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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.

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  • $\begingroup$ Do you know the Sylow theorems? $\endgroup$ – Dietrich Burde Nov 24 '20 at 21:48
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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.

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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.

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