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Difference between revisions of "Sylow Theorems"

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== The Theorems ==
 
== The Theorems ==
  
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Throughout this article, <math>p</math> will be a prime.
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Throughout this article, <math>p</math> will be an arbitrary prime.
  
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First, we show a lemma.
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The three Sylow theorems are as follows:
 +
:'''Theorem.'''  Every finite group contains a [[Sylow p-subgroup |Sylow <math>p</math>-subgroup]].
 +
:'''Theorem.'''  In every finite group, the Sylow <math>p</math>-subgroups are [[conjugate (group theory) |conjugates]].
 +
:'''Theorem.'''  In every finite group, the number of Sylow <math>p</math>-subgroups is equivalent to 1 (mod <math>p</math>).
  
−
'''Lemma.'''  Let <math>n=p^rm</math>, where <math>r</math> is an nonnegative integer and <math>m</math> is a nonnegative integer not divisible by <math>p</math>.  Then
+
As <math>0\neq 1 \pmod{p}</math>, the third theorem implies the first.
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<cmath> \binom{n}{p^r} \not\equiv 0 \pmod{p} . </cmath>
 
  
−
''Proof.''  Let <math>G</math> be a group of order <math>p^r</math> (e.g., <math>\mathbb{Z}/p^r\mathbb{Z}</math>, and let <math>S</math> be a [[set]] of size <math>m</math>.  Let <math>G</math> act on the set <math>G \times T</math> by the law <math>g(\alpha, x) \mapsto (g\alpha, x)</math>; extend this action canonically to the subsets of <math>G \times S</math> of size <math>p^r</math>.  Evidently, there are <math>\binom{n}{p^r}</math> such subsets.
+
Before proving the third theorem, we show some preliminary results.
  
−
Evidently, a subset of <math>A \times T</math> is stable under this action if and only if <math>A= G</math>.  Thus the fixed points of the action are exactly the subsets of the form <math>G \times \{x\}</math>, for <math>x\in S</math>.  Then there are <math>m</math> fixed points.  Then
+
'''Lemma 1.''' Let <math>r</math> and <math>m</math> be nonnegative integers.  Then
−
<cmath> \binom{n}{p^r} \equiv m \not\equiv 0 \pmod{p}, </cmath>
+
<cmath> \binom{p^r m}{p^r} \equiv m \pmod{p} . </cmath>
−
since the <math>p</math>-group <math>G</math> operates on a set of size <math>\binom{n}{p^r}</math> with <math>m</math> fixed points.  <math>\blacksquare</math>
 
  
−
Let <math>G</math> be a finite group.
+
''Proof.''  Let <math>G</math> be a group of order <math>p^r</math> (e.g., <math>\mathbb{Z}/p^r\mathbb{Z}</math>, and let <math>S</math> be a [[set]] of size <math>m</math>.  Let <math>G</math> act on the set <math>G \times T</math> by the law <math>g(\alpha, x) \mapsto (g\alpha, x)</math>; extend this action canonically to the subsets of <math>G \times S</math> of size <math>p^r</math>.  There are <math>\binom{p^r m}{p^r}</math> such subsets.
  
−
'''Theorem 1.''' For every prime <math>p</math>, every finite group contains a [[Sylow p-subgroup |Sylow <math>p</math>-subgroup]].
+
Evidently, a subset of <math>A \times T</math> is stable under this action if and only if <math>A= G</math>.  Thus the fixed points of the action are exactly the subsets of the form <math>G \times \{x\}</math>, for <math>x\in S</math>.  Then there are <math>m</math> fixed points.  Therefore
 +
<cmath> \binom{p^r m}{p^r} \equiv m, </cmath>
 +
since the <math>p</math>-group <math>G</math> operates on a set of size <math>\binom{p^r m}{p^r}</math> with <math>m</math> fixed points. <math>\blacksquare</math>
  
−
''Proof.''  Let <math>G</math> be a finite group of [[order (group theory) |order]] <math>n = p^r m</math>, for some positive integer <math>m</math>, not divisible by <math>p</math>.  Let <math>\mathfrak{P}</math> denote the set of subsets of <math>G</math> of size <math>p^r</math>.  Consider the action of <math>G</math> by left translation on the elements of <math>\mathfrak{P}</math>.  There are <math>\binom{n}{p^r}</math> such subsets.  Since
+
Let <math>G</math> be a finite group, and let its order be <math>n= p^r m</math>, for some integer <math>m</math> not divisible by <math>p</math>.
−
<cmath> \binom{n}{p^r} \not\equiv 0 \pmod{p}, </cmath>
 
−
some [[orbit]] <math>\mathfrak{H}</math> of <math>\mathfrak{P}</math> must have cardinality not divisible by <math>p</math>.  Since <math>\lvert \mathfrak{H} \rvert \mid n</math>, it follows that <math>\lvert \mathfrak{H} \rvert \mid m</math>; in particular, <math>\lvert \mathfrak{H} \rvert \le m</math>.  Since each element of <math>G</math> must be contained in one of the elements of <math>\mathfrak{H}</math>, it follows that the elements of <math>\mathfrak{H}</math> must be disjoint.
 
  
−
Consider now the equivalence relation <math>R(x,y)</math> on elements of <math>G</math>, defined as "<math>x</math> and <math>y</math> are in the same element of <math>\mathfrak{H}</math>".  Then <math>R</math> is compatible with left translation by <math>G</math>; since the elements of <math>\mathfrak{H}</math> are disjoint, <math>R</math> is an [[equivalence relation]].  Thus the [[equivalence class]] of the identity is a subgroup <math>H</math> of <math>G</math>, which must have order <math>p^r</math>. <math>\blacksquare</math>
+
'''Lemma 2.''' Let <math>\mathfrak{P}</math> be the set of subsets of <math>G</math> of size <math>p^r</math>, and let <math>G</math> act on <math>\mathfrak{P}</math> by left translation.  Suppose <math>\mathfrak{H}</math> is an [[orbit]] of <math>\mathfrak{P}</math> such that <math>p</math> does not divide <math>\lvert \mathfrak{H} \rvert</math>.  Then <math>\mathfrak{H}</math> has <math>m</math> elements, and they are disjoint.
  
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'''Theorem 2.'''  The Sylow <math>p</math>-subgroups of <math>G</math> are [[conjugate (group theory) |conjugates]].
+
''Proof.''  Since <math>\lvert \mathfrak{H} \rvert</math> divides <math>n = p^r m</math> but is relatively prime to <math>p^r</math>, it follows that <math>\lvert \mathfrak{H} \rvert</math> divides <math>m</math>; in particular, <math>\lvert \mathfrak{H} \rvert \le m</math>.  Since every element of <math>G</math> is included in some element of <math>\mathfrak{H}</math>,
 +
<cmath> \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, </cmath>
 +
with equality only when the elements of <math>\mathfrak{H}</math> are disjoint and when <math>\mathfrak{H}=m</math>.  Since equality does occur, both these conditions must be true. <math>\blacksquare</math>
  
−
'''Theorem 3.''' The number of Sylow <math>p</math>-subgroups of <math>G</math> is equivalent to 1 (mod <math>p</math>).
+
'''Theorem 1.''' The number of finite subgroups of <math>G</math> is equivalent to 1 (mod <math>p</math>).
 +
 
 +
''Proof.''  Consider the action of <math>G</math> on <math>\mathfrak{P}</math>, as described in Lemma 2.  Since
 +
<cmath> \binom{n}{p} \equiv m \pmod{p}, </cmath>
 +
and every orbit <math>\mathfrak{H}</math> of <math>\mathfrak{P}</math> for which <math>p</math> does not divide <math>\lvert \mathfrak{H} \rvert</math> has <math>m</math> elements, it follows that the number of such orbits is equivalent to 1 (mod <math>p</math>).  It thus suffices to show that every such orbit contains exactly one Sylow <math>p</math>-subgroup, and that every Sylow <math>p</math>-subgroup is contained in exactly one such orbit.
 +
 
 +
To this end, let <math>\mathfrak{H}</math> be an orbit of <math>\mathfrak{P}</math> for which <math>p</math> does not divide <math>\mathfrak{H}</math>.  Consider the equivalence relation <math>R(x,y)</math> on elements of <math>G</math>, defined as "<math>x</math> and <math>y</math> are in the same element of <math>\mathfrak{H}</math>".  Then <math>R</math> is compatible with left translation by <math>G</math>; since the elements of <math>\mathfrak{H}</math> are disjoint, <math>R</math> is an [[equivalence relation]].  Thus the [[equivalence class]] of the identity is a subgroup <math>H</math> of <math>G</math>, which must have order <math>p^r</math>.  Since this is the only element of <math>\mathfrak{H}</math> that contains the identity, it is the only group in <math>\mathfrak{H}</math>.
 +
 
 +
Conversely, if <math>H</math> is a Sylow <math>p</math>-subgroup, then its orbit is its set of [[coset |left cosets]], which has size <math>m</math>, which <math>p</math> does not divide.  Since orbits are disjoint, <math>H</math> is contained in exactly one orbit of <math>\mathfrak{P}</math>.  <math>\blacksquare</math>
 +
 
 +
'''Theorem 2.'''  The Sylow <math>p</math>-subgroups of <math>G</math> are conjugates.
  
 
== See also ==
 
== See also ==

Revision as of 20:27, 2 June 2008

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 1. 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 $\mathfrak{P}$. $\blacksquare$

Theorem 2. The Sylow $p$-subgroups of $G$ are conjugates.

See also

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