View source

The '''Sylow theorems''' are a collection of results in the theory of [[finite]] [[group]]s.  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, <math>p</math> will be an arbitrary prime.

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>).

As <math>0\neq 1 \pmod{p}</math>, the third theorem implies the first.

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

'''Lemma 1.'''  Let <math>r</math> and <math>m</math> be nonnegative integers.  Then
<cmath> \binom{p^r m}{p^r} \equiv m \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>.  There are <math>\binom{p^r m}{p^r}</math> such subsets.

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>

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>.

'''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.

''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 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>

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

'''Theorem 4.'''  Let <math>H</math> be a Sylow <math>p</math>-subgroup of <math>G</math>, and let <math>K</math> be a <math>p</math>-subgroup of <math>G</math>.  Then <math>K</math> is a subgroup of some conjugate of <math>H</math>.

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

'''Corollary 5 (second Sylow theorem).'''  The Sylow <math>p</math>-subgroups of <math>G</math> are conjugates.

'''Corollary 6.'''  Every subgroup of <math>G</math> that is a <math>p</math>-group is contained in a Sylow <math>p</math>-subgroup of <math>G</math>.

'''Corollary 7.'''   Let <math>P</math> be a Sylow <math>p</math>-subgroup of <math>G</math>, let <math>N</math> be its [[normalizer]], and let <math>M</math> be a subgroup of <math>G</math> that contains <math>M</math>.  Then <math>M</math> is its own normalizer.

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

'''Corollary 8.'''  Let <math>G_1</math> and <math>G_2</math> be finite groups, and <math>f</math> a [[homomorphism]] of <math>G_1</math> into <math>G_2</math>.  Let <math>P_1</math> be a Sylow <math>p</math>-subgroup of <math>G_1</math>.  Then there exists a Sylow <math>p</math>-subgroup <math>P_2</math> of <math>G_2</math> such that <math>f(P_1) \subseteq P_2</math>.

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

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

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

'''Corollary 10.''' Conversely, if <math>P'</math> is a Sylow <math>p</math>-subgroup of <math>G</math> and <math>H</math> is a [[normal subgroup]] of <math>G</math>, then <math>H \cap P'</math> is a Sylow <math>p</math>-subgroup of <math>H</math>.

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

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

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

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

== See also ==

* [[p-group |<math>p</math>-group]]
* [[Sylow p-subgroup]]
* [[Lagrange's Theorem]]
* [[Nilpotent group]]

[[Category:Group theory]]