Any group of 3 elements is isomorphic to Z3

[kathrynag]

Homework Statement

Prove that any group with three elements is isomorphic to Z_{3}

Homework Equations

The Attempt at a Solution

Let G be the group of three elements
We have an isomorphism if given c:G--->Z_{3},
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

 

[micromass]

Take a group of three elements {e,a,b}. Since the order of every element must be three, we have that b=a². Thus the group is {e,a,a²}.

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

It is easily checked that this is indeed an iso.

 

[kathrynag]

So c(a)=c(b)
c(a)=1
c(b)=2
1=2 not true, but that means it's not an isomorphism

 

[micromass]

Uh what? c(a) doesn't equal c(b)? does it?

 

[kathrynag]

I thought for the one to one part, you assume c(a)=c(b)

 

[micromass]

Yes... never mind...

 

[kathrynag]

but I assumed that but that amounts to 1=2. How does that work?

 

[micromass]

You assumed that c(a)=c(b), and from that assimption followed that 1=2. So your assumption is wrong, and thus c(a)\neq c(b)

 

[kathrynag]

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]

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...