Group of order $3^3$ must have a normal subgroup

Question

How can I show that a group of order $27=3^3$ must contain a normal subgroup? I’m not sure what Sylow’s Theorem tells me

Answer 1

A group of order $27$ is a $3$-group, so it has a non-trivial center, which is a normal subgroup.

Answer 2

Sylow's theorem won't tell you anything since the order of your group has a prime power.

I suppose you want your subgroup to be different from $\{ 1\}$ and $G$.

If $G$ is abelian, pick the subgroup generated by an element of order $p$.

For the non-abelian case, a classical result that if $G$ is a non trivial $p$-group (that is $\vert G\vert= p^n, n\geq 1$, where $p$ is prime), then its center $Z(G)$ is a non trivial subgroup. But the center is a normal subgroup, different from $G$ if $G$ is non abelian.

To prove this result, let $G$ acts on itself by conjugation. Since the order of $G$ is a nontrivial power of $p$, an order has either one element or a cardinality multiple of $p$. The first case corresponds to elements of the center. We then get $\vert G\vert\equiv \vert Z(G)\vert [p]$, and then In particular, $\vert Z(G)\vert\equiv \ 0 [p]$ by assumption on $G$, and $Z(G)$ cannot be trivial.