1. Isomorphisms This is when seemingly different things seem to be governed by the same rules. Example You know that Arg(z 1  z 2 ) = Arg(z 1 ) + Arg(z 2 ) Arg(z 1 / z 2 ) = Arg(z 1 ) - Arg(z 2 ) Arg(z n ) = nArg(z) These are the same as the rules for logarithms. The rules for logarithms are isomorphic to the rules for arguments. Finite isomorphic groups Two finite groups are isomorphic if their tables are the same apart from different symbols being used. You prove isomorphism by finding the 1 to 1 correspondence between the elements.

2. {a,b,c,d} under “ is isomorphic to {%,^,&,*} under @ since a=%, b=^, c=&, d=* Ex1 and

3. Ex2 Show that {1, -1, i, -i} under  is isomorphic to {1, 2, 3, 4} under  modulo 5? At first it seems not because the headings imply 1 = 1, -1 = 2, i = 3, -i = 4 but the (centre) of the body of the table requires 1 = 4 and –i = 1. So Reorder the elements so that the identity element is first and the other self-inversive element is next in both tables:

4. The groups are isomorphic, since there is a 1 to 1 correspondence: 1 = 1, -1 = 4, i = 2, -i =3 Identity is 1 for both tables Table 1: –1 is self inversive Table 2: 4 is self inversive

5. Ex3 Prove (Z 3, + 3 ) is isomorphic to {0, 2, 4}, + 6 ) + 3 = addition in modulo 3 Clearly the two tables are isomorphic as 0replaces 0 2replaces1 4replaces2 Strategy To be isomorphic, groups must have the same order (obvious) and the same number of self-inversive elements. The order of the elements must match. You can often show that groups are not isomorphic from this.

6. b and c are self inversive All the elements are self inversive The groups therefore cannot be isomorphic. Ex 11 pg. 412 No. 1, 3 Ex 11 Misc No.s 1, 2, 7, 5, 6, 3