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Let $\alpha = (1 \ 6 \ 3) (2 \ 9) (4 \ 8 \ 10) \in S_{10}$ be a permutation. Write $\alpha$ as a product of transpositions, i.e. of cyclic permutations of order 2. Note that transpositions do not need to be disjunked.

Really don't know how to "go/walk" on this.

I think that $$1->6, 6->3$$

$$2->9$$ $$5->8, 7->10$$ So something like $$(1 6)(6 3)(3 2)(2 9)(9 4)(4 8)(8 10)$$

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marked as duplicate by gt6989b, Matt Samuel, Lord Shark the Unknown, user10354138, Brahadeesh Nov 26 '18 at 8:24

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  • $\begingroup$ Can you write $(163)$ as a product of transpositions? $\endgroup$ – gt6989b Nov 25 '18 at 16:36
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Hint: $(a_1a_2\dots a_n)=(a_1a_2)(a_2a_3)\dots(a_{n-1}a_n)$.

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  • $\begingroup$ α=(1 6 3)(2 9)(4 8 10) with your hint we got; $\alpha_1$=1, $\alpha_2$=6, $\alpha_3$=3, $\alpha_4$=2, $\alpha_5$=9, $\alpha_6$=4, $\alpha_7$=8, $\alpha_8$=10, or? $\endgroup$ – soetirl13 Nov 25 '18 at 17:01
  • $\begingroup$ Well, just split $(163)=(16)(63)$. I leave $(4 8 10)$ for you. $\endgroup$ – Chris Custer Nov 25 '18 at 17:09
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    $\begingroup$ ahh (1 6)(6 3)(2 9)(4 8)(8 10) $\endgroup$ – soetirl13 Nov 25 '18 at 18:24

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