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Any group of 3 elements is isomorphic to Z3

by kathrynag
Tags: elements, isomorphic
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kathrynag
#1
Nov1-10, 07:40 PM
P: 603
1. The problem statement, all variables and given/known data
Prove that any group with three elements is isomorphic to [tex]Z_{3}[/tex]


2. Relevant equations



3. The attempt at a solution
Let G be the group of three elements
We have an isomorphism if given c:G--->[tex]Z_{3}[/tex],
if c is one-to -one and onto and c(ab)=c(a)c(b)

First, we check one-to-one
We want c(a)=c(b) to imply a=b
My problem here is how to define c(a), c(b).
Onto:
We want c(a)=x and want to solve for a?
c(ab):
Same problem with not knowing what c(ab) is
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micromass
#2
Nov1-10, 07:45 PM
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Take a group of three elements {e,a,b}. Since the order of every element must be three, we have that b=aČ. Thus the group is {e,a,aČ}.

Define the map G --> Z3 by
e ---> 0
a ---> 1
b ---> 2

It is easily checked that this is indeed an iso.
kathrynag
#3
Nov1-10, 07:50 PM
P: 603
So c(a)=c(b)
c(a)=1
c(b)=2
1=2 not true, but that means it's not an isomorphism

micromass
#4
Nov1-10, 07:53 PM
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Any group of 3 elements is isomorphic to Z3

Uh what? c(a) doesnt equal c(b)??? does it?
kathrynag
#5
Nov1-10, 07:59 PM
P: 603
I thought for the one to one part, you assume c(a)=c(b)
micromass
#6
Nov1-10, 08:02 PM
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Yes... never mind...
kathrynag
#7
Nov1-10, 08:06 PM
P: 603
but I assumed that but that amounts to 1=2. How does that work?
micromass
#8
Nov1-10, 08:15 PM
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You assumed that c(a)=c(b), and from that assimption followed that 1=2. So your assumption is wrong, and thus [tex]c(a)\neq c(b) [/tex]
kathrynag
#9
Nov1-10, 08:33 PM
P: 603
Then I guess I don't see how to show 1-1

onto
y=c(x)
Do I just take any element, say a
y=c(a)=1
y=1, but we want to solve for x I thought

c(a)c(b)
1*2
2=c(ab)
micromass
#10
Nov1-10, 08:35 PM
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You show 1-1 by a simple proof by contradiction. I'm sorry, but a math major really should be able to do such a thing...


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