Show group of order $4563=3^3\cdot13^2$ is not simple.

The next things to try are often:

• to consider the intersections of Sylow groups, and
• to use the fact that groups of order $p^2$, $p$ a prime, are always abelian.

Let's see. Assume that we have two Sylow $13$-subgroups that intersect non-trivially, say $P_1\cap P_2$ and $x\neq 1_G, x\in P_1\cap P_2$ with $|P_1|=|P_2|=13^2$. Then $x$ is centralized by both $P_1$ and $P_2$. This means that $C_G(x)$ has order that is a proper multiple of $13^2$. The size of the conjugacy class of $x$, $[G:C_G(x)]$, is thus a proper factor of $[G:P_1]=27$, and therefore at most nine. If $x$ is in the center, $G$ cannot be simple. Otherwise we get a non-trivial homomorphism from $G$ to $S_9$ by the conjugation action on the conjugacy class of $x$. Again, we are done.

So let's assume that all those twenty-seven Sylow $13$-subgroups intersect trivially. In that case their union contains $(27\cdot 168)+1=|G|-26$ elements. Now we can look at Sylow $3$-subgroups. Their non-identity elements must all be among those $26$ elements not belonging to any Sylow $13$. This means that there is room for only a single Sylow $3$-subgroup, which is then normal concluding the proof.