I thought that I'll consolidate the comments of @Jykhri and it misses some fine details. So, my adding it might answer your question and hence one less question for math.SE.
Note that $Q$ is a well-defined map since by defining the effect of $Q$ on the generator of the domain (which is a cyclic group), we have defined its effect on the whole of the domain. (You can infact prove that the effect of the map on generator uniquely determines the homomrphism on cyclic groups.)
Now note that $Q(1)=(1,4,2,5,7,6)$ since the cycles are not disjoint, you could always write them together. I assume that you read permutations from right to left.
So, $Q(1)$ has order $6$. This means that $Q(1)^6=Id$ where $Id$ is the identity permutation. Since, $Q$ is a homomorphism, $Q(6)=Id$. Now, you can prove that the kernel of $Q$, whose notion you have got right, is $Ker(Q)=\{6n| n \in \mathbb{Z}\}$ (Probably, this does not require any proof!!)
For $Q(21)$, note that $Q(21)=Q(18)\circ Q(3)$, which as you have already seen is, $Q(1)\circ Q(1) \circ Q(1)$. This again is $(1,5)(4,7)(2,6)$.
You now have answered your question. As a more general question, you learn that the homomorphic image of a cyclic group is cyclic.