A function f : AB is said to be surjective (or onto) if for every yB there exists xA such that f(x) = y. A function f is said to be injective (or one to one) if, whenever f(x) = f(y), we have x = y.

### Isomorphism

First, let's consider the word itself. The prefix iso, from the Greek, means equal, identical, or similar. The root word, morph, also from the Greek, indicates a specified form, shape, or structure. The suffix, ism indicates an action, practice or process. So, an isomorphism can be interpreted as the process of two structures being of essentially equal form.

Let us now consider the group G of symmetries of an equilateral triangle. As we saw earlier, this group is the dihedral group D3. Recall that D3 is a group of order 6, generated by a counterclockwise rotation R through an angle of 2/3 and a flip F about the X-axis. The elements of D3 are then

{I, R, R2, F, FR, FR2}.

The multiplication table for D3, was seen to be as follows:

 I R R2 F FR FR2 I I R R2 F FR FR2 R R R2 I FR2 F FR R2 R2 I R FR FR2 F F F FR FR2 I R R2 FR FR FR2 F R2 I R FR2 FR2 F FR R R2 I

On the other hand, the symmetric group on three letters is also a group of order 6, since

Let us set

Then,

Therefore, if we denote the identity of S3 by I, we see that

S3 = {I, A, A2, B, BA, BA2}.

Moreover, a mildly tedious calculation shows that the multiplication table for S3 is given by

 I A A2 B BA BA2 I I A A2 B BA BA2 A A A2 I BA2 B BA A2 A2 I A BA BA2 B B B BA BA2 I A A2 BA BA BA2 B A2 I A BA2 BA2 B BA A A2 I

We now observe a curious phenomenon. Consider the function :S3D3 defined by

(I) = I,   (B) = F,
(1)
(A) = R,   (BA) = FR,
(A2) = R2,   (BA2) = FR2.

If we apply to every element in the multiplication table for S3, we get the multiplication table for D3. Let us pause for a moment and ponder the significance of the existence of the function . The function just replaces all A's by R's and all B's by F's. Therefore, the structures of the multiplication tables for S3 and D3 are identical for the two tables differ only in which letters we have chosen to call the elements. But the multiplication table of a group tells us everything there is to know about the group. Therefore, since D3 and S3 have the "same" multiplication table, we should regard them as the same group. For every property of D3, there is a corresponding property of S3, and for every property of S3 there is a corresponding property of D3.

Since it is clumsy to speak of two groups having the "same" multiplication table, let us examine more closely just what this entails. We started out with the function :S3D3 which transformed the multiplication table of S3 into the multiplication table of D3. In order for to accomplish this, it must preserve multiplication. That is, if a,b S3, then

(2)
(a · b) = (a) · (b).

Note that the product a · b is taken in S3, whereas the product (a) · (b) is taken in D3. Equation (2) does not reflect the only important property of . Indeed, it is clear that is also an injection. Therefore, we are led to make the following definition:

Definition 1: Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection :G1G2 such that

(a · b) = (a) · (b)  (a,b G1).

If there exists an isomorphism :G1G2 which is surjective, then we say that G1 and G2 are isomorphic (denoted G1G2).

It is clear that the two groups S3 and D3 are isomorphic with respect to the isomorphism which was defined by (1). Moreover, isomorphic groups have the "same" multiplication table, in the sense which we meant in our example, and therefore, in the theory of groups, isomorphic groups can be regarded as the same. Usually, it is not sufficient to say that two groups are isomorphic; rather one must specify the isomorphism. For it is the isomorphism which allows one to translate, directly, properties of one group onto the other.

The following properties of isomorphism are more or less obvious:

(a)GG; (b) if G1G2, then G2G1; (c) if G1G2 and G2G3, then G1G3. Thus, isomorphism is an equivalence relation.

An extremely difficult problem in group theory is to determine all finite groups having given order n. Of course, without the notion of isomorphism, such a problem would be hopeless. For there are infinitely many groups isomorphic to a given group. (Just keep changing the names of the elements.) Thus, there are infinitely many groups having a given order n. However, let us rephrase the problem to read: Determine all non-isomorphic groups, having order n. There are only finitely many for a given n. For let us choose fixed names for the elements of the group, say A1,...,An. Then the group will be completely determined by its multiplication table, which is an n × n array whose entries are chosen from among A1,...,An. It is clear that there are at most nn2 such arrays. Of course very few of these arrays will satisfy the group axioms. But nevertheless, we can conclude:

Theorem 2: There are at most nn2 non-isomorphic groups of order n.

Using the next result, we will be able to list all non-isomorphic groups of order n for n < 7. As far as the general question of determining all finite groups of a given order, it seems to be far beyond our present state of knowledge. In fact determination of a satisfactory answer to this question can be viewed as one of the basic objectives of the theory of groups. For the moment, let us be content with the following modest result:

Theorem 3: Let n be a positive integer. Then every cyclic group G of order n is isomorphic to Zn. Therefore, any two cyclic groups of order n are isomorphic.

Proof: Let G be a cyclic group of order n and let a be a generator of G. Then an = 1, and G = {1, a,..., an-1}. Let f :ZnG be defined by f(i) = ai (0 < i < n-1). It is clear that f is a bijection. We assert that f is an isomorphism. Suppose that i and j both satisfy 0 < i,j < n-1, and let p and q be positive integers such that

i + j = np + q     (0 < q < n-1).

p and q exist by the division algorithm. Then f(i + j) = f(q) = aq, by the definition of f, and thus

f(i+j) = ai+j - np
= aiaj(an)-p
= aiaj    (since an = 1)
= f(i) f(j)

Thus f is an isomorphism of Zn onto G, so that G is isomorphic to Zn.

Let us now determine a list of all non-isomorphic groups of at most 7. For each positive integer n, there exists a group of order n, the cyclic group of order n. Moreover, if n is prime, then a group of order n is cyclic, so that by isomorphism, there is only one group of order n if n is prime, the cyclic group of Zn. This takes care of groups of order 2, 3, 5, and 7. There is only one group of order 1, the group G = {1}. Thus, it remains to consider only the cases n = 4 and n = 6. It will turn out that there are two non-isomorphic groups for each of these orders.

Let G = {1, a, b, c} be a noncyclic group of order 4. Then G cannot contain an element of order 4, since otherwise G would be cyclic. But the orders of a, b, c must divide 4 and must be greater than 1, so that a, b, and c are all of order 2: a2=b2=c2=1. We assert that ab = c. If ab = 1, then a = b-1; but b-1=b, since b2 = 1. Therefore, if ab = 1, we see that a = b, which contradicts the fact that G is of order 4. Thus ab 1. If ab = a, then b = 1, a contradiction. Thus, ab a. Similarly, ab b. The only possibility is ab = c. Thus, we have shown that if G is a noncyclic group of order 4, then G is of the form G = {1, a, b, ab} where a2 = b2 = (ab)2 = 1. It is easy to see that from these facts, the product of any two elements of G is completely specified, and G has the multiplication table pictured in Table 2.

 1 a b ab 1 1 a b ab a a 1 ab b b b ab 1 a ab ab b a 1
Table 2

Thus, we see if G is a noncyclic group of order 4, then G is uniquely determined up to isomorphism and its multiplication table is given by table 2. One could check that Table 2 defines a group. But verification of associativity is somewhat laborious. It is much easier to observe that Z2 × Z2 is a group of order 4 which is not cyclic, and thus, is an example of a group having a multiplication table as in Table 2. In particular, Table 2 does define a group. This group is called Klein's 4-Group and is denoted V4. Thus we have shown that a group of order 4 is isomorphic to either Z4 or V4

Next, let G be a noncyclic group of order 6. The order of an element of G is either 1,2,3,or 6. Since G is noncyclic, G does not contain an element of order 6. Thus, if x is an element of G other than 1G, then x has an order of 2 or 3. Assume for the moment that G has no elements of order 3 and let a,b 1g. Then by reasoning as in the case of groups of order 4, we can see that {1G, a, b, ab} is a subgroup of G of order 4. But since 4 6, we see that this contradicts Lagrange's theorem. Thus, G has at most one element of order 2 and therefore G has an element a of order 3. Choose b G, so that b 1G, a, a2. Then it is easy to see that G = {1G, a, a2, b, ab, a2b}. The order of b must be 2 or 3. Let us show that it must be 2, If not, then b3 = 1G, and b21G. Therefore, b2 = a, a2, b, ab, a2b. The last three possibilities are impossible, since b 1G, a, a2. Thus either b2 = a or b2 = a2. In the first case, b3 = ab = 1G, which is a contradiction. In the second case b3 = a2b = 1G, Thus, we see that the order of b cannot be 3 and b must have order 2. Now what are the possibilities for ba? It is trivial to see that ba cannot be 1G, a, a2, or b, since a has order 3 and b has order 2. Thus, either ba = ab or ba = a2b. In the first case, G is abelian. Moreover it is easy to check that ab has order 6, so G is cyclic, a contradiction. Thus, we see that ba = a2b. The reader should have no problem convincing himself that once ba is determined, the entire multiplication table for G is completely determined. Thus, if a noncyclic group of order 6 exists, then it is uniquely determined up to isomorphism. Does such a group exist? Of course. We have seen that S3 is a non-abelian group of order 6. In particular, S3 is not cyclic. Thus, we have proved that if G is a group of order 6, then G is isomorphic to either Z6 or S3.

Definition 4: An isomorphism of G onto itself is called an automorphism of G.

The set of all automorphisms of G forms a group and is denoted Aut(g)