Difference between revisions of "Sylow Theorems"
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== The Theorems == | == The Theorems == | ||
− | Throughout this article, <math>p</math> will be | + | 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{ | + | <cmath> \binom{p^r m}{p^r} \equiv m \pmod{p} . </cmath> |
− | |||
− | Let <math>G</math> be a | + | ''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 | + | '''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 21:27, 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, will be an arbitrary prime.
The three Sylow theorems are as follows:
- Theorem. Every finite group contains a Sylow
-subgroup.
- Theorem. In every finite group, the Sylow
-subgroups are conjugates.
- Theorem. In every finite group, the number of Sylow
-subgroups is equivalent to 1 (mod
).
As , the third theorem implies the first.
Before proving the third theorem, we show some preliminary results.
Lemma 1. Let and
be nonnegative integers. Then
Proof. Let be a group of order
(e.g.,
, and let
be a set of size
. Let
act on the set
by the law
; extend this action canonically to the subsets of
of size
. There are
such subsets.
Evidently, a subset of is stable under this action if and only if
. Thus the fixed points of the action are exactly the subsets of the form
, for
. Then there are
fixed points. Therefore
since the
-group
operates on a set of size
with
fixed points.
Let be a finite group, and let its order be
, for some integer
not divisible by
.
Lemma 2. Let be the set of subsets of
of size
, and let
act on
by left translation. Suppose
is an orbit of
such that
does not divide
. Then
has
elements, and they are disjoint.
Proof. Since divides
but is relatively prime to
, it follows that
divides
; in particular,
. Since every element of
is included in some element of
,
with equality only when the elements of
are disjoint and when
. Since equality does occur, both these conditions must be true.
Theorem 1. The number of finite subgroups of is equivalent to 1 (mod
).
Proof. Consider the action of on
, as described in Lemma 2. Since
and every orbit
of
for which
does not divide
has
elements, it follows that the number of such orbits is equivalent to 1 (mod
). It thus suffices to show that every such orbit contains exactly one Sylow
-subgroup, and that every Sylow
-subgroup is contained in exactly one such orbit.
To this end, let be an orbit of
for which
does not divide
. Consider the equivalence relation
on elements of
, defined as "
and
are in the same element of
". Then
is compatible with left translation by
; since the elements of
are disjoint,
is an equivalence relation. Thus the equivalence class of the identity is a subgroup
of
, which must have order
. Since this is the only element of
that contains the identity, it is the only group in
.
Conversely, if is a Sylow
-subgroup, then its orbit is its set of left cosets, which has size
, which
does not divide. Since orbits are disjoint,
is contained in exactly one orbit of
.
Theorem 2. The Sylow -subgroups of
are conjugates.