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Let $H \leq S_{10}$ such that $H$ has at least one 9-cycle, one 2-cycle, and acts transitively on $\{1,2,...,10\}$. Find the minimum size of $H$.

So, I am really stuck on this problem. I think the 9-cycle can be rearranged by an operation on itself, so that we get all the 9-cycles containing those 9 elements. I don't know how to go from there.

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marked as duplicate by Namaste, Community Mar 9 '18 at 19:21

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We'll prove that $H= S_{10}$.

WLOG let the $9$-cycle be $(1,2,\dots,9)$. Then it's enough to show that we have a transpostion, having $10$ as an element, as then conjugation by the $9$-cycle will yield the generating set of $S_{10}$ $\{(i,10)| i\le 9\}$.

Now let $(i,j)$ be the transposition in $G$. Then by the transitivity there exists an element $\sigma$ of it, sending $i$ to $10$. Conjugating the transposition by $\sigma$ we obtain the element $(\sigma(i),\sigma(j)) = (10,\sigma(j))$ and $\sigma(j) \not = 10$. Therefore the claim is proven.

In fact this can be generalized and you can show that if a transitive subgroup of $S_n$ contains a $(n-1)$-cycle and a transposition, then it's $S_n$ itself.

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    $\begingroup$ I think you mean $(10,\sigma(j))$ and $\sigma(j)\ne10$ in the penultimate para. $\endgroup$ – almagest Mar 9 '18 at 19:09
  • $\begingroup$ @almagest Yeah, you are right. Fixed. $\endgroup$ – Stefan4024 Mar 9 '18 at 19:12

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