# Homomorphisms

• A function from a group G to a group H is a homomorphism (or a group map) if for all .
• A homomorphism is an isomorphism if it is bijective --- equivalently, if it has an inverse.
• If G and H are groups, G and H are isomorphic if there is an isomorphism . Isomorphic groups are the same as groups.
• If is a group map, then and .
• The kernel of a group map is . is a subgroup of G.
• The image of a group map is . is a subgroup of H.
• Groups G and H are not isomorphic if they have different orders, or if one has a group-theoretic property that the othe doesn't. For example, two groups are not isomorphic if one is abelian and the other is not; two groups are not isomorphic if the orders of elements of one are not the same as the orders of elements of the other.

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

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

Remarks. 1. You have seen patterns like this before; for example, "The derivative of a sum is the sum of the derivatives".

2. Group homomorphisms are to groups as linear transformations are to vector spaces. Consider the definitions:

Example. ( The identity map and inclusion maps are group maps) If G is a group, the identity map given by and the constant map given by are homomorphisms.

Moreover, if , the inclusion map given by is a homomorphism.

Example. ( Constant maps are usually not group maps) In general, constant maps aren't homomorphisms. Consider the group under addition, and look at given by for all n. Then

Example. ( Logs and exponentials) is a homomorphism from the reals under addition to the positive reals under multiplication.

The operation on the domain is addition, while the operation on the range is multiplication. Therefore, to show is a homomorphism, I have to show that

But and , so the equation to be verified comes down the to familiar identity . Thus, is a homomorphism.

Notice that is not a homomorphism. In this case, I'm using addition as the operation on both the domain and range, so the homomorphism property would say " " --- in other words, " ". This is not an identity; for example, if and , , while , and .

In basic math courses, people often get sloppy and refer to "the function ". As this example shows, a function isn't just a {\it rule}; it's a rule together with a domain and a range. In many basic math situations, the domain and range don't play a large role; in this situation they do.

Example. ( Checking whether a function is a group map) Define by

To show that f is a homomorphism, let . Since the operation on is addition, I must show that . Check it:

Therefore, f is a homomorphism.

To show that a function is not a homomorphism, give a specific counterexample. For example, define by

To show that g is not a homomorphism, I must find such that . I'll pick two values at random, say and . Try it:

Since , g is not a homomorphism.

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

Now

Since , it follows that

Therefore, is a homomorphism.

Example. ( Group maps and linear transformations) I observed earlier that group homomorphisms are analogous to linear transformations. In fact, a vector space is an abelian group under vector addition. Thinking of a vector space in this way, a linear transformation is a group homomorphism.

Remember that the definition of a linear transformation requires that

for all vectors and . This means that T is a group homomorphism.

Here's a specific example. is a 2-dimensional vector space. If you think about the axioms for a vector space, you can see that is an abelian group under vector addition.

Consider the linear transformation defined by

In matrix form, this is

Since matrix multiplication distributes over vector addition,

This gives a direct proof that T is a group homomorphism.

Example. ( A group map involving multiplication and addition) Let be the group of rational numbers under addition; let be the group of positive rational numbers under multiplication. Define by

I'll check that f is a homomorphism. Note that since the operation on is addition (+) and the operation on is multiplication ( ), I must show that

Here's the computation:

Therefore, f 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.

If I cancel off both sides, I obtain .

Now 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. The function given by

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

On the other hand, consider given by

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

Since , g is not a homomorphism.

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 is

(b) The image of is (as usual)

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

Warning. The crummy notation does not imply that the inverse of exists. is simply the set of inputs which 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 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. ( A bijective group map) For the homomorphism , and . In fact, this map is bijective; the inverse is .

Example. ( Finding the kernel and image) Let

Make into a group under multiplication of complex numbers. Each element can be uniquely written in the form

The identity element is 1; the inverse of is .

Define by

By the remarks above, . To see that is a homomorphism, note that

The kernel of is

Using , you can see that .

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? A precise definition would be 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, 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 {\it 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 --- it is an isomorphism by the first part of the lemma.

Example. ( A group isomorphism) Consider the exponential map given by . By an earlier example, is a homomorphism from the reals under addition to the positive reals under multiplication.

The natural logarithm inverts :

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.

Define a map by

Clearly, is invertible: Its inverse is

I'll show 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.

What does it mean to say that isomorphic groups have the same group-theoretic properties? Here are some examples.

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

Proof. Let . I must show that . Since is onto, there exist such that and . Then

Therefore, H is abelian.

Example. ( Non-isomorphic groups) and are both groups of order 6. However, 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. (More specifically, isomorphic groups have the same cardinality.

Proof. This is trivial, since puts G and H in 1-1 correspondence.

Example. ( Groups of different cardinality aren't isomorphic) and cannot be isomorphic, since the integers are countable, while the reals are uncountable.

Note that two groups with the same order are not necessarily isomorphic. In the previous example, I showed that and are not isomorphic, even though both of them have order 6.

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 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 infinitely many properties that will be shared by isomorphic groups. In fact, you might say that a group-theoretic property is one that is necessarily 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) and are not isomorphic. Both groups have 4 elements; however, every element of has order 1 or 2, while 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.

Similarly, , , and are all abelian groups of order 8. Look at the orders of the elements.

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.

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