Any group of 3 elements is isomorphic to Z3

**kathrynmath** · November 1st 2010, 05:42 PM

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)

**Drexel28** · November 1st 2010, 05:56 PM

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.

**kathrynmath** · November 1st 2010, 06:11 PM

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

**Drexel28** · November 1st 2010, 06:15 PM

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 $G=\{e,a,b\}$ then $\langle a\rangle$ is a subgroup of $G$ . But, by Lagrange's theorem $|\langle a\rangle|\mid 3$ and since $|\langle a\rangle|\geqslant$ (since $e,a\in\langle a\rangle$ ) it follows that $|\langle a\rangle|=3$ and thus $\langle a\rangle=G$ . 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.

**kathrynmath** · November 1st 2010, 06:23 PM

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.

**Drexel28** · November 1st 2010, 06:34 PM

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 $G=\langle g\rangle$ is a cyclic group of order $3$ then $G\cong \mathbb{Z}_3$ via the isomorphism $f:\mathbb{Z}_3\to\langle g\rangle:k\mapsto g^k$ as you can clearly check.

But, what I showed above proves that every group of order $3$ (and in fact any prime) is cyclic...sooo

**kathrynmath** · November 1st 2010, 07:00 PM

then it's isomorphic to Z3?

**Drexel28** · November 1st 2010, 07:18 PM

Originally Posted by kathrynmath

then it's isomorphic to Z3?

Yes, via the mapping $\theta:\mathbb{Z}_3\to\langle g\rangle$ given by (maybe this will help)

$\begin{array}{c|c}k & \theta(k)\\ \hline 1 & g\\ 2 & g^2\\ 3 & g^3\end{array}$

**HallsofIvy** · November 1st 2010, 07:46 PM

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 $Z_3$ 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)

**kathrynmath** · November 1st 2010, 08:10 PM

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?

**HallsofIvy** · November 2nd 2010, 05:43 AM

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 $Z_3$ ! The elements of $Z_3$ 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)!

**kathrynmath** · November 2nd 2010, 06:30 AM

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

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