Prove or disprove: a group with 3 different elements of order 6 can't be cyclic.

Question

I have a question of prove or disprove as above: a group $G$ with $3$ different elements of order $6$ can't be cyclic.
I tried to find a group of order $6*n$ such that it has 3 different elements, but having trouble doing so. I know it must be isomorphic to $Z_{6n}$, and the elements $n, 5n$ are both of order 6. However, I can't find a third different element (it can be a numeric example i.e. $120$ and $20/100$). That's why I think it may be true. However, I can't think of a way to prove it.
Any help will be appreciated!

Answer 1

As you noted, if a group is cyclic and has at least an element of order $6$, then $6$ is a divisor of the cardinal of the group. There is exactly one subgroup of order $6$. And it happens, that the cyclic group of order $6$ does not have three distinct generators, only two.

Answer 2

A cyclic group of order $n$ has exactly one subgroup of any order dividing $n$.

Answer 3

Note that there is no group with an odd number of elements of order $n$ for any $n\neq 1,2$. This is because for each $x\in G$ of order $n$, $x^{-1}$ has order $n$, and this partitions the set of elements of order $n$ into pairs. Thus if there are finitely many of them there is an even number of them. In particular, a group that doesn't exist cannot be cyclic.

If $G$ has even order, this means that the number of elements of order $2$ is odd.