Skip to main content Link Search Menu Expand Document (external link)

The Sylow Theorems

Background

Sylow’s Theorem was proved in 1872.

The Theorems

Theorem: (Cauchy) Let $G$ be a group of order $n$ and suppose $p\divides n$. Then $G$ has an element of order $p$.

Proof: We prove this by induction on the order of $G$. We have proved this if $G$ is abelian, and in particular if $\lvert G\rvert =p$. Suppose $G$ is not abelian. Consider its class equation

\[\lvert G\rvert = \lvert Z(G)\rvert + \sum_{g}[G:C_{G}(g)]\]

where the sum is over representatives for the conjugacy classes of $G$ of size greater than one. Since $G$ is nonabelian there is at least one such class. Since the left side of this equation is divisible by $p$, so is the right side. If $\lvert Z(G)\rvert$ is divisible by $p$ then, since $Z(G)$ is abelian, it contains an element of order $p$. Otherwise, at least one of $[G:C_{G}(g)]$ is not divisible by $p$. Therefore by Lagrange’s theorem $\lvert C_{G}(g)\rvert$ is divisible by $p$. Since $C_{G}(g)$ is smaller then $G$, it contains an element of order $p$ by induction.

Theorem: Let $G$ be a finite group of order $n$ and let $p$ be a prime number.

  1. There exists a subgroup $P$ of $p$-power order such that $[G:P]$ is prime to $p$. Such a subgroup is called a Sylow $p$-subgroup of $G$.
  2. If $P$ is any Sylow $p$-subgroup of $G$, and $H$ is any subgroup of $G$ of prime power order, then some conjugate of $H$ is contained in $P$. In particular, all Sylow $p$-subgroups are conjugate.
  3. Let $n_p$ be the number of Sylow $p$-subgroups in $G$. Then $n_p\equiv 1 \pmod{p}$ and $n_p\vert n$.

Note: One proof is given in DF, Chapter 4, page 139-140. We follow Keith Conrad’s approach.

We will construct our Sylow $p$-subgroup by constructing an increasing sequence of groups $H_1\subset H_2\subset\cdots$ where $H_{i}$ has order $p^{i}$ and the process stops when the index of $H_{i}$ in $G$ becomes prime to $p$.

We recall this lemma about group actions.

Lemma: Suppose $X$ is a set on which a group $G$ acts transitively.
Let $x\in X$ and let $H$ be the stabilizer of $x$ in $G$. Let $N=N_{G}(H)$. Then $N$ permutes the fixed points of $H$ transitively, so the number of such fixed points is $[N:H]$.

Proof: Suppose $H$ fixes $x$. If $H$ also fixes $x’$, write $x’=gx$ for some $g\in G$. Then $gHg^{-1}$ fixes $x’$ so $gHg^{-1}=H$ and $g\in N_{G}(H)$. Conversely, if $g\in N_{G}(H)$, then $gHg^{-1}$ fixes $gx$ so $H$ fixes $gx$.

We will need the following lemma about the action of $p$ groups on finite sets.

Lemma: Suppose $X$ is a finite set with an action of a finite $p$-group $H$. Let $\mathrm{Fix}_{H}(X)$ be the set of points in $X$ that are fixed by $H$. Then \(\order{X} \equiv \lvert \mathrm{Fix}_{H}(X)\rvert\pmod{p}.\)

Proof: Under the action of $H$, $X$ breaks up into orbits of varying sizes; these sizes divide the order of $H$, which is a power of $p$. The orbits of size bigger than one all of size divisible by $p$. The orbits of size one are the fixed points. So the number of elements in $X$ is the number of fixed points plus a sum of powers of $p$.

Proof of Sylow:

Part 1.

  1. By Cauchy there is a subgroup $H$ of order $p$ in $G$.
  2. Suppose now that $H$ is a subgroup of order $p^{i}$. If $[G:H]$ is prime to $p$, we are done.
  3. Consider the action of $H$ on the homogeneous space of cosets $X=G/H$.
  4. $X$ breaks up into orbits under the action of $H$, each orbit being of size $1$ or a power of $p$.
  5. By the second Lemma above, the number of elements of $X$ fixed by $H$ is divisible by $p$.
  6. By the first lemma, the index $[N_{G}(H):H]$ is divisible by $p$.
  7. By Cauchy’s theorem, the group $N_{G}(H)/H$ has a subgroup of order $p$.
  8. By the isomorphism theorems, the subgroups of $N_{G}(H)/H$ correspond to the subgroups of $N_{G}(H)$ containing $H$. Therefore there is a subgroup of $N_{G}(H)$ of order $p^{i+1}$.
  9. We have shown that if $p^{k}$ is the exact power of $p$ dividing $n$, and $i<k$, then any subgroup of order $p^{i}$ is contained in a subgroup of order $p^{i+1}$. Thus there must be a subgroup of order $p^{k}$.

Part 2.

  1. Let $Q$ be a group of $p$-power order, and let $P$ be a Sylow $p$-subgroup. The group $Q$ acts on the homogeneous space $G/P$, which has prime-to-$p$ order.
  2. Since $Q$ is a $p$-group, we know that \(\vert G/P\vert \equiv \vert \mathrm{Fix}_{Q}(G/P)\pmod{p}.\)
  3. Since the left side isn’t zero, there must be a coset $kP$ which is stabilized by $Q$.
  4. This means $kQk^{-1}$ stabilizes $P$, which means $Q\subset P$.

Part 3.

  1. Let $P$ be a Sylow $p$-subgroup and let $P$ act on the set of Sylow $p$-subgroups by conjugation.
    $P$ fixes $Q$ provided $gQg^{-1}=Q$ for all $g\in P$, or, in other words, if $P\subset N_{G}(Q)$.
  2. Now $Q$ is also in $N_{G}(Q)$, and both $P$ and $Q$ are Sylow $p$-subgroups in $N_{G}(Q)$.
  3. That means $P$ and $Q$ are conjugate in this group by part (2) of Sylow’s theorems.
  4. On the other hand $Q$ is normal in $N_{G}(Q)$, which means $P=Q$.
  5. We’ve shown that $P$ is the only fixed point under conjugation by $P$, and the other orbits under conjugation have size divisible by $p$.
  6. Therefore the number of conjugates is 1 mod $p$.
  7. On the other hand, since $G$ permutes the Sylow $p$-subgroups transitively under conjugation by part (2), we know that the number of such subgroups (being the size of an orbit) is a divisor of $n$.