Prove that if $G$ is a finite simple group containing a subgroup $H$ of order $27$, then $|G : H| \ge 9.$

Question

Prove that if $G$ is a finite simple group containing a subgroup $H$ of order $27$, then $|G : H| \ge 9.$

I have so far been solving this by looking at $|G:N|=1, 2, 3...$ etc., all the way up to $8$ and showing that each case is not possible. This is a method that I believe will work in solving the problem if I continue using it (I have shown it does not work for $1, 2$, and $3$).

However, I was wondering if there was a more efficient way to go about a problem like this.

Answer 1

If $|G:H|<9$, then since $G$ is a simple group, it is isomorphic to a subgroup of the symmetric group $S_8$. But the group $S_8$ has a Sylow $3$-subgroup of order $9.$ We have a contradiction.

Answer 2

$G$ acts by left multiplication on the left quotient $G/H$. If $G$ is simple, then $G$ embeds into $S_{[G:H]}$ and hence $|G|$ divides $[G:H]!$. But $|G|=27[G:H]$, so $|G|$ divides $[G:H]!$ if and only if $3^3 \mid ([G:H]-1)!$ if and only if $[G:H]\ge 10$. Therefore, a stronger condition holds, actually.