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Difference between revisions of "Sylow Theorems"

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Latest revision as of 11:25, 9 April 2019

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 $P' = \phi(a^{-1} P a)$. $\blacksquare$

See also

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