# 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.

• Well the fact that each subgroup has $18$ elements distinct to it (not included in any other group) might not be true. – Stefan4024 Oct 8 '16 at 19:25
• @Stefan4024 The minimum of $|\bigcup_{i=1}^{37} H_i|$ occur in $H_i\cap H_j=K$ for any $i\neq j$, where $|K|=9$. In that case, $|\bigcup_{i=1}^{37} H_i|=n_{37}\cdot |H_i\backslash K|+|K|=37\cdot (37-9)+9=675$. – bfhaha Oct 9 '16 at 8:02
• But look at this $H_1$ intersects with $H_2$ and have $9$ common elements. Now $H_1$ intersects with $H_3$ and have $9$ common elements (except the identity), other than those included in $H_2$. $H_1$ intersects with $H_4$ and another of the $9$ elements in $H_1$ are the intersection. Now what if intersection of $H_3$ and $H_4$ isn't the identity? Aren't we doublecounting? – Stefan4024 Oct 9 '16 at 8:12
• You say that the minimum occurs when each pair of subgroups intersect in the same subgroup $K$ of order $9$. How do you justify that claim? – Derek Holt Oct 9 '16 at 16:47
• The group $A_7$ has $70$ Sylow $3$-subgroups of order $9$. According to your reasoning there should be at least $6 \times 70 + 3 = 423$ elements of order $3$ or $9$. But in fact there are only $350$. – Derek Holt Oct 9 '16 at 16:52

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.

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 $10989=3^3\cdot 11\cdot 37$).

1. 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.
2. Show that $G$ contains either a normal Sylow $37$-subgroup or a normal Sylow $3$-subgroup.
3. Explain briefly why (in all cases) $G$ has a normal Sylow $11$-subgroup.
4. 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.)