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§ 3.4 Isomorphisms: Lab Questions

1. The aim of this problem is to show that Table 3.3.2 on page 103 of the text and the table shown by Groups15 for the cyclic group Z6 actually describe the same group. Rewrite Table 3.3.2, using A=e, B=a, C=a2, D=a3, E=a4, and F=a5. Check that the new table is the same as the one listed by Groups15 for Z6.

Solution: Rewriting Table 3.3.2 as suggested produces the following table.

  ×     A     B     C     D     E     F  
  A     A     B     C     D     E     F  
  B     B     C     D     E     F     A  
  C     C     D     E     F     A     B  
  D     D     E     F     A     B     C  
  E     E     F     A     B     C     D  
  F     F     A     B     C     D     E  

You can check that this table is the same as the one listed by Groups15 for Z6. You will notice that the tables in Groups15 do not list the top row and left column that are shown above and in Table 3.3.2.

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2. The aim of this problem is to see whether Table 3.3.3 on page 104 of the text and the table shown by Groups15 for group S3 describe the same group. Rewrite Table 3.3.3, using A=e, B=a, C=a2, D=b, E=ab, and F=a2b. Check whether the new table is the same as the one listed by Groups15 for S3. If not, search for a different correspondence that will produce the same table. This will show that the two groups are isomorphic.

Solution: Rewriting Table 3.3.3 as suggested produces the following table.

  ×     A     B     C     D     E     F  
  A     A     B     C     D     E     F  
  B     B     C     A     E     F     D  
  C     C     A     B     F     D     E  
  D     D     F     E     A     C     B  
  E     E     D     F     B     A     C  
  F     F     E     D     C     B     A  

The pattern in this table is definitely not the same as the pattern in the table for S3 given in Groups15. In fact, in this table we have BD=E, while BD=F in the table in Groups15. This suggests that we should interchange E and F and try again.

  ×     A     B     C     D     F     E  
  A     A     B     C     D     F     E  
  B     B     C     A     F     E     D  
  C     C     A     B     E     D     F  
  D     D     E     F     A     C     B  
  F     F     D     E     B     A     C  
  E     E     F     D     C     B     A  

Finally, we need to interchange the last two columns, and then interchange the last two rows, so that we have the same order as in the Groups15 table.

  ×     A     B     C     D     E     F  
  A     A     B     C     D     E     F  
  B     B     C     A     F     D     E  
  C     C     A     B     E     F     D  
  D     D     E     F     A     B     C  
  E     E     F     D     C     A     B  
  F     F     D     E     B     C     A  

Now the table is the same as the one in Groups15.

Summary: If G1 = {1,a,a2,b,ab,a2b} is the group described by Table 3.3.3 in the text, and G2 = {A,B,C,D,E,F} is the group described by the table for S3 in Groups15, then the function : G1 -> G2 defined by
   (1) = A
   (a) = B
   (a2) = C
   (b) = D
   (ab) = F
   (a2b) = E
is a group isomorphism.


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