Prove that any group of order $27$ is not simple I'm stuck on this problem from my abstract algebra course:

Prove that if $G$ is a group with $|G|=27$, then $G$ is not simple.

First I noticed $|G|=27=3^3$. I thought I can use a statement I saw on the text book:

*

*Given $H\leq G$ with $G$ finite and $|G:H|=p$ being $p$ the minimum prime number that divides $|G|$, then $H\unlhd G.$
This would prove that $G$ has a non-trivial normal subgroup, and that would mean $G$ is not simple. But in order to use this I need to prove first that my group $G$ has some subgroup of order $3^2$ (If I'm not wrong, this isn't trivial). So if my reasoning is right, I need to prove that any group of order $27$ has some subgroup of order $3^2$, and my problem will be solved. Am I right? How can I prove this last statement? Any help will be appreciated, thanks in advance.

A well-known fact is that any $p$-group of order $p^n$ contains subgroups of order $p^i$ for each $i\le n$.
This is more than enough, since we can then take a subgroup of index $p$, which, by another well-known fact must be normal.
Thus no $p$-group with $n\gt1$ is simple.
There's an induction proof of the first fact that relies on the additional well-known fact that $p$-groups have nontrivial center (which can be seen by looking at the class equation).  Couple this with the fourth well-known fact that the center is always a normal subgroup, and we have a different proof.
