Prove that any group with three elements is isomorphic to Z3




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


We need to show 1-1, onto, and the homomorphism property
I'm having problems with all of the properties
Assume c(a)=c(b)
This implies 1=2, so c(a) is not equal to c(b).
Does this show 1-1?

Onto:
Need to show y=c(x) and solve for x
y=c(a)=1
y=1
I'm confused on this

Homomorphism:
c(a)c(b)
1*2
2=c(ab)
I know this should equal c(ab), but I didn't see why 2=c(ab)

Originally Posted by kathrynmath

Prove that any group with three elements is isomorphic to Z3




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


We need to show 1-1, onto, and the homomorphism property
I'm having problems with all of the properties
Assume c(a)=c(b)
This implies 1=2, so c(a) is not equal to c(b).
Does this show 1-1?

Onto:
Need to show y=c(x) and solve for x
y=c(a)=1
y=1
I'm confused on this

Homomorphism:
c(a)c(b)
1*2
2=c(ab)
I know this should equal c(ab), but I didn't see why 2=c(ab)

How much group theory do you know? Because this isn't necessarily the easiest route.

This is my first semester of abstract algebra. Basically, I know groups and subgroups, and a little about cyclic groups.

Originally Posted by kathrynmath

This is my first semester of abstract algebra. Basically, I know groups and subgroups, and a little about cyclic groups.

Do you know Lagrange's theorem? If so, note that if then is a subgroup of . But, by Lagrange's theorem and since (since ) it follows that and thus . If you haven't done Lagrange's I'm not sure there's an easier method than to just write out the possible Cayley tables.

I have done Lagranges Theorem, but I'm not sure how that would show an isomorphism. Doesn't
an isomorphism exist if it is 1-1, onto, and a homomorphism? I don't see how those properties follow.

Originally Posted by kathrynmath

I have done Lagranges Theorem, but I'm not sure how that would show an isomorphism. Doesn't
an isomorphism exist if it is 1-1, onto, and a homomorphism? I don't see how those properties follow.

Ok, so clearly if is a cyclic group of order then via the isomorphism as you can clearly check.

But, what I showed above proves that every group of order (and in fact any prime) is cyclic...sooo

then it's isomorphic to Z3?

Originally Posted by kathrynmath

then it's isomorphic to Z3?

Yes, via the mapping given by (maybe this will help)

Originally Posted by kathrynmath

Prove that any group with three elements is isomorphic to Z3




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


We need to show 1-1, onto, and the homomorphism property
I'm having problems with all of the properties
Assume c(a)=c(b)
This implies 1=2, so c(a) is not equal to c(b).
Does this show 1-1?

Onto:
Need to show y=c(x) and solve for x
y=c(a)=1
y=1
I'm confused on this

You already have that c(e)= 0, c(a)= 1, and c(b)= 2. That proves both "one-to-one" and "onto" immediately- two different members of {3, a, b} are not mapped to the same thing and there is something mapped to each of {0, 1, 2}

Homomorphism:
c(a)c(b)
1*2
2=c(ab)

No, the operation in is addition (modulo 3) NOT multiplication. 1+ 2= 0 in Z_3. You need to prove that c(ab)= 0 and that is true because, since this is a group
1) ab cannot be equal to a since if a= ab then e= b which is not true.
2) ab cannot be equal to b since if b= ab then e= a which is not true.
That leaves only ab= e and c(e)= 0.

I know this should equal c(ab), but I didn't see why 2=c(ab)

Can I prove the homomorphism property with c(ab)=c(a)c(b)
c(a)=1
c(b)=2
1+2
3
e
c(ab)


So the 1-1, onto is just based on how I defined the group. i don't have to do any of the assuming c(a)=c(b) stuff?

Originally Posted by kathrynmath

Can I prove the homomorphism property with c(ab)=c(a)c(b)
c(a)=1
c(b)=2
1+2
3

There is NO "3" in ! The elements of are 0, 1, 2 and 1+ 2= 0.

e
c(ab)


So the 1-1, onto is just based on how I defined the group. i don't have to do any of the assuming c(a)=c(b) stuff?

Not how you defined the group but how you defined the isomorphism: you said c(e)= 0, c(a)= 1, c(b)= 2. Clearly c(a) is NOT equal to c(b)!

c(a)c(b)
=1+2
=0
=c(e)
=c(ab)