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IsomorphismFirst, 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 D_{3}. Recall that D_{3} is a group of order 6, generated by a counterclockwise rotation R through an angle of
{I, R, R^{2}, F, FR, FR^{2}}.
The multiplication table for D_{3}, was seen to be as follows:
On the other hand, the symmetric group on three letters is also a group of order 6, since Let us set Then,
S_{3} = {I, A, A^{2}, B, BA, BA^{2}}.
Moreover, a mildly tedious calculation shows that the multiplication table for S_{3} is given by
We now observe a curious phenomenon. Consider the function
(I) = I, (B) = F,
(1)
(A) = R, (BA) = FR,
(A^{2}) = R^{2}, (BA^{2}) = FR^{2}.
If we apply to every element in the multiplication table for S_{3}, we get the multiplication table for D_{3}. 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 S_{3} and D_{3} 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 D_{3} and S_{3} have the "same" multiplication table, we should regard them as the same group. For every property of D_{3}, there is a corresponding property of S_{3}, and for every property of S_{3} there is a corresponding property of D_{3}. 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 :S_{3}D_{3} which transformed the multiplication table of S_{3} into the multiplication table of D_{3}. In order for to accomplish this, it must preserve multiplication. That is, if
(a · b) = (a) · (b).
Definition 1: Let G_{1} and G_{2} be groups. An isomorphism from G_{1} to G_{2} is an injection
(a · b) = (a) · (b) (a,b G_{1}).
If there exists an isomorphism It is clear that the two groups S_{3} and D_{3} 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) 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 nonisomorphic 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 A_{1},...,A_{n}. Then the group will be completely determined by its multiplication table, which is an Theorem 2: There are at most n^{n2} nonisomorphic groups of order n. Using the next result, we will be able to list all nonisomorphic groups of order n for Theorem 3: Let n be a positive integer. Then every cyclic group G of order n is isomorphic to Z_{n}. 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
i + j = np + q (0 < q < n1).
p and q exist by the division algorithm. Then
f(i+j) = a^{i+j  np}
= a^{i}a^{j}(a^{n})^{p}
= a^{i}a^{j} (since a^{n} = 1)
= f(i) f(j)
Thus f is an isomorphism of Z_{n} onto G, so that G is isomorphic to Z_{n}. Let us now determine a list of all nonisomorphic 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 Z_{n}. This takes care of groups of order 2, 3, 5, and 7. There is only one group of order 1, the group Let
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 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 1_{G}, then x has an order of 2 or 3. Assume for the moment that G has no elements of order 3 and let 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 
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