Prove that any group of order $27$ is not simple

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

It suffices to prove that $G$ is not centerless. By contradiction, suppose it is; then, every noncentral (i.e. nontrivial) element has centralizer of order $3$ or $9$, and hence the class equation yields: $$27=1+9k+3l$$ for non-negative integers $k,l$: a contradiction because $3\nmid 26$.

This argument holds, verbatim, for every group of order $p^n$, $p$ prime, as $p\nmid p^n-1$.

Answer 4

Here's an alternative proof using elemental character theory.
If $G$ is abelian the result follows from Cauchy's theorem. We may assume then that $G$ is nonabelian. Recall that $G'\unlhd G$, where $G'$ is the derived subgroup of $G$.
If $\chi\in\text{Irr}(G)$ is an irreducible character of $G$ then $\chi(1)\mid|G|$. Moreover, $$\sum_{\chi\in\text{Irr}(G)}\chi(1)^2=27,$$ which implies that $\chi(1)\in\{1,3\}$ for each $\chi\in\text{Irr}(G)$. If we denote $$\text{Lin}(G)=\{\chi\in\text{Irr}(G):\chi(1)=1\},$$ then $|G:G'|=|\text{Lin}(G)|$. Using the previous identity it follows that $$|G:G'|+9|\text{Irr}(G)\setminus\text{Lin}(G)|=27.$$ Now, since $G$ is nonabelian we have that $|\text{Irr}(G)\setminus\text{Lin}(G)|\geq1$. Hence, $$27\geq|G:G'|+9\implies18\geq|G:G'|.$$ This forces that $|G:G'|\in\{3,9\}$, since in other case it would hold $$1+9|\text{Irr}(G)\setminus\text{Lin}(G)|=27,$$ which is not possible. As a consequence $|G'|\in\{3,9\}$, proving the statement.