There is a unique homomorphism θ : ZZ6 → S3 such that θ([1]) = (1 2 3)

There is a unique homomorphism θ : ZZ6 → S3 such that θ([1]) = (1 2 3). Determine
θ([k]) for each [k] ∈ ZZ6. Which elements are in Ker θ?

It's question #12 in this link
http://www.maths.tcd.ie/~evd/251/Notes/problemsheet2.pdf

We can actually write out all of the elements of θ.
Since θ is a homomorphism, multiplication by n in Z6 corresponds to nth power in S3.

θ([0]) = θ(0 * [1]) = (1 2 3)^0 = (1)
θ([1]) = (1 2 3)
θ([2]) = θ(2 * [1]) = (1 2 3)^2 = (1 3 2)
θ([3]) = θ(3 * [1]) = (1 2 3)^3 = (1)
θ([4]) = θ(4 * [1]) = (1 2 3)^4 = (1 2 3)
θ([5]) = θ(5 * [1]) = (1 2 3)^2 = (1 3 2)

Hence, ker θ = {[0], [3]}, being the set of elements mapped to the identity of S3.

I hope this helps!

That should be '5' for the exponent (even though it still equals (1 3 2)).

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Thanks!

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