TITLE: Group of order $315$ has an order $5$ group in its center
QUESTION [6 upvotes]: This question is based off of the Fall 2019 UCLA Algebra qualifying exam:
I am trying to show that any group $G$ of order $315=3^2\cdot5\cdot 7$ has a normal subgroup of order 5. Since the automorphism group of $\mathbb{Z}/5\mathbb{Z}$ is of order 4, which is prime to 315, this would prove that there is a group of order 5 in the center of $G$.
By the Sylow theorems, the number of 5-Sylow subgroups is $\equiv 1\mod{5}$, and by the orbit-stabilizer theorem, the number of 5-Sylow subgroups must divide $3^2\cdot 7$, which implies that the number of 5-Sylow subgroups $n_5$ is either 21 or 1. Using the same method, we see that $n_7$ is either 15 or 1 and $n_3$ is either 1 or 7.
I at least want to prove that one of the Sylow sugroups is normal. Using a counting argument doesn't help me here, since there are at most $21\cdot4=84$ elements of order 5 and there are at most $15\cdot 6=90$ elements or order 7. I am unsure of how to bound the elements of order 3 or 9, but adding the maximum number of elements of order 3 or 9 with $84+90$ won't get us a number higher than 315, so we won't get a contradiction.
Another possible way to prove there exists a normal sylow subgroup is by looking at their normalizers. Suppose that none of the Sylow groups are normal, so $n_5=21$, $n_7=15$, and $n_3=7$. Then there exists normalizers of certain Sylow groups of order $15$, $21$, and $45$. I am unsure how to use this fact to find a contradiction.
I am not looking for a detailed solution, but any small hints to this problem are appreciated!

REPLY [5 votes]: If $n_3 = 7$ and $P \in {\rm Syl}_3(G)$, then $|N_G(P)|=45$.
Neither of the the two possible types of groups of order $9$ ($C_9$ and $C_3 \times C_3$) has an automorphism of order $5$, so there exists $Q \in {\rm Syl}_5(G)$ with $Q \le C_G(P)$, and hence $P \le N_G(Q)$, and now we can prove that $n_5=1$.
I will leave the case $n_3=1$ to you.