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=aČ. Thus the group is {e,a,aČ}.
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)
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.
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}
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 groupHomomorphism:
c(a)c(b)
1*2
2=c(ab)
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)
There is NO "3" in ! The elements of are 0, 1, 2 and 1+ 2= 0.
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)!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?