Existence of a $\mathbb{Z} _{15}$ in a group $G$ where $|G|=3 \cdot 5 \cdot 17$ 
Existence of a $\mathbb{Z} _{15}$ in a group $G$ where $|G|=3 \cdot 5 \cdot 17$.

This is my proof:
we know (from a previous point of the exercise) that $\mathbb{Z}_{17}$ is a characteristic subgroup of $G$. For Cauchy, we have that $\exists \mathbb{Z}_3,\ \mathbb{Z}_5 <G$. If we demonstrate that $\mathbb{Z}_3$ or $\mathbb{Z}_5$ is normal in $G$, then we have that $\mathbb{Z}_3\mathbb{Z}_5$ (or $\mathbb{Z}_5\mathbb{Z}_3$) is a subgroup of $G$ of order 15, therefore is isomorphic to $\mathbb{Z}_{15}$.
For Sylow, we have that $n_5 \equiv 1 \pmod{5}$ and $n_5 \mid 51$ where $n_5$ is the number of $5$-Sylow in $G$, so $n_5=1,51$.
If $n_5=1$ then the only $5$-Sylow in $G$ is normal.
If $n_5=51$ then there are $51 \cdot 4$ elements of order $5$ in $G$, and of the remaining $51$ elements there are $16$ of order $17$ (from the only $\mathbb{Z}_{17}$) and one of order $1$, that is the identity. So we have $34$ elements left.
At this point, I thought about two different conclusions, but I am not really sure which one is the correct one:

*

*All the $34$ elements have to be of order $3$, therefore there are $17$ $3$-Sylow, but this is impossible because $17 \not\equiv 1 \pmod{3}$. So we can conclude that $n_5=1$.


*We have $34$ elements, so we can have at most $17$ $3$-Sylow (in other words, $n_3\le 17$), and this means that $n_3$ can only be $1$. So if $n_5 \neq 1$ then $n_3=1$.
 A: $\textbf{Hint}:$ let's take a look at the number of groups in $Syl_{17}(G)$ and $Syl_3(G)$.
We know that $n_{17}\equiv1\text{ (mod 17)}\implies n_{17}\in \{1\not3,\not5,\not{15}\}\implies \exists!P_{17}\trianglelefteq G$.
While $n_3\equiv 1\text{ (mod 3)}\implies n_3\in\{1,\not5,\not17,85\}.$
If we now consider the action of conjugation of $G$ on its Sylow's groups we have that $$|Syl_{17}(G)|=|G:N_G(P_{17})|\implies|N_G(P_{17})|=|G|.$$
Furthermore $$\dfrac{|N_G(P_{17})|}{|C_G(P_{17})|}\bigg||\operatorname{Aut}(P_{17})|=\varphi(17)=16\implies\dfrac{3\cdot 5\cdot 17}{|C_G(P_{17})|}\big |2^4\iff|C_G(P_{17})|=|G|.$$
$P_{17}$ is abelian, so $P_{17}\le N_G(P_3)\implies17\big||N_G(P_{3})|$. We also have $\{1,85\}\ni n_3=|G:N_G(P_3)|$, so $n_3=1$. You can do something similar for $P_5$.
A: As also indicated in the comments, there is a fast argument based on counting elements of order $3$ and $5$. It has also been used as a first step here.
By Sylow, $n_3\in \{1,85\}$ and $n_5\in\{1,51\}$.
Assume that $n_3 = 85$ and in the same time $n_5 = 51$. Then $G$ contains $85\cdot 2 = 170$ elements of order $3$ and $51 \cdot 4 = 204$ elements of order $5$, which is not possible since $G$ contains only $255$ elements.
So $n_3 = 1$ or $n_5 = 1$. Let $P_3$ be a $3$-Sylow subgroup and $P_5$ a $5$-Sylow subgroup. Then at least one of the groups $P_3$ and $P_5$ is normal in $G$, implying that $P_3 P_5$ is a subgroup of $G$. Since $P_3$ and $P_5$ have trivial intersection, $\#(P_3 P_5) = 3\cdot 5 = 15$. Now by the classification of $pq$-groups and $5\not\equiv 1\bmod 3$, the subgroup $P_3 P_5$ is cyclic of order $15$.
(In fact, much more can be shown, namely that every group of order $3\cdot 5\cdot 17 = 255$ is abelian and hence cyclic (by the classification of finite abelian groups, since 255 is square-free). Therefore, there is a unique subgroup of order $15$, which is cyclic.)
