Prove that a group $G$ containing $20$ elements of order $19$ is not cyclic.

Question

I am trying to prove that a group $G$ containing $20$ elements of order $19$ is not cyclic.

With just this information I'm not sure how to go about finding a generator and am looking at proving this more generally. I was thinking of using the fact that a group is not cyclic if there are two subgroups of the same order but not completely sure how to go about that.

Answer 1

Choose an element $x\in G$ of order 19. As the subgroup generated by $x$ has cardinality $|\langle x\rangle |$ = 19 and there are 20 elements of order 19, we find an element $y \in G \setminus \langle x\rangle$ of order 19. Now, $\langle x\rangle$ and $\langle y\rangle$ are two distinct subgroups of $G$ that are both of the same order. Hence $G$ is not cyclic.

Answer 2

For a finite $G$, it follows from McKay’s proof of Cauchy’s Theorem that the number of elements of order $19$ must be $\equiv -1$ mod $19$. Hence the indicated group does not exist.

Answer 3

At least $19$ such elements is a sufficient condition to get the claim, as it would imply that there are more than one cyclic subgroup of order $19$ (each containing exactly $18$ elements of order $19$). And elements of order $19$ can come in multiple of $18$, only, because each such an element lie in one and only one subgroup of order $19$.