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In my study of group homomorphism I find a question like this.

Let $Q\colon \mathbb Z\to S_8$ be group homomorphism such that $Q(1)=(1,4,2,6)(2,5,7)$. Then find $\ker(Q)$ and $Q(21)$.

In my mind I know that $\ker(Q)$ are those elements of $\mathbb Z$ which map to the identity of $S_8$ .How I can work with this concept to get those elements? Much more difficult I get is to find $Q(21)$.Thank again

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    First find the smallest positive integer $n$ such that $Q(n)$ is the identity permutation. Remember that the group operation is addition of integers on one side and composition of permutations on the other. $Q(2)=Q(1+1)$,$Q(3)=Q(2+1),\ldots$2011-12-25

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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.

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    +1: Somehow I had assumed that the cycles would be disjoint. Thanks for drawing attention to that!2011-12-25
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we have $Q(1)=\pi=(1426)(257)=(142576)$ (unless there was a typo and one of the 2's was a 3 or 8) and we know that $\pi^6=1$. note that $21\equiv 3\mod 6$ so that $Q(21)=Q(3)=\pi^3$. now calculate (or read off) $\pi^3=(15)(26)(47)$. to find the kernel, note that $\pi^6=1$ since $\pi$ is a six-cycle, so that the kernel of $Q$ is $6\mathbb{Z}$

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    sorry, fixed it up, merry christmas2011-12-25