Is my proof correct? Let $G$ be a group of order $10989$. Prove that $G$ contains either a normal Sylow $37$-subgroup or a normal Sylow $3$-subgroup Let $G$ be a group of order $10989=3^3\cdot 11\cdot 37$. Prove that $G$ contains either a normal Sylow $37$-subgroup or a normal Sylow $3$-subgroup
It is easy to show that $n_3\in \{1, 37\}$ and $n_{37}\in \{1, 3^3\cdot 11\}$ by Sylow 3rd Theorem.
Suppose that $n_{37}=3^3\cdot 11$.
By Lagrange's Theorem, 
every non-identity element in a Sylow $37$-subgroup is of order $37$.
Hence, 
there are $n_{37}\cdot (37-1)=3^3\cdot 11\cdot 36=10692$ elements of order $37$.
There are $10989-10692=297$ remaining.
If $n_3=37$.
Let $H_1, H_2, ..., H_{37}$ be all the Sylow $3$-subgroups of $G$.
There are at least $37\cdot 18+9=675$ elements in $\bigcup_{i=1}^{37}H_i$.
The remaining $297$ elements can't form these Sylow $3$-subgroups.
Therefore, $n_3=1$.
There is another solution. 
But my method is more basic. 
But I have never see this method.
So I worry about that there are some mistake I am not aware.

This was answered by Derek Holt. 
Consider the alternating group $A_7$.
Consider the following $6$ Sylow $3$-subgroups.
A triangle stand for a Sylow $3$-subgroup.
(I omitted some product of two $3$-cycles.)

There are $4\cdot 6+2\cdot 6+1=37$ elements of order $3$.
(No element in $A_7$ is of order $9$.)
But according to my thinking. 
There are at least $6*6+3=39$ elements of order $3$ in these $6$ Sylow $3$-subgroups.
