Let $G$ be a group of order $72$. Show that if $G$ is not cyclic, then for any $g \in G$ either $g^{24} = 1$ or $g^{36} = 1$

Question

I have the following problem and could not understand why there are just two cases.

Problem: Let $G$ be a group of order $72$. Show that if $G$ is not cyclic, then for any $g \in G$ either $g^{24} = 1$ or $g^{36} = 1$

I know that the order of an element in a group divides the order of the group, and $g^{72} = 1$ cannot be the case because $G$ is not cyclic but, for example, why are we not considering of $g^{12} = 1$ case?

I would be appreciated if anyone help me.

Regards.

Answer 1

Let $g\in G$.

Since $|G|=72$, we must have, by Lagrange's Theorem, that $|g|$ divides $72$.

But $G$ is not cyclic, so the order of $g$ cannot be $72$ (despite the fact that $g^{72}=1$).

Therefore,

$$|g|\in\{ 1,2,3,6,8,9,12,18,24,36\}.$$

The result follows; for example, if $|g|=9$, then $g^9=1$ implies $g^{36}=(g^9)^{4}=1^{4}=1$.