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How to show that strictly less than $n-1$ involutions (of which the transpositions are a special case) could not generate $S_{n}$ for $n > 3$?

I know that $n-1$ transpositions are sufficient, or that $S_n$ could be generated by the two elements $(1 ~ 2), (1 ~ 2 ~ \ldots ~ n)$, but what about less than $n-1$ involutions?

StefanH
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This is not true. For example, $S_{10}$ can be generated by the three involutions (2,3), (3,5)(4,7)(6,9)(8,10), (1,2)(3,4)(5,6)(7,8).

Of course for transpositions the answer is easy, as you need $n-1$ transpositions to be transitive on $n$ points.

ahulpke
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    In fact $S_n$ can be generated by three involutions for any $n$. Choose two involutions that generate the dihedral group of order $2n$ containing the $n$-cycle $(1,2,3,\ldots,n)$, together with $(1,2)$. – Derek Holt Oct 03 '15 at 15:29
  • @DerekHolt Yes, of course. – ahulpke Oct 03 '15 at 15:45
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It is not true. $S_3$ can be generated by two involution for example.

$<\{(1,2),(2,3)\}> =S_3$.

mesel
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  • Yes, sorry, I wanted to exclude that case, which is somewhat special and I guess to be the only one. I edited the question for supposing that $n > 3$. – StefanH Oct 03 '15 at 15:04
  • It is not the only, $S_4$ can be generated by $3$ involution. – mesel Oct 03 '15 at 15:07
  • Yes, sorry again, $S_n$ could always be generated by the $n-1$ involutions $(1 ~ 2), (1 ~ 3), \ldots, (1 ~ n)$, I meant less than $n-1$! – StefanH Oct 03 '15 at 15:12