Group of order 231, show the existence of a normal subgroup of order 21

Question

Let G be a group of order $231=3 \cdot 7 \cdot 11$, by Sylow theorems we know that there exists $P_{3}\in Syl_{3}(G)$ and $P_{7}\in Syl_{7}(G)$ with $P_{7}\trianglelefteq G$.

Clearly by defining $N=P_{3}P_{7}$ we obtain a subgroup (as $P_{7}$ is normal so the product subgroup is indeed a subgroup) of G of order 21 (intersection of Sylow subgroups is trivial) but I've no idea on how to show that N is normal.

Thanks in advance

Answer 1

Hints: Show that $G/P_7$ is abelian. Use this to conclude that $N = P_3P_7$ is normal in $G$.