Cyclic group of order 24

Question

I have been asked to prove or disprove the following:

If $G$ is a group of order $24$ such that $a\in G$ , $a^{12}\ne e$ and $a^{8}\ne e$ where $e$ is the identity of $G$, then the group is cyclic.

Can anyone assist?

Answer 1

By Lagrange's theorem, order of $a$ is a divisor of $24$. Since $a^{12}$ and $a^{8}$ are not identity, this order does not divide $12$ or $8$. Consider the divisors of $24$ and eliminate each divisor that divides $12$ or $8$. You are left with a single divisor which must be the order of $a$. What is that number?