$f:\mathbb{Z}^2\rightarrow S_6$ is a homomorphism such that $f(1,0)=(1\ 2\ 3\ 4\ 5\ 6)$. Find $f(0,1)$.

Question

I have the following question :

$f:\mathbb{Z}^2\rightarrow S_6$ is a homomorphism such that $f(1,0)=(1\ 2\ 3\ 4\ 5\ 6)$. Find $f(0,1)$.

$$f(0,1)= \left(\begin{array}{ccc} 1 & 2 & 3 & 4 & 5 & 6 \\ ? & ? & 6 & ? & ? & ?\\ \end{array}\right) $$

I have no idea how to approach this problem, I guess there is a smarter method than checking for each option if $f(a,b)+f(c,d)=f(a+c,b+d)$.

Any ideas how to approach this?

Any help will be appreciated.

Answer 1

Expanding Matt B's solution:

the permutation $(1,2,3,4,5,6)$ only commutes with its own powers.

This is because if $\sigma$ commutes with $(1,2,3,4,5,6)$ we need $(\sigma(1),\sigma(2),\sigma(3),\sigma(4),\sigma(5),\sigma(6))=\sigma(1,2,3,4,5,6)\sigma^{-1}=(1,2,3,4,5,6)$.

So $\sigma$ must permute the elements $(1,2,3,4,5,6)$ cyclically.

Therefore there are exactly $6$ such morfisms.

Answer 2

$\mathbb{Z}^2$ is abelian so $f(0,1)$ must commute with $f(1,0)$. Otherwise these are completely independent so anything satisfying this will do since you can define a homomorphism on a generating set (provided it satisfies the relations which in this case is just commutativity.

As an example, you can take $f(0,1)=f(1,0)^3=(14)(25)(36)$ which will trivially satisfy commutativity as the image is isomorphic to $C_6$.