# Sylow Theorems

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The Sylow theorems are a collection of results in the theory of finite groups. They give a partial converse to Lagrange's Theorem, and are one of the most important results in the field. They are named for P. Ludwig Sylow, who published their proof in 1872.

## The Theorems

Throughout this article, $p$ will be an arbitrary prime.

The three Sylow theorems are as follows:

Theorem. Every finite group contains a Sylow $p$-subgroup.
Theorem. In every finite group, the Sylow $p$-subgroups are conjugates.
Theorem. In every finite group, the number of Sylow $p$-subgroups is equivalent to 1 (mod $p$).

As $0\neq 1 \pmod{p}$, the third theorem implies the first.

Before proving the third theorem, we show some preliminary results.

Lemma 1. Let $r$ and $m$ be nonnegative integers. Then $$\binom{p^r m}{p^r} \equiv m \pmod{p} .$$

Proof. Let $G$ be a group of order $p^r$ (e.g., $\mathbb{Z}/p^r\mathbb{Z}$, and let $S$ be a set of size $m$. Let $G$ act on the set $G \times T$ by the law $g(\alpha, x) \mapsto (g\alpha, x)$; extend this action canonically to the subsets of $G \times S$ of size $p^r$. There are $\binom{p^r m}{p^r}$ such subsets.

Evidently, a subset of $A \times T$ is stable under this action if and only if $A= G$. Thus the fixed points of the action are exactly the subsets of the form $G \times \{x\}$, for $x\in S$. Then there are $m$ fixed points. Therefore $$\binom{p^r m}{p^r} \equiv m,$$ since the $p$-group $G$ operates on a set of size $\binom{p^r m}{p^r}$ with $m$ fixed points. $\blacksquare$

Let $G$ be a finite group, and let its order be $n= p^r m$, for some integer $m$ not divisible by $p$.

Lemma 2. Let $\mathfrak{P}$ be the set of subsets of $G$ of size $p^r$, and let $G$ act on $\mathfrak{P}$ by left translation. Suppose $\mathfrak{H}$ is an orbit of $\mathfrak{P}$ such that $p$ does not divide $\lvert \mathfrak{H} \rvert$. Then $\mathfrak{H}$ has $m$ elements, and they are disjoint.

Proof. Since $\lvert \mathfrak{H} \rvert$ divides $n = p^r m$ but is relatively prime to $p^r$, it follows that $\lvert \mathfrak{H} \rvert$ divides $m$; in particular, $\lvert \mathfrak{H} \rvert \le m$. Since every element of $G$ is included in some element of $\mathfrak{H}$, $$\lvert G \rvert = \biggl\lvert \bigcup_{H \in \mathfrak{H}} H \biggr\rvert \le \lvert \mathfrak{H} \rvert \cdot p^r \le m \cdot p^r,$$ with equality only when the elements of $\mathfrak{H}$ are disjoint and when $\mathfrak{H}=m$. Since equality does occur, both these conditions must be true. $\blacksquare$

Theorem 3. The number of finite subgroups of $G$ is equivalent to 1 (mod $p$).

Proof. Consider the action of $G$ on $\mathfrak{P}$, as described in Lemma 2. Since $$\binom{n}{p} \equiv m \pmod{p},$$ and every orbit $\mathfrak{H}$ of $\mathfrak{P}$ for which $p$ does not divide $\lvert \mathfrak{H} \rvert$ has $m$ elements, it follows that the number of such orbits is equivalent to 1 (mod $p$). It thus suffices to show that every such orbit contains exactly one Sylow $p$-subgroup, and that every Sylow $p$-subgroup is contained in exactly one such orbit.

To this end, let $\mathfrak{H}$ be an orbit of $\mathfrak{P}$ for which $p$ does not divide $\mathfrak{H}$. Consider the equivalence relation $R(x,y)$ on elements of $G$, defined as "$x$ and $y$ are in the same element of $\mathfrak{H}$". Then $R$ is compatible with left translation by $G$; since the elements of $\mathfrak{H}$ are disjoint, $R$ is an equivalence relation. Thus the equivalence class of the identity is a subgroup $H$ of $G$, which must have order $p^r$. Since this is the only element of $\mathfrak{H}$ that contains the identity, it is the only group in $\mathfrak{H}$.

Conversely, if $H$ is a Sylow $p$-subgroup, then its orbit is its set of left cosets, which has size $m$, which $p$ does not divide. Since orbits are disjoint, $H$ is contained in exactly one orbit of $\mathfrak{P}$. $\blacksquare$

We now prove a more general theorem that implies the second Sylow theorem.

Theorem 4. Let $H$ be a Sylow $p$-subgroup of $G$, and let $K$ be a $p$-subgroup of $G$. Then $K$ is a subgroup of some conjugate of $H$.

Proof. Consider the left operation of $K$ on $G/H$, the set of left cosets of $G$ modulo $H$. Since the order of $K$ is some power of $p$, say $p^s$, the size of each orbit must divide $p^s$; since there are $m$ cosets, and $p \nmid m$, it follows that some orbit must have size 1, i.e., there must be some $g\in G$ such that $gH$ is stable under the operation by $K$. Then for all $k \in K$, $$g^{-1}kgH = g^{-1}gH = H,$$ i.e., $g^{-1}kg$ stabilizes $H$. Thus $g^{-1}kg \in H$. It follows that $g^{-1}Kg \subseteq H$, or $K\subseteq gHg^{-1}$. $\blacksquare$

Corollary 5 (second Sylow theorem). The Sylow $p$-subgroups of $G$ are conjugates.

Corollary 6. Every subgroup of $G$ that is a $p$-group is contained in a Sylow $p$-subgroup of $G$.

Corollary 7. Let $P$ be a Sylow $p$-subgroup of $G$, let $N$ be its normalizer, and let $M$ be a subgroup of $G$ that contains $M$. Then $M$ is its own normalizer.

Proof. Let $g$ be an element of $G$ such that $gMg^{-1} \subseteq M$. Then $gPg^{-1}$ is a Sylow $p$-subgroup of $M$. Since the Sylow $p$-subgroups of $M$ are conjugates, there exists $m\in M$ such that $mPm^{-1} = gPg^{-1}$. It follows that $g^{-1}m$ normalizes $P$, so $g^{-1}m \in N \subseteq M$. Hence $g\in M$. Thus the normalizer of $M$ is a subset of $M$. Since the opposite is true in general, $M$ is its own normalizer. $\blacksquare$

Corollary 8. Let $G_1$ and $G_2$ be finite groups, and $f$ a homomorphism of $G_1$ into $G_2$. Let $P_1$ be a Sylow $p$-subgroup of $G_1$. Then there exists a Sylow $p$-subgroup $P_2$ of $G_2$ such that $f(P_1) \subseteq P_2$.

For $f(P_1)$ is a $p$-subgroup of $G_2$; this corollary then follows from Corollary 6.

Corollary 9. Let $H$ be as subgroup of $G$ and let $P$ be a Sylow $p$-subgroup of $H$. Then there exists a Sylow $p$-subgroup $P'$ of $G$ such that $P = H \cap P'$.

Proof. There must be some Sylow $p$-subgroup $P'$ of $H$ that contains $P$. Since $P' \cap H$ is a $p$-subgroup of $H$, it can be no larger than $P$. $\blacksquare$

Corollary 10. Conversely, if $P'$ is a Sylow $p$-subgroup of $G$ and $H$ is a normal subgroup of $G$, then $H \cap P'$ is a Sylow $p$-subgroup of $H$.

Proof. Let $s$ be the greatest integer such that $p^s$ divides the order of $H$. Let $\phi$ be the canonical homomorphism from $G$ to $G/H$. Then $H \cap P'$ is the kernel of the restriction of $\phi$ to $P'$. Since $P'/(H \cap P')$ is isomorphic to $P'/H$, a $p$-subgroup of $G/H$ of order not exceeding $p^{r-s}$, it follows from Lagrange's Theorem that $H \cap P'$ has order at least $p^s$. Thus $p \nmid (H : H\cap P')$; since $H\cap P'$ is a $p$-subgroup of $H$, it follows that it is a Sylow $p$-subgroup of $H$. $\blacksquare$

Note that this is not true in general if $H$ is not normal. For instance $\{e, (12) \}$ is a Sylow 2-subgroup of the symmetric group $S_3$ and $\{e,(23)\}$ is a subgroup of $S_3$, but their intersection, $\{e\}$, is evidently not a Sylow 2-subgroup of $\{e,(23)\}$.

Corollary 11. Let $N$ be a normal subgroup of $G$, let $\phi : G \to G/N$ be the canonical homomorphism. Then the image of every Sylow $p$-subgroup of $G$ under $\phi$ is a Sylow $p$-subgroup of $G/N$; furthermore, every Sylow $p$-subgroup of $G/N$ is the image of a Sylow $p$-subgroup of $G$ under $\phi$.

Proof. Let $P$ be a Sylow $p$-subgroup of $G$; let the order of $N$ be $p^sa$, where $a$ is a positive integer not divisible by $p$. By Corollary 9, $P \cap N$ is a group of order $p^s$; it follows that $\phi(P)$ is isomorphic to $P/(P \cap N)$, a group of order $p^{r-s}$, so $\phi(P)$ is a Sylow $p$-subgroup of $G/N$. Now, let $P'$ be a Sylow $p$-subgroup of $G/N$; then there exists $\alpha \in G/N$ such that $P' = \alpha^{-1} \cdot \phi(P) \alpha$. Let $a \in G$ be such that $\phi(a) = \alpha$; then $P' = \phi(a^{-1} P a)$. $\blacksquare$