Let $G$ have exactly $24$ elements of order $6$. Why are there $12$ cyclic subgroups of $G$ of order $6$

Question

This Exercise 8.47 of Gallian's "Contemporary Abstract Algebra".

Answers that use material from the textbook prior to the question are preferred.

Let $G$ be a group with exactly $24$ elements of order $6$. Why are there $12$ cyclic subgroups of $G$ of order $6$?

The question is phrased differently in the textbook. It asks for the number of such cyclic subgroups and provides $12$ as an answer in the back.

Context:

I've been thinking about this on & off for about a week but have got nowhere. This is particularly frustrating because I think I should be able to do it myself.

Here is a similar question of mine:

Elements and cyclic subgroups of order $15$ in $\Bbb Z_{30}\times \Bbb Z_{20}.$

The kind of answer I'm looking for is a general approach (with the preference above). The reason why is that it's evident that my understanding is lacking here. Perhaps some easier exercises to this end might help too.

Please help 🙂

Answer 1

I'll make a try:

We have $24$ elements of order $6$. Every cyclic group of order $6$ has $2$ generators. So we have at least $\dfrac{24}{2}=12$ cyclic groups of order $6$.