Having a Normal Sylow 3 subgroup

Question

Suppose that a group $G$ of order $1575 = 3^2* 5^2 * 7$ has a normal Sylow $3$ subgroup, then how do I show that $G$ contains a normal Sylow $5$ and Sylow $7$ subgroup? I know that $G/P_{3}$ ($P_{3}$ denotes the sylow $3$ subgroup) has a normal sylow $5$ and sylow $7$ subgroup $M_{5}$ and $M_{7}$ by doing some arithmetic, and that the correspondence theorem tells me that $M_{5}P_{3} = \{ab : a\in M_{5}, b \in P_{3}\}$ and $M_{7}P_{3}$ are normal in $G$ and that normal Sylow $p$ subgroups are characteristic. Yet I am still unable to figure out how to proceed.

Answer 1

In $M_{5}P_{3}$ the number of conjugates of $M_{5}$ is $\frac {225}{|N(M_{5})|}$ and is therefore 1, 3 or 9. By Sylow's Theorems we know this number is $1+5k$ and so it must equal 1. Hence $M_{5}$ is a characteristic subgroup of a normal subgroup of $G$ and is therefore normal in $G$.

Similarly, the number of conjugates of $M_{7}$ in $M_{7}P_{3}$ is 1, 3 or 9 and is also $1+7k$. Thus $M_{7}$ is normal in $G$.