Finding kernel of homomorphism $f:\mathbb Z \to S_8$ such that $f(1)=(1426)(257)$

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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 $\mathrm{ker}(f)$?