IsomorphismsThis is when seemingly different things seem to be governed by the same rules.Example You know that Arg(z1 z2) = Arg(z1) + Arg(z2) Arg(z1 / z2) = Arg(z1) - Arg(z2) Arg(zn) = nArg(z)

These are the same as the rules for logarithms. The rules for logarithms are isomorphic to the rules for arguments.

Finite isomorphic groupsTwo 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.

{a,b,c,d} under “ is isomorphic to {%,^,&,*} under @

since a=%, b=^, c=&, d=*

“ a b c d a a b c d b b d a c c c a d b d d c b a

@ % ^ & * % % ^ & * ^ ^ * % & & & % * ^ * * & ^ %

Ex1 and

Ex2 Show that {1, -1 , i, -i} under is isomorphic to {1, 2, 3, 4} under modulo 5?

x 1 -1 i -i 1 1 -1 i -i -1 -1 1 -i i i i -i -1 1 -i -i i 1 -1

x 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1

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. SoReorder the elements so that the identity element is first and the other self-inversive element is next in both tables:

x 1 -1 i -i

1 1 -1 i -i -1 -1 1 -i i i i -i -1 1 -i -i i 1 -1

x 1 4 2 3

1 1 4 2 3 4 4 1 3 2 2 2 3 4 1 3 3 2 1 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 tablesTable 1: –1 is self inversiveTable 2: 4 is self inversive

Ex3 Prove (Z3, +3) is isomorphic to {0, 2, 4}, +6) +3 = addition in modulo 3

+ mod 3 0 1 2 0 0 1 2 1 1 2 0 2 2 0 1

+ mod 6 0 2 4 0 0 2 4 2 2 4 0 4 4 0 2

Clearly the two tables are isomorphic as 0 replaces 0 2replaces 1

4 replaces 2Strategy 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.

• a b c d a b d a c b d c b a c a b c d d c a d b

b and c are self inversive

• 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1

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