---
title: The Sylow Theorems
module: Group Actions and Sylow Theory
moduleNumber: 4
lessonNumber: 3
order: 403
summary: >
  Lagrange's theorem forbids subgroups whose order fails to divide the group
  order; Sylow's theorems supply a partial converse for prime powers. A Sylow
  p-subgroup always exists, all of them are conjugate, and their count satisfies
  two congruence-and-divisibility constraints tight enough to prove many groups
  non-simple from their order alone.
topics: [Group Actions and Sylow Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 4 Group Actions; §4.5 The Sylow Theorems"
  - book: Judson
    ref: "Ch. 15 The Sylow Theorems; §15.1 The Sylow Theorems, §15.2 Examples and Applications"
draft: false
---

[Lagrange's theorem](/abstract-algebra/subgroups-and-quotients/cosets-lagrange-and-normal-subgroups)
says the order of a subgroup divides the order of the group. Its converse fails:
$A_4$ has order $12$ but no subgroup of order $6$. The Sylow theorems recover as
much of the converse as is true — for every prime power dividing $|G|$ to its
maximal extent, a subgroup of that order exists — and then say precisely how many
there are.[^df-sylow] The count is constrained enough that, for many orders, it
alone forces a normal subgroup, which is the standard route to proving a group is
not simple.

## Sylow subgroups

> **Definition ($p$-group and Sylow $p$-subgroup).** Let $p$ be a prime. A finite
> group of order $p^a$ ($a \ge 1$) is a **$p$-group**; a subgroup that is a
> $p$-group is a **$p$-subgroup**. If $|G| = p^a m$ with $p \nmid m$, a subgroup of
> order $p^a$ — the largest possible $p$-power — is a **Sylow $p$-subgroup**.

Write $\mathrm{Syl}_p(G)$ for the set of Sylow $p$-subgroups and $n_p = |\mathrm{Syl}_p(G)|$
for their number. The definition isolates the $p$-part of the order: factor $|G| =
p^a m$ once, and a Sylow $p$-subgroup accounts for the entire $p^a$.

$$
% caption: With $|G| = p^a m$ and $p\nmid m$, a Sylow $p$-subgroup $P$ has order
% $p^a$; it sits inside its normalizer, whose index in $G$ is the count $n_p$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[draw=black, thick] (0,0) rectangle (8.2,4.0);
\node[anchor=north east, font=\footnotesize, text=black] at (8.1,3.9) {$G$, order $p^a m$};
\draw[draw=black, fill=acc!5] (0.6,0.5) rectangle (6.0,3.1);
\node[anchor=north east, font=\footnotesize, text=black] at (5.9,3.0) {$N_G(P)$};
\draw[draw=acc, fill=acc!16] (1.2,1.0) rectangle (4.2,2.4);
\node at (2.7,1.7) {$P$, order $p^a$};
% index labels
\draw[<->, acc] (6.3,1.7) -- (7.9,1.7);
\node[font=\footnotesize, text=acc, anchor=south] at (7.1,1.75) {index $n_p$};
\node[font=\footnotesize, text=black, anchor=north] at (7.1,1.65) {(divides $m$)};
\end{tikzpicture}
$$

## The theorems

> **Theorem (Sylow).** Let $|G| = p^a m$ with $p$ prime and $p \nmid m$.
>
> - **Existence.** $\mathrm{Syl}_p(G) \ne \varnothing$.
> - **Conjugacy and domination.** If $P$ is a Sylow $p$-subgroup and $Q$ is any
>   $p$-subgroup, then $Q \subseteq g P g^{-1}$ for some $g \in G$. In particular
>   any two Sylow $p$-subgroups are conjugate.
> - **Count.** $n_p \equiv 1 \pmod p$, and $n_p = [G : N_G(P)]$, so $n_p \mid m$.

Two independent constraints bound $n_p$: the congruence $n_p \equiv 1 \bmod p$
and the divisibility $n_p \mid m$. Their intersection is often a very short list.

Existence and the congruence both rest on a lemma controlling how a $p$-subgroup
meets a Sylow normalizer.

> **Lemma.** Let $P \in \mathrm{Syl}_p(G)$ and let $Q$ be any $p$-subgroup. Then
> $Q \cap N_G(P) = Q \cap P$.

If $H = N_G(P) \cap Q$, then $PH$ is a subgroup (because $H$ normalizes $P$) of
order $|P||H| / |P \cap H|$, a power of $p$; since $P \le PH$ and $P$ already has
the maximal $p$-power order, $PH = P$, forcing $H \le P$.[^df-lemma19]

### The congruence

Let $G$ act by conjugation on the set $S = \{P_1, \dots, P_r\}$ of _conjugates_ of
one Sylow $p$-subgroup $P$. Restrict the action to a $p$-subgroup $Q$ and split $S$
into $Q$-orbits. By orbit-stabilizer and the lemma, the orbit of $P_i$ has size
$[Q : Q \cap P_i]$. Taking $Q = P_1$: the orbit of $P_1$ is a single point (since
$P_1 \cap P_1 = P_1$), while every other orbit has size $[P_1 : P_1 \cap P_i] > 1$,
a positive power of $p$. Summing,
$$
r = 1 + (\text{multiples of } p) \equiv 1 \pmod p.
$$
The conjugacy part then shows $S$ is _all_ of $\mathrm{Syl}_p(G)$, so $n_p = r
\equiv 1 \pmod p$, and orbit-stabilizer for the full group gives $n_p = [G :
N_G(P)]$.[^df-thm18] Existence is a separate induction on $|G|$ using the
[class equation](/abstract-algebra/group-actions-and-sylow/conjugation-and-the-class-equation):
if $p \mid |Z(G)|$, quotient by a central subgroup of order $p$; otherwise some
non-central centralizer has index prime to $p$ and supplies the Sylow subgroup by
induction.

$$
% caption: The conjugacy part: $G$ permutes its Sylow $p$-subgroups transitively
% by conjugation, so there is a single orbit of size $n_p = [G:N_G(P)]$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\foreach \i/\ang in {1/90, 2/18, 3/{-54}, 4/{-126}, 5/162} {
  \node[draw=acc, fill=acc!12, circle, minimum size=8mm] (P\i) at (\ang:1.9) {$P_{\i}$};
}
\node[font=\footnotesize, text=black] at (0,0) {one orbit};
\foreach \i/\j in {1/2, 2/3, 3/4, 4/5, 5/1} {
  \draw[->, acc, bend left=18] (P\i) to node[font=\scriptsize, text=black, inner sep=1pt] {} (P\j);
}
\node[font=\footnotesize, anchor=west, text=black] at (2.5,0) {$G$ conjugates each Sylow};
\node[font=\footnotesize, anchor=west, text=black] at (2.5,-0.5) {subgroup onto another};
\end{tikzpicture}
$$

### Normality from uniqueness

> **Corollary (Unique Sylow is normal).** For $P \in \mathrm{Syl}_p(G)$ the
> following are equivalent: $n_p = 1$; $P \trianglelefteq G$; $P$ is characteristic
> in $G$.

If $n_p = 1$ then $g P g^{-1} \in \mathrm{Syl}_p(G) = \{P\}$ for all $g$, so $P$ is
normal. Conversely a normal $P$ is the only Sylow $p$-subgroup, since all are
conjugate to it. A unique subgroup of its order is characteristic, hence
normal.[^df-cor20] Proving $n_p = 1$ therefore proves normality, the target of
every non-simplicity argument below.

## Reading the count in examples

Factor the order, list the candidate $n_p$, and let the two constraints prune.

| Group | $\lvert G\rvert$ | $n_2$ | $n_3$ | Notes |
| --- | --- | --- | --- | --- |
| $S_3$ | $6 = 2\cdot 3$ | $3$ | $1$ | unique (normal) Sylow $3$-subgroup $A_3$ |
| $A_4$ | $12 = 2^2\cdot 3$ | $1$ | $4$ | normal Sylow $2$-subgroup $V_4$; $n_3 \equiv 1 \bmod 3$ |
| $S_4$ | $24 = 2^3\cdot 3$ | $3$ | $4$ | Sylow $2$-subgroups $\cong D_8$ |
| $\mathbb{Z}/n$ | $n$ | $1$ | $1$ | abelian: one Sylow per prime |

$$
% caption: Sylow constraints prune the candidate counts for order $12$: divisors of
% $m$ are boxed and those congruent to $1$ modulo $p$ are shaded, leaving
% $n_3\in\{1,4\}$ and $n_2\in\{1,3\}$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[font=\footnotesize, anchor=east, text=black] at (-0.4,1.4) {$n_3$};
\node[draw=acc, fill=acc!16, minimum size=8mm] at (0,1.4) {$1$};
\node[draw=black, fill=black!4, text=black, minimum size=8mm] at (1.3,1.4) {$2$};
\node[draw=acc, fill=acc!16, minimum size=8mm] at (2.6,1.4) {$4$};
\node[font=\footnotesize, anchor=east, text=black] at (-0.4,0) {$n_2$};
\node[draw=acc, fill=acc!16, minimum size=8mm] at (0,0) {$1$};
\node[draw=acc, fill=acc!16, minimum size=8mm] at (1.3,0) {$3$};
\node[font=\footnotesize, text=acc, anchor=west] at (3.6,1.4) {keep 1 and 4};
\node[font=\footnotesize, text=black, anchor=west] at (3.6,0) {keep 1 and 3};
\end{tikzpicture}
$$

For $A_4$: $n_3 \mid 4$ and $n_3 \equiv 1 \pmod 3$ leave $n_3 \in \{1, 4\}$, and
the four subgroups $\langle(1\,2\,3)\rangle, \langle(1\,2\,4)\rangle,
\langle(1\,3\,4)\rangle, \langle(2\,3\,4)\rangle$ realize $n_3 = 4$; meanwhile $n_2
\mid 3$ and $n_2 \equiv 1 \pmod 2$ force $n_2 = 1$, the normal Klein four-group.
The largest prime tends to give the tightest constraint, since $n_p \equiv 1 \bmod
p$ rules out more values as $p$ grows.

## Proving non-simplicity

> **Definition (Simple group).** A group $G \ne 1$ is **simple** if its only normal
> subgroups are $1$ and $G$.

If some $n_p = 1$, the corresponding Sylow subgroup is a proper nontrivial normal
subgroup, so $G$ is not simple. Three techniques cover most small orders.

**Direct pruning.** Suppose $|G| = pq$ with primes $p < q$. Then $n_q \mid p$ and
$n_q \equiv 1 \pmod q$; since $p < q$ the only option is $n_q = 1$, so the Sylow
$q$-subgroup is normal and $G$ is not simple. Any group of order $pq$ has a normal
Sylow subgroup for the larger prime.[^judson-apps]

**Element counting.** When the constraints allow $n_p > 1$, count elements of each
prime order. Take $|G| = 30 = 2 \cdot 3 \cdot 5$ and suppose neither the Sylow
$5$- nor the Sylow $3$-subgroup is normal, so $n_5 = 6$ and $n_3 = 10$. Distinct
Sylow $5$-subgroups meet only in the identity, so they contribute $6 \cdot 4 = 24$
elements of order $5$; the ten Sylow $3$-subgroups contribute $10 \cdot 2 = 20$
elements of order $3$. That is $44$ non-identity elements in a group of $30$ — a
contradiction. One of the Sylow subgroups must be normal.

$$
% caption: The order-$30$ contradiction: assuming no normal Sylow $5$- or
% $3$-subgroup forces $24+20=44$ elements of prime order into a group of size $30$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{alt}{HTML}{9A6A3A}
% bar for elements of order 5
\draw[fill=acc!16, draw=acc] (0,1.2) rectangle (4.8,1.9);
\node[font=\footnotesize] at (2.4,1.55) {$24$ elements of order $5$};
% bar for elements of order 3
\draw[fill=alt!16, draw=alt] (0,0.3) rectangle (4.0,1.0);
\node[font=\footnotesize] at (2.0,0.65) {$20$ elements of order $3$};
% capacity line
\draw[black, thick, dashed] (6.0,0.1) -- (6.0,2.1);
\node[font=\footnotesize, text=black, anchor=south, rotate=90] at (6.25,1.1) {capacity $30$};
\node[font=\footnotesize, text=black, anchor=west] at (6.5,1.1) {$44 > 30$: impossible};
\end{tikzpicture}
$$

**Changing primes and using cosets.** For borderline orders the
[coset action](/abstract-algebra/group-actions-and-sylow/actions-and-cayleys-theorem)
supplies the last step. If $H \le G$ has index $k$ and $G$ is simple, then $G$
embeds in $S_k$ (the kernel of the coset action is a proper normal subgroup, hence
trivial), so $|G| \mid k!$. This bounds how small an index a simple group can
have, and combined with element counting it classifies the simple group of order
$60$ as $A_5$.

> **Worked example.** No group of order $20$ is simple. Factor $20 = 2^2 \cdot 5$,
> so $m = 4$ for the prime $5$. The count $n_5$ divides $4$ and satisfies $n_5
> \equiv 1 \pmod 5$; of the divisors $1, 2, 4$ only $n_5 = 1$ is congruent to $1$
> modulo $5$. The Sylow $5$-subgroup is therefore unique, hence normal, and $G$ is
> not simple.[^judson-apps]

> **Worked example.** No group of order $56$ is simple. Factor $56 = 2^3 \cdot 7$.
> Then $n_7 \mid 8$ and $n_7 \equiv 1 \pmod 7$ leave $n_7 \in \{1, 8\}$. If $n_7 =
> 1$, the Sylow $7$-subgroup is normal. If $n_7 = 8$, the eight Sylow
> $7$-subgroups have prime order $7$ and meet pairwise in the identity, so they
> account for $8 \cdot 6 = 48$ elements of order $7$; the remaining $56 - 48 = 8$
> elements are exactly enough for a single Sylow $2$-subgroup of order $8$, forcing
> $n_2 = 1$. Either branch yields a normal Sylow subgroup.[^judson-apps]

### Worked example: groups of order 12

The three techniques together settle the structure of every group of order
$12 = 2^2 \cdot 3$. The claim is that such a $G$ either has a normal Sylow
$3$-subgroup or is isomorphic to $A_4$.[^judson-apps]

Suppose the Sylow $3$-subgroup is _not_ normal, so $n_3 \ne 1$. The constraints
$n_3 \mid 4$ and $n_3 \equiv 1 \pmod 3$ leave $n_3 = 4$. The four Sylow
$3$-subgroups have prime order $3$, so they intersect pairwise in the identity and
account for $4 \cdot 2 = 8$ elements of order $3$. Since $[G : N_G(P)] = n_3 = 4$,
the normalizer $N_G(P)$ equals $P$ itself. Now let $G$ act by conjugation on its
four Sylow $3$-subgroups; the resulting homomorphism $\varphi : G \to S_4$ has
kernel $K = \bigcap_g N_G(P_g) \le N_G(P) = P$, and since $P$ is not normal, $K =
1$. So $\varphi$ is injective and $G$ embeds in $S_4$. The $8$ elements of order
$3$ map to the $8$ three-cycles of $S_4$, all of which lie in $A_4$, so the image
meets $A_4$ in a subgroup of order at least $8$; both groups have order $12$, forcing
$\varphi(G) = A_4$ and $G \cong A_4$. In that case the Sylow $2$-subgroup — the
Klein four-group of double transpositions — is the unique one, hence normal, so $G$
still has _some_ normal Sylow subgroup. The order-$12$ groups are thereby sorted
into those with a normal Sylow $3$-subgroup and the single exception $A_4$.

The same $|G| \mid k!$ bound settles order $60$. A simple group of order $60$ can
have no subgroup of index less than $5$, since an index $k \in \{2,3,4\}$ would
embed $G$ in $S_k$ with $60 \nmid k!$. Running the possible values $n_2 \in \{3, 5,
15\}$ through this bound and an element count forces the Sylow $2$-normalizer to
have index $5$; the coset action then embeds $G$ in $S_5$ as a subgroup that must
coincide with $A_5$. Every simple group of order $60$ is isomorphic to $A_5$.

```algorithm
caption: $\textsc{TestNonSimple}(G)$ — the standard order-based argument
factor $|G| = p_1^{a_1}\cdots p_t^{a_t}$
for each prime $p_i$ do
  list candidates: $n_{p_i}$ with $n_{p_i} \equiv 1 \pmod{p_i}$ and $n_{p_i} \mid m_i$
  if the only candidate is $n_{p_i} = 1$ then
    return "not simple" // unique Sylow subgroup is normal
for each prime with $n_{p_i} > 1$ forced do
  count elements of order $p_i$ across distinct Sylow subgroups
  if the counts exceed $|G|$ then
    return "not simple" // some Sylow subgroup must be normal
embed $G$ in $S_k$ via a small-index coset action; check $|G| \mid k!$
```

[^df-sylow]: **Dummit & Foote**, §4.5 — The Sylow Theorems: the definitions of $p$-group and Sylow $p$-subgroup and the statement of Sylow's theorem as a partial converse to Lagrange's theorem.
[^df-thm18]: **Dummit & Foote**, §4.5, Theorem 18: existence, conjugacy/domination, and the count $n_p \equiv 1 \pmod p$ with $n_p = [G:N_G(P)]$; the orbit-counting proof of the congruence.
[^df-lemma19]: **Dummit & Foote**, §4.5, Lemma 19: $Q \cap N_G(P) = Q \cap P$ for a Sylow $p$-subgroup $P$ and any $p$-subgroup $Q$.
[^df-cor20]: **Dummit & Foote**, §4.5, Corollary 20: $n_p = 1$ is equivalent to normality and to being characteristic.
[^judson-apps]: **Judson**, §15.2 — Examples and Applications: groups of order $pq$, element-counting arguments, and non-simplicity of groups of small order.