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.
Best Answer
Every group of prime power order $p^n$ has a non-trivial center. If the group is not abelian, the center thus is a non-trivial proper normal subgroup and hence such a group cannot be simple.
There are many different ways to argue here, see also this post:
Classify groups of order 27
It shows that all subgroups of index $p=3$ are normal.