# 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.

• Do you know the Sylow theorems? – Dietrich Burde Nov 24 '20 at 21:48

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.

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.