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.
 A: 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.
A: This is a question of the algebra Ph.D. qualifying exam on Jan. 13, 2009 of the university of Vermont.
The original question is:
Let $G$ be a group of order (note that $10989=3^3\cdot 11\cdot 37$).


*

*Compute the number, $n_p$, of Sylow $p$-subgroups permitted by Sylow's Theore for each of $p=3, 11$, and $37$;
for eachof these $n_p$ give the order of the normalizer of a Sylow $p$-subroup.

*Show that $G$ contains either a normal Sylow $37$-subgroup or a normal Sylow $3$-subgroup.

*Explain briefly why (in all cases) $G$ has a normal Sylow $11$-subgroup.

*Deduce that the center of $G$ is nontrivial.


The correct solution of 3. is as following.
We already know that $n_3\in \{1, 37\}$ and $n_{37}\in \{1, 297\}$.
Suppose that $n_3=37$ and $n_{37}=297$.
We prove that there is a contradiction.
Since $n_3=37$,
let $H$ be a Sylow $3$-subgroup,
then $[G:N(H)]=n_3=37$ and $N(H)=3^3\cdot 11$.
Since $n_{37}=297$,
there are $36\cdot 297=10692$ elements of order $37$.
There are $10989-10692=3^3\cdot 11$ elements remaining.
Thus, these remaining elements form the unique subgroup $N(H)$ of order $3^3\cdot 11$.
Since $gN(H)g^{-1}$ all are subgroup of order $3^3\cdot 11$ for all $g\in G$,
by the uniqueness of $N(H)$,
$gN(H)g^{-1}=N(H)$ and $N(H)\lhd G$ and $N(N(H))=G$.
Since $H$ is a Sylow $3$-subgroup,
we have $N(N(H))=N(H)$.
Which is a contradiction $N(H)=N(N(H))=G$.
(Or one can show that $H$ is the unique subgroup of order $3^3$ in $N(H)$ by the Sylow's Theorem.
Since $\sigma(H)$ is also a subgroup of order $3^3$ for all automorphism on $N(H)$,
by the uniqueness of $H$,
$\sigma(H)=H$ and $H$ is characteristic in $N(H)$.
Thus, $H\stackrel{\text{char.}}{\leq}N(H)\lhd G$ and $H\lhd G$,
a contradiction.)
