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Let $n$ be odd and not a multiple of $3$. Do the cycle $\sigma:=(1, 2, \dots, n)$ and any cycle of length $3$ generate $A_n$?

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$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$No.

$$ \Span{(1,2,3,4,5,6,\dots,35), (1,6,11)} $$ is not the alternating group on $35$ letters.

The reason is that it is imprimitive. A block is $$ \Set{1,6,11,16,21,26,31} $$

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    $\begingroup$ Is this the smallest counterexample? $\endgroup$ – David Hill Apr 22 '16 at 15:52
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    $\begingroup$ If the group $G$ generated by the two permutations is primitive, Jordan's theorem kicks in. So we need $n$ to be composite, in order to have blocks. Come to think of it, $n = 25$ would do as well, and that would be smallest possible. Will leave the example as it is, though. And many thanks for the useful question. $\endgroup$ – Andreas Caranti Apr 22 '16 at 15:54
  • $\begingroup$ Great. Thank you for the explanation. $\endgroup$ – David Hill Apr 22 '16 at 15:59
  • $\begingroup$ Thank you. I won't forget that trick! $\endgroup$ – doubleCat Apr 22 '16 at 15:59
  • $\begingroup$ @doubleCat, thanks, I appreciate. $\endgroup$ – Andreas Caranti Apr 22 '16 at 16:00

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