I am preparing for an exam and am stuck on the following problem :
Let $G$ be a group of order $2555 = 5 \cdot 7 \cdot 73$, show that $G$ is cyclic.
It is not hard to show that the Sylow-73 subgroup is normal in $G$ and that at either the Sylow-5 or the Sylow-7 subgroup is normal. My thought was to prove that both the the Sylow-5 and Sylow-7 subgroups are normal, because then the claim would follow, but I am unsure how to proceed.
Perhaps using the fact that $G$ mod the Sylow-73 subgroup is cyclic? Any hint would be appreciated.
EDIT: So it suffices to show that $G$ is abelian because there is only one abelian group of order 2555, namely the cyclic group of order $2555$. Let $P_{73}$ denote the Sylow 73-subgroup. It is easy to see that $G/P_{73}$ is cyclic. Further we have that $G$ is abelian if $P_{73} \subseteq Z(G)$.
I believe that $P_{73} \subseteq Z(G)$ can (may?) be shown by letting $G$ act on $P_{73}$ via conjugation. Then by the class equation, $73 = \sum_{i =1}^n \text{cl}(x_{i})$ where $x_{1}, \dots, x_{n}$ is a set of representatives of each conjugacy class. I think that one can show that $\text{cl}(x_{i})$ is trivial, which implies that $gx_{i}g^{-1} = x_{i}$ for all $g \in G$ or, equivalently, that $gx_{i} = x_{i}g$. This would show that each element of $P_{73}$ is an element of the center of $G$ and then the claim would follow.