[Math] Show by Lagrange’s theorem that a group $G$ of order $27$ should have a subgroup of order $3$.

Question

Show by Lagrange's theorem that a group $G$ of order $27$ should have a subgroup of order $3$.

Attempt

We have $o(a)|o(G)$ for all $a\in G$. Then the possibilities of orders for the elements og $G$ are $1, 3, 9, 27$.

Let $e\neq x\in G$ then $x^{27}=e$ then $(x^9)^3=e$. Now let us consider a cyclic subgroup $H=<x^9>=\{e, x^9, x^{18}\}$. $H$ is a subgroup of $G$ of order $3$.

Is that proof error free? Is there any other method to solve the problem? Please use elementary tools (not use Sylow's theorem, Normal subgroup etc).

Answer 1

Let $x\in G$ with $x\neq e$. By Lagrange theorem, the order of $x$ is $3$, $9$ or $27$.

• If $x$ has order $3$, take $H = <x>$.
• If $x$ has order $9$, then $x^3$ has order $9/3$. Take $H = <x^3>$.
• If $x$ has order $27$, then $x^9$ has order $27/9$. Take $H = <x^9>$.