Below is my attempt at proving that every group of order $595$ is cyclic. I would really appreciate feed back on its validity. Thanks in advance!
Proof:
Let $G$ be a group and $|G| = 595$. The prime factorization of $595$ is $5\cdot7\cdot17$. Let $n_{p}$ the number Sylow $p$-subgroups. By Sylow's Theorem, $n_{5} \equiv 1\pmod{5}$ and $n_{5} \mid 119$. The factors of $119$ are $1,7,17,$ and $109$. Only $1$ is congruent to $1$ modulo $5$, so $n_{5} = 1$. Since there is a unique Sylow $5$-subgroup, it is therefore normal, denote it $H_{5} = \langle x \rangle$.
We also have that $n_{17} \equiv 1\pmod{17}$ and $n_{17} \mid 35$. The factors of $35$ are $1,5,7,$ and $35$. Hence, $n_{17} = 1$ or $n_{17} = 35$. Suppose $n_{17} = 35$. Since $17^{2} \nmid 595$, those $85$ subgroups contain $35\cdot16+1=561$ distinct elements. Let $K_{17}$ be one of those subgroups, then $K_{17} = \langle k \rangle$ for some element $k \in G$ of order $17$. We can then construct a cyclic subgroup generated by $\langle xk \rangle$ which has order $17\cdot5=85$. Since there are $64$ numbers less than $85$ that are coprime with $85$, there are $64$ elements that generate this subgroup and thus $64$ elements of order $85$. Since the order of an element divides the order of a subgroup, they are not contained in any of the $35$ Sylow $17$-subgroups. Therefore we have at least $561+64=625$ elements, which contradicts $|G|=595$. Therefore $n_{17}=1$ and there is a normal Sylow $17$-subgroup, denote it $H_{17} =\langle y\rangle$.
Similarly, $n_{7} \equiv 1\pmod{7}$ and $n_{7} \mid 85$. The factors of $85$ are $1,5,17$, and $85$. As such, $n_{7}=1$ or $n_{7} = 85$. Suppose $n_{7} = 85$. Then since $7^{2} \nmid 595$, those $85$ Sylow $7$-subgroups would contain $85\cdot6+1=511$ distinct elements. Let $K_{7}$ be one of those Sylow $7$-subgroups, then $K_{7} = \langle l \rangle$ for some element $l \in G$ with order $7$. We can construct a cyclic subgroup generated by $\langle ly\rangle $ which has order $7 \cdot 17 = 119$. Since there are $96$ numbers less than $119$ that are coprime with $119$, there are $96$ elements of this cyclic subgroup that generate it and hence $96$ elements of order $119$. These elements are not contained in any of the Sylow $7$-subgroups since $119 \nmid 7$. As such, $G$ would have at least $511 + 96 = 607$ elements which contradicts $|G| = 595$. Therefore, $n_{7} = 1$ and there is a normal Sylow 7-subgroup, denote it $H_{7} = \langle z \rangle$.
Since $H_{5}$, $H_{7}$, and $H_{17}$ are all of prime order, each pair of them intersect trivially, and so:
\begin{align*} |H_{5}H_{7}H_{17}| = \frac{|H_{5}||H_{7}||H_{17}|}{|H_{5} \cap H_{7}||H_{5} \cap H_{17}||H_{7} \cap H_{17}|} = \frac{5 \cdot 7 \cdot 17}{1\cdot 1\cdot 1} = 595 \end{align*}
And since $|G| = |H_{5}H_{7}H_{17}|$, each intersect trivially, and they are each normal, we have: $G = H_{5}H_{7}H_{17} \cong H_{5} \times H_{7} \times H_{17} \cong \mathbb{Z}/595\mathbb{Z}$. Hence, $G$ is cyclic.
