Some Simple Applications of the Sylow Theorems

Proposition 1. Any group $G$ of order $20449 = 11^2 \cdot 13^2$ must be abelian.

Proof: First, the number of $11$-Sylow subgroups of $G$ must be a divisor of $13^2 = 169$ and must also be congruent to $1$ modulo $11$. Among
$$
\{1,13,169\},
$$
the only number congruent to $1$ modulo $11$ is $1$. Hence $G$ has exactly one $11$-Sylow subgroup, so it is normal. Denote it by $A$. Then $|A| = 11^2$.

Similarly, the number of $13$-Sylow subgroups of $G$ must be a divisor of $11^2 = 121$ and congruent to $1$ modulo $13$. Hence it must also be $1$. So $G$ has exactly one $13$-Sylow subgroup, which is therefore normal. Denote it by $B$. Then $|B| = 13^2$.

Since every group of order $p^2$ is abelian, both $A$ and $B$ are abelian.

Next, note that $(|A|,|B|)=1$. Therefore
$$
A \cap B = \{e\},
$$
and hence
$$
|AB| = \frac{|A||B|}{|A \cap B|} = 11^2 \cdot 13^2 = |G|.
$$
Since $AB \subseteq G$, we must have
$$
AB = G.
$$

Finally, since $A$ and $B$ are both normal subgroups of $G$ and $A \cap B = \{e\}$, we have
$$
AB = BA.
$$
Combining this with the fact that both $A$ and $B$ are abelian, we conclude that $G$ is also abelian.

Proposition 2. Any group $G$ of order $56$ is not simple.

Proof: Since
$$
56 = 2^3 \cdot 7,
$$
the number of $7$-Sylow subgroups of $G$ must be a divisor of $8$ and congruent to $1$ modulo $7$. Among
$$
\{1,2,4,8\},
$$
the possible values are therefore $1$ or $8$.

We now divide into two cases.

If $G$ has exactly one $7$-Sylow subgroup, then it is normal, so $G$ is not simple.

If $G$ has exactly $8$ distinct $7$-Sylow subgroups, then each such subgroup has $7-1=6$ nonidentity elements of order $7$. Since distinct $7$-Sylow subgroups intersect trivially, the total number of elements of order $7$ is
$$
8 \times (7-1) = 48.
$$
Since
$$
56 - 48 = 8,
$$
the remaining $8$ elements must form the unique $2$-Sylow subgroup of $G$, because any $2$-Sylow subgroup has order $8$. Hence $G$ has a normal $2$-Sylow subgroup, so $G$ is not simple.

In either case, $G$ is not simple.

Proposition 3. Any group of order $108$ is not simple.

Proof: Since
$$
108 = 2^2 \cdot 3^3,
$$
the number of $3$-Sylow subgroups of $G$ must be a divisor of $4$ and congruent to $1$ modulo $3$. Among
$$
\{1,2,4\},
$$
the possible values are therefore $1$ or $4$.

We now divide into two cases.

If $G$ has exactly one $3$-Sylow subgroup, then it is normal, so $G$ is not simple.

If $G$ has exactly four $3$-Sylow subgroups, let $A$ and $B$ be two distinct $3$-Sylow subgroups of $G$. Then
$$
|AB| = \frac{|A||B|}{|A \cap B|}
= \frac{27 \times 27}{|A \cap B|}
\le |G| = 108,
$$
which implies
$$
|A \cap B| \ge \frac{27}{4}.
$$
Since $A \cap B$ is a subgroup of $A$, by Lagrange’s theorem its order must divide $27$. Therefore
$$
|A \cap B| = 9,
$$
and hence
$$
|AB| = 81.
$$

We now prove that $A \cap B$ is a normal subgroup of $G$. It suffices to show that
$$
N(A \cap B) = G.
$$

From the previous article, we know that any subgroup of order $p^{n-1}$ in a group of order $p^n$ is normal. Hence
$$
A \cap B \lhd A
\qquad \text{and} \qquad
A \cap B \lhd B.
$$
Therefore,
$$
A \subseteq N(A \cap B)
\qquad \text{and} \qquad
B \subseteq N(A \cap B).
$$
By closure of subgroups under the group operation, we have
$$
AB \subseteq N(A \cap B),
$$
so
$$
|N(A \cap B)| \ge 81.
$$
Since $N(A \cap B)$ is a subgroup of $G$, its order must divide $108$ by Lagrange’s theorem. Hence
$$
|N(A \cap B)| = 108 = |G|,
$$
so
$$
N(A \cap B) = G.
$$
Therefore $A \cap B$ is a normal subgroup of $G$, and thus $G$ is not simple.

Proposition 4. Any group of order $63$ is not simple.

Proof: Since
$$
63 = 7 \cdot 3^2,
$$
the number of $7$-Sylow subgroups must be a divisor of $9$ and congruent to $1$ modulo $7$. Among
$$
\{1,3,9\},
$$
the only possibility is $1$. Hence the $7$-Sylow subgroup is normal, so the group is not simple.

Proposition 5. Any group of order $148$ is not simple.

Proof: Since
$$
148 = 37 \cdot 2^2,
$$
the number of $37$-Sylow subgroups must be a divisor of $4$ and congruent to $1$ modulo $37$. Among
$$
\{1,2,4\},
$$
the only possibility is $1$. Hence the $37$-Sylow subgroup is normal, so the group is not simple.

Proposition 6. Any group of order $200$ is not simple.

Proof: Since
$$
200 = 5^2 \cdot 2^3,
$$
the number of $5$-Sylow subgroups must be a divisor of $8$ and congruent to $1$ modulo $5$. Among
$$
\{1,2,4,8\},
$$
the only possibility is $1$. Hence the $5$-Sylow subgroup is normal, so the group is not simple.

Proposition 7. The $11$-Sylow subgroup of a group $G$ of order $231$ is contained in the center of $G$.

Proof: Note that
$$
231 = 11 \times 7 \times 3.
$$
The number of $11$-Sylow subgroups must divide $21$ and be congruent to $1$ modulo $11$. Among
$$
\{1,3,7,21\},
$$
the only possibility is $1$. Hence the $11$-Sylow subgroup is normal; denote it by $A$.

Similarly, the $7$-Sylow subgroup $B$ and the $3$-Sylow subgroup $C$ are also both normal.

We now prove that
$$
G = ABC.
$$

Since $(11,7)=1$, we have
$$
A \cap B = \{e\},
$$
and hence
$$
|AB| = \frac{|A||B|}{|A \cap B|} = 77.
$$
Furthermore, since $(77,3)=1$, we have
$$
AB \cap C = \{e\},
$$
and therefore
$$
|ABC| = \frac{|AB||C|}{|AB \cap C|} = 231 = |G|.
$$
Thus
$$
G = ABC.
$$

Now, from $A \cap B = \{e\}$, we know that for all $x \in A$ and $y \in B$,
$$
xy = yx.
$$
Similarly, for all $x \in A$ and $z \in C$,
$$
xz = zx.
$$
Since $A$ is cyclic, it follows that every element of $A$ commutes with every element of $G$. Therefore
$$
A \subseteq Z(G),
$$
that is, the $11$-Sylow subgroup of $G$ is contained in the center of $G$.

Proposition 8. Any group of order $36$ is not simple.

Proof: Since
$$
36 = 2^2 \times 3^2,
$$
the number of $3$-Sylow subgroups must be a divisor of $4$ and congruent to $1$ modulo $3$. Among
$$
\{1,2,4\},
$$
the possible values are therefore $1$ or $4$.

We divide into two cases.

If $G$ has exactly one $3$-Sylow subgroup, then it is normal, so $G$ is not simple.

If $G$ has exactly four $3$-Sylow subgroups, let $A$ and $B$ be two distinct $3$-Sylow subgroups of $G$. Then
$$
|AB| = \frac{|A||B|}{|A \cap B|}
= \frac{9 \times 9}{|A \cap B|}
\le |G| = 36,
$$
which implies
$$
|A \cap B| \ge \frac{9}{4}.
$$
Since $A \cap B$ is a subgroup of $A$, by Lagrange’s theorem its order must divide $9$. Therefore
$$
|A \cap B| = 3,
$$
and hence
$$
|AB| = 27.
$$

We now prove that $A \cap B$ is a normal subgroup of $G$. It suffices to show that
$$
N(A \cap B) = G.
$$

From the previous article, we know that any subgroup of order $p^{n-1}$ in a group of order $p^n$ is normal. Hence
$$
A \cap B \lhd A
\qquad \text{and} \qquad
A \cap B \lhd B.
$$
Therefore,
$$
A \subseteq N(A \cap B)
\qquad \text{and} \qquad
B \subseteq N(A \cap B).
$$
By closure of subgroups under the group operation, we have
$$
AB \subseteq N(A \cap B),
$$
so
$$
|N(A \cap B)| \ge 27.
$$
Since $N(A \cap B)$ is a subgroup of $G$, its order must divide $36$ by Lagrange’s theorem. Hence
$$
|N(A \cap B)| = 36 = |G|,
$$
so
$$
N(A \cap B) = G.
$$
Therefore $A \cap B$ is a normal subgroup of $G$, and thus $G$ is not simple.

The cover image of this article was taken at Lake Baikal in Russia.