group of order 27 must have a subgroup of order 3

Question

How to prove a group of order 27 must have a subgroup of order 3

Answer 1

By Lagrange's theorem, the order of every element is a divisor of $27$.

Take any $g\in G$ with $g\ne1$.

If $ord(g)=27$, then $g^9$ has order $3$.

If $ord(g)=9$, then $g^3$ has order $3$.

If $ord(g)=3$, then $g$ has order $3$.

The same argument proves this:

If $p$ is prime, then every group of order $p^n$ has an element of order $p$.

Indeed, take any $g\in G$ with $g\ne1$. Let $ord(g)=p^m$. Then $g^{p^{m-1}}$ has order $p$.