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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 D3. Recall that D3 is a group of order 6, generated by a counterclockwise rotation R through an angle of
{I, R, R2, F, FR, FR2}.
The multiplication table for D3, 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,
S3 = {I, A, A2, B, BA, BA2}.
Moreover, a mildly tedious calculation shows that the multiplication table for S3 is given by
We now observe a curious phenomenon. Consider the function
(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) = (a) · (b).
Note that the product
Definition 1: Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection
(a · b) = (a) · (b) (a,b G1).
If there exists an isomorphism 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) 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 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 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
i + j = np + q (0 < q < n-1).
p and q exist by the division algorithm. Then
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 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 1G, 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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