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On this page:

• Introduction
• Sylow’s Theorems
• Applications of Sylow
• Worked Exercises

Introduction

A classical theorem theorem in group theory is Lagrange’s theorem. In particular, it says that if $H$ is a subgroup of a finite group $G$, then the order of $H$ must be a factor of $G$. A very natural question is to ask the converse: given a particular divisor of $|G|$, does there exist a subgroup $H \le G$ with order being that divisor? The answer is no but there are partial converses that are true as we will see.

Sylow’s Theorems

[definition] [deftitle]Definition %counter%[/deftitle]

Let $G$ be a group and let $p \in \ZZ_{> 0}$ be prime.

1. A group of order $p^n$ for some $n \in \ZZ_{> 0}$ is called a $p$-group. Subgroups of $G$ that are $p$-groups are called $p$-subgroups.
2. If $G$ is a group of order $p^n m$ where $p$ and $m$ are relatively prime, then a subgroup of order $p^n$ is called a Sylow $p$-subgroup of $G$.
3. The set of Sylow $p$-subgroups of $G$ will be denoted by $\Syl_p(G)$ and $n_p(G) = |\Syl_p(G)|$. If $G$ is clear from the context, we may just write $n_p \coloneqq G$. [/definition]

Let us now immediately give the Sylow Theorems. We won’t prove them, but we will show some applications of them.

[theorem] [thmtitle]Theorem %counter% (Sylow’s Theorems)[/thmtitle]

Let $G$ be a group of order $p^n m$ where $p$ and $m$ are relatively prime.

1. $G$ has at least one Sylow $p$-subgroup (i.e. $\Syl_p(G)$ is nonempty).
2. All pairs of Sylow $p$-subgroups are conjugate in $G$.
3. $n_p \equiv 1 \pmod p$. Futhermore, $n_p$ is the index of $N_G(P)$ in $G$ for any Sylow $p$-subgroup $P$. Thus, $n_p \mid m$. [/theorem]

[example] [extitle]Example %counter%[/extitle]

Let $G$ be a finite group and let $p$ be a prime integer. If the order of $G$ is $p^n$ for some positive integer $n$, then the $G$ is the unique Sylow $p$-subgroup of $G$. [/example]

[example] [extitle]Example %counter%[/extitle]

Let $G$ be a finite abelian group. By the classification of finitely-generated modules over a PID, we know that $G$ can be realized as

where $C_{p_i^{e_i}}$ is the cyclic group of order $p_i^{e_i}$ written multiplicatively. So, then a finite abelian group has a unique Sylow $p$-subgroup for each prime $p$. [/example]

Applications of Sylow

Sylow’s theorem is typically used to show that groups of particular orders are not simple or sometimes even to show that there are no subgroups of a particular order.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $G$ be a group of order $pq$ where $p$ and $q$ are primes with $p < q$. Let $P \in \Syl_p(G)$ and let $Q \in \Syl_q(G)$. Then $Q$ is normal in $G$. Moreoever, if $P$ is also normal in $G$, then $G$ is cyclic. [/theorem]

[proof] First by Sylow’s theorem, we know that $P$ and $Q$ actually exist to begin with.

(Normality of $Q$) It suffices to show that $n_q = 1$. By Sylow III, we know that

and also that $n_p \mid q$ and $n_q \mid p$ (because $G$ is a group of order $pq$). So then either $n_q = 1$ or $n_q = p$. Note that it cannot be $p$ since $p < q$ which would contradict the fact that $n_q \equiv 1 \pmod q$. Thus, $n_q = 1$. This means that there is only one subgroup of order $q$ in $G$ and Sylow II tells us that all subgroups of order $q$ are conjugate to each other. Therefore, $Q \unlhd G$.

(Cyclicity of $G$) Now suppose that $P$ is normal in $G$. The normality of $P$ implies that $n_p = 1$. Consider $G \setminus P\cup Q$. All elements of $G$ must have order something other than $1$ (the identity is in both $P$ and $Q$), $p$, and $q$ (these elements generate $P$ and $Q$). Thus, if this set is nonempty, there must be an element of order $pq$. So it suffices to show that said set is actually nonempty. As

and $p < q$ implies

it follows that $(p+q)^2 \le 4q^2$ which means that

Thus, $p + q - < pq = |G|$ so $G$ has an element of order $pq$. [/proof]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $G$ be a group of order $pqr$ where $p < q < r$ are prime integers. Then $G$ has a least one normal Sylow subgroup. [/theorem]

[proof] By Sylow III, we know that $n_r \equiv 1 \pmod r$ and also that $n_r \mid pq$. This implies that $n_r \in \{1, p, q, pq\}$. If $n_r = 1$, we are done. Now, we know that $n_r = p$ and $n_r = q$ is impossible since this would violate the claim that $n_r \equiv 1 \pmod r$ as $p < q < r$.

For $n_r = pq$ case, we know that $n_q \in \{1, p, r, pr\}$. Clearly, if $n_q = 1$, we are done. It’s also clear that $n_q = p$ is impossible by Sylow III and if $n_q = r$, then we would have at least

in $G$. Notice that these are all distinct as well because each Sylow subgroup has prime order and so all pairs of (distinct) Sylow subgroups have trivial intersection. But since

it follows that we would have more than $pqr$ elements in $G$, contradiction. Obviously, if $n_q = r$ is impossible, then we certainly cannot have $n_q = pr$ so the claim follows. [/proof]

Worked Exercises

[example] [extitle]TODO: Cite[/extitle]

Let $G$ be a group of order $462 = 2\cdot 3 \cdot 7 \cdot 11$. Prove that $G$ is not simple. [/example]

[proof] We claim that $n_{11} = 1$. By Sylow III, we know that $n_{11}$ is a divisor of $2\cdot 3 \cdot 7$. The list of divisors of $2\cdot 3 \cdot 7$ is

The only number in this list congruent 1 modulo $11$ is just $1$ so it follows immediately that $n_{11} = 1$ which means that the Sylow $11$-subgroup is normal by Sylow II. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Prove that if $|G| = 312 = 2^3 \cdot 3 \cdot 13$, then $G$ has a normal Sylow $p$-subgroup for some $p$. [/example]

[proof] We claim that $n_{13} = 1$. By Sylow III, we know that $n_{13}$ is a divisor of $2^3 \cdot 3$. The list of divisors of $2^3 \cdot 3$ is

The only number in this list congruent 1 modulo $13$ is $1$ so the claim follows. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Prove that if $|G| = 132 = 2^2 \cdot 3 \cdot 11$, then $G$ is not simple. [/example]

[proof] By Sylow III, we know that:

• $n_{11}\mid 2^2\cdot 3$ and $n_{11}\equiv 1\pmod{11}$. The integers satisfying this constraint are $1$ and $12$.
• $n_{3}\mid 2^2\cdot 11$ and $n_{3}\equiv 1\pmod{3}$. The integers satisfying this constraint are $1$, $4$, and $22$.
• $n_{2}\mid 3\cdot 11$ and $n_{2}\equiv 1\pmod{2}$. The integers satisfying this constraint are $1$, $3$, $11$, and $33$.

In the case where neither $n_{11}$ and $n_{3}$ are equal to one, then we have at least $12\cdot (11 - 1) = 120$ elements of order $11$ and $4\cdot(3 - 1) = 8$ elements of order $3$. So the remaining

elements come from a unique Sylow $2$-subgroup by Sylow I and Sylow II implies that such a subgroup must be normal. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Let $G$ be a group of order $105$. Prove that if a Sylow $3$-subgroup of $G$ is normal then $G$ is abelian. [/example]

[proof] First, note that $105 = 3\cdot 5\cdot 7$ and so the normality assumption implies that $n_3 = 1$. Sylow III implies that

• $n_5 \mid 3\cdot 7$ and $n_5\equiv 1 \pmod 5$. So $n_5$ is either $1$ or $21$.
• $n_7 \mid 3 \cdot 5$ and $n_7 \equiv 1 \pmod 7$. So $n_7$ is either $1$ or $15$.

As each Sylow subgroup has prime order, they are each cyclic. So they intersect trivially which implies that there are

elements in the union of the Sylow subgroups. If $n_5 = 21$ and $n_7 = 15$, then we have more than $105$ elements total which is not possible. So either $n_5 = 1$ or $n_7 = 1$.

Let $P_3 \in \Syl_3(G)$ and $P_5 \in \Syl_5(G)$. Note that since $P_3$ is normal in $G$, it is certainly normal in $P_3P_5$. By the second isomorphism theorem,

Thus, $P_3P_5$ has $15$ elements since, by Lagrange,

Furthermore, groups of order $15$ are cyclic so it follows $P_5$ is normal in $P_3P_5$. This implies that $P_3$ is a subgroup of $N_G(P_5)$. Since we also have $P_5$ is a subgroup of $N_G(P_5)$, it follows that $15$ divides the order of $N_G(P_5)$. By Lagrange,

So either $N_G(P_5)$ has order $15$ or it has order $105$. In the case where it is $105$, there is nothing to show since $n_5 = 1$. In the case where it is $15$, $n_5 = 7$ which contradicts our work above showing that $n = 21$ or $n_5 = 1$. Thus, $n_5 = 1$. We can use the same reasoning to show that $n_7 = 1$ as well.

So we have shown that the Sylow subgroups $P_3, P_5, P_7$ (where $P_i \in \Syl_i(G)$) are all normal. Furthermore, we can use Lagrange to show that $P_3P_5 \cap P_7 = 1$. This implies that $P_3P_5P_7$ is a subgroup of $G$ of order $3\cdot 5\cdot 7 = 105$. Thus, $P_3P_5P_7 = G$. As $P_3P_5P_7$ is isomorphic to $P_3\times P_5\times P_7$, it follows that $G$ is isomorphic to $P_3 \times P_5 \times P_7$ and we are done. [/proof]