There is no simple group of order $2025$.
If $G$ is simple then by the Sylow theorem, there must be $81$ Sylow $5$-subgroups and $25$ Sylow $3$-subgroups. Also, I can see that if Sylow $5$ subgroups intersect only on the identity element then $1944$ elements have order $5$ or $25,$ from here, Sylow $3$ subgroup must be unique hence normal and the assumption that $G$ is simple must be false. But I couldn't find a contradiction for distinct Sylow $5$-subgroups $P$ and $Q$ where $|P \cap Q|=5. $
My question is: How can I prove the rest of this argument? And could we solve the same problem if we work on Sylow 3-subgroups using the same argument?