The complete question is the following:

For $G$ a simple group containing an element of order $22$. Show that every proper subgroup of $G$ has index at least $13$.

I think I am supposed to use Sylow's Theorems to show this is true, but I don't know exactly what to do. Is it easier to show contradiction?


Hint: Show that if $G$ is a simple group with a subgroup of index $n>1$, then $G$ injects into the symmetric group $S_n$.

  • $\begingroup$ Thanks a lot!!! $\endgroup$ – J.Summer Oct 5 '17 at 18:59
  • $\begingroup$ @Ted As I see it we have that $22|n! \Rightarrow n\ge11$. Why is it $n\ge13$? Thanks in advance! $\endgroup$ – 1123581321 Sep 1 '20 at 7:27
  • $\begingroup$ $S_{12}$ has no element of order $22$. $\endgroup$ – Derek Holt Sep 1 '20 at 7:34
  • $\begingroup$ Yes right! Thanks! $\endgroup$ – 1123581321 Sep 1 '20 at 7:38

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