Show that there exist a cyclic normal subgroup $N$ of $G$ whose order is $255$ such that $G/N$ is cyclic

Question

I was practicing for my Group Theory exam and I could not solve the following question:

Let $G$ be a group of order $255=3\cdot 5\cdot 17$. Find a cyclic normal subgroup $N$ of $G$ such that $G/N$ is cyclic.

Since the number of Sylow $17$-subgroups of $G$ is $1$, it should be normal. I defined $N$ to be that subgroup. Since $|N|=17$, $N$ is isomorphic to $\Bbb Z/17\Bbb Z$, so it is cyclic. Now, how can I show that $G/N$ is cyclic?

Thanks in advance.

Answer 1

Since $[G:N]=15$ the quotient must be isomorphic to $\mathbb{Z}_{15}$, because that's the only subgroup of order $15$

To prove it simply since the p-Sylow subgroups of a group of order $15$ are unique, there are 4 elements of order 5, 2 elements of order 3, and of course, a single element of order 1 (the identity). Since $4+2+1=7<15$, there exists $x∈G$ of a different order. Thus, the order of $x$ must be 15, and therefore $G≅⟨x⟩≅\mathbb{Z}_{15}$

Answer 2

HINT:

$|G/N|=15$ and any group of order $15$ is cyclic.