.

Every element of has order 1 or 2. For if , then

Therefore, the order of divides 2, and the only positive divisors of 2 are 1 and 2.

Every element of has order 1, 2, or 4. For if , then

Therefore, the order of divides 4, and the only positive divisors of 2 are 1, 2, and 4. Note that is an element of order 4. This means that can't be isomorphic to , since the latter has no elements of order 4.

has elements of order 8. (1 has order 8, for example.) Therefore, it can't be isomorphic to or to , since these two groups have no elements of order 8.

Therefore, the three groups aren't isomorphic.

Copyright 2018 by Bruce Ikenaga