ehrenfest
Mar5-08, 05:36 PM
1. The problem statement, all variables and given/known data
How many homomorphism are there of a free abelian group of rank 2 into a) Z_6 and b) S_3.
2. Relevant equations
3. The attempt at a solution
Since the images of the generators completely determine a homomorphism, the upper bound for both is 36.
Now a free abelian group of rank 2 is isomorphic to Z \times Z, which has basis {(1,0),(0,1)} and by the definition of a free abelian group, every nonzero element in the group can be uniquely expressed in the form a(1,0)+b(0,1) with a,b in Z. Therefore, given two arbitrary elements x and y in Z_6, we define phi((0,1)) = x and phi((1,0)) = y and for an arbitrary nonzero c = a(1,0)+b(0,1) in Z \times Z, we define \phi(c) = a\cdot x+b\cdot y. And then define \phi(0) = 0.
Now, why is that a homomorphism? Because if d and e are arbitrary nonzero elements in Z \times Z, such that d = a_1(1,0)+b_1(0,1) and e = a_2(1,0)+b_2(0,1), then we have
\phi(d+e) = \phi((a_1+a_2)(1,0)+(b_1+b_2)(0,1)) = (a_1+a_2)x+(b_1+b_2)y = (a_1x+b_1y)+(a_2x+b_2y) = \phi(d)+\phi(e)
where in the second-to-last step I have used the fact that Z_6 is abelian. If d is 0 then we have \phi(d+e)=\phi(e)=\phi(e)+0=\phi(e)+\phi(0).
Therefore the answer to part a is 36. Please confirm that proof. I have absolutely no idea how to do part b since that group is not abelian.
How many homomorphism are there of a free abelian group of rank 2 into a) Z_6 and b) S_3.
2. Relevant equations
3. The attempt at a solution
Since the images of the generators completely determine a homomorphism, the upper bound for both is 36.
Now a free abelian group of rank 2 is isomorphic to Z \times Z, which has basis {(1,0),(0,1)} and by the definition of a free abelian group, every nonzero element in the group can be uniquely expressed in the form a(1,0)+b(0,1) with a,b in Z. Therefore, given two arbitrary elements x and y in Z_6, we define phi((0,1)) = x and phi((1,0)) = y and for an arbitrary nonzero c = a(1,0)+b(0,1) in Z \times Z, we define \phi(c) = a\cdot x+b\cdot y. And then define \phi(0) = 0.
Now, why is that a homomorphism? Because if d and e are arbitrary nonzero elements in Z \times Z, such that d = a_1(1,0)+b_1(0,1) and e = a_2(1,0)+b_2(0,1), then we have
\phi(d+e) = \phi((a_1+a_2)(1,0)+(b_1+b_2)(0,1)) = (a_1+a_2)x+(b_1+b_2)y = (a_1x+b_1y)+(a_2x+b_2y) = \phi(d)+\phi(e)
where in the second-to-last step I have used the fact that Z_6 is abelian. If d is 0 then we have \phi(d+e)=\phi(e)=\phi(e)+0=\phi(e)+\phi(0).
Therefore the answer to part a is 36. Please confirm that proof. I have absolutely no idea how to do part b since that group is not abelian.