(adsbygoogle = window.adsbygoogle || []).push({}); 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 [itex]\phi(c) = a\cdot x+b\cdot y[/itex]. 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

[tex]\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)[/tex]

where in the second-to-last step I have used the fact that Z_6 is abelian. If d is 0 then we have [itex]\phi(d+e)=\phi(e)=\phi(e)+0=\phi(e)+\phi(0)[/itex].

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Homomorphisms from a free abelian group

**Physics Forums | Science Articles, Homework Help, Discussion**