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Let $f:\mathbb Z \to S_8$ be a homomorphism such that $f(1)=(1426)(257)$ , then how to compute $\ker(f)$ and $f(20)$? I know that $f(n)=f^n(1)$ but this seems too tedious; please help

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you have a product disjoint cycles, so the order of the image is just the lcm of their orders. – Adam Hughes Jul 31 '14 at 4:32
@Adam they're not disjoint! – blue Jul 31 '14 at 4:33
Perhaps that should be the first thing OP should tackle. Rewrite what you have as a product of disjoint cycles. – Kaj Hansen Jul 31 '14 at 4:37
@blue well that's rather underhanded of the problem writer to make one work out the actual disjoint cycle decomposition! – Adam Hughes Jul 31 '14 at 5:00
I wish you would stop creating new tags with almost every question you post. The site existed for four years, and tons of effort have been put into designing the tag system. If something has not been made a tag, perhaps there is a reason for that. In this case, [group-homomorphism] is a useless tag: it's not a subfield of group theory, homomorphisms are used all over the theory. – ᴡᴏʀᴅs Aug 3 '14 at 20:05

In $S_8$, the permutation $\sigma = (1426)(257) = (142576)$ has order $6$. Now, $$f(20) = f^{20}(1) = (142576)^{20} = (127)(456),$$ where we used the fact that $(142576)^{18} = (1)$ since $|(142576)| = 6$. For $n$ to be in $\mathrm{ker}(f)$, we need $(142576)^n = (1)$, which happens precisely when $n$ is a multiple of $6$. What can you conclude about $\mathrm{ker}(f)$?

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The kernel of any homomorphism $f$ out of $\Bbb Z$ is generated by the order of $f(1)$.

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Note that $S_8$ is a finite group, so if you look at powers of $f$ you will find that it eventually repeats. Perhaps you can reason about how long it takes to repeat itself without explicitly drawing everything out.

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