# The index of every proper subgroup of a simple group containing an order 22 element

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 $S_n$.
• @Ted As I see it we have that $22|n! \Rightarrow n\ge11$. Why is it $n\ge13$? Thanks in advance! – 1123581321 Sep 1 '20 at 7:27
• $S_{12}$ has no element of order $22$. – Derek Holt Sep 1 '20 at 7:34