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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.

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    By homomorphic you mean that $f$ is a surjective morphism? – 2017-01-28
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    To make sure, you are given that $f(0, 1)$ is a permutation that sends $3$ to $6$? – 2017-01-28
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    @Mathematician42 He probably means "homomorphism", since there are no surjective homomorphisms from an abelian group onto a non-abelian group. – 2017-01-28
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    @Mathematician42 Not surjective just homomorphic meaning that $f(a,b)+f(c,d)=f(a+c,b+d)$ – 2017-01-28
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    Okay, because surjective is indeed impossible by the first isomorphism theorem. – 2017-01-28
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    @pjs36 Yes its given in the question – 2017-01-28
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    I'll edit with this the right term (homomorphism) thought it means the same thing in English. – 2017-01-28
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    No, it's my fault I was confused. The term homomorphic image is sometimes used to say that something is the image of a morphism, but it has nothing to do with surjectivity. – 2017-01-28

2 Answers 2

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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.

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    Can you explain how do you know that (123456) is only commute with its own power? Because otherwise I'd have to check of 5! options (to fill up $f(0,1)$ – 2017-01-28
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    I explained it, it is because $(\sigma(1),\sigma(2),\sigma(3),\sigma(4),\sigma(5),\sigma(6))=(1,2,3,4,5,6)$. this means that he elements in the first cycle must be a cyclic permutation of $(1,2,3,4,5,6)$. In other words $\sigma$ is a power of the permutation $(1,2,3,4,5,6)$. – 2017-01-28
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$\mathbb{Z}^2$ 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 $C_6$.