Difference between revisions of "Sylow Theorems"
(started article) |
Hashtagmath (talk | contribs) (→See also) |
||
(3 intermediate revisions by one other user not shown) | |||
Line 3: | Line 3: | ||
== 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 3.''' The number of Sylow <math>p</math>-subgroups of <math>G</math> is | + | '''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 == | == See also == | ||
Line 35: | Line 80: | ||
* [[Sylow p-subgroup]] | * [[Sylow p-subgroup]] | ||
* [[Lagrange's Theorem]] | * [[Lagrange's Theorem]] | ||
+ | * [[Nilpotent group]] | ||
[[Category:Group theory]] | [[Category:Group theory]] | ||
+ | [[Category: Theorems]] |
Latest revision as of 12:25, 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, 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 3. 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
.
We now prove a more general theorem that implies the second Sylow theorem.
Theorem 4. Let be a Sylow
-subgroup of
, and let
be a
-subgroup of
. Then
is a subgroup of some conjugate of
.
Proof. Consider the left operation of on
, the set of left cosets of
modulo
. Since the order of
is some power of
, say
, the size of each orbit must divide
; since there are
cosets, and
, it follows that some orbit must have size 1, i.e., there must be some
such that
is stable under the operation by
. Then for all
,
i.e.,
stabilizes
. Thus
. It follows that
, or
.
Corollary 5 (second Sylow theorem). The Sylow -subgroups of
are conjugates.
Corollary 6. Every subgroup of that is a
-group is contained in a Sylow
-subgroup of
.
Corollary 7. Let be a Sylow
-subgroup of
, let
be its normalizer, and let
be a subgroup of
that contains
. Then
is its own normalizer.
Proof. Let be an element of
such that
. Then
is a Sylow
-subgroup of
. Since the Sylow
-subgroups of
are conjugates, there exists
such that
. It follows that
normalizes
, so
. Hence
. Thus the normalizer of
is a subset of
. Since the opposite is true in general,
is its own normalizer.
Corollary 8. Let and
be finite groups, and
a homomorphism of
into
. Let
be a Sylow
-subgroup of
. Then there exists a Sylow
-subgroup
of
such that
.
For is a
-subgroup of
; this corollary then follows from Corollary 6.
Corollary 9. Let be as subgroup of
and let
be a Sylow
-subgroup of
. Then there exists a Sylow
-subgroup
of
such that
.
Proof. There must be some Sylow -subgroup
of
that contains
. Since
is a
-subgroup of
, it can be no larger than
.
Corollary 10. Conversely, if is a Sylow
-subgroup of
and
is a normal subgroup of
, then
is a Sylow
-subgroup of
.
Proof. Let be the greatest integer such that
divides the order of
. Let
be the canonical homomorphism from
to
. Then
is the kernel of the restriction of
to
. Since
is isomorphic to
, a
-subgroup of
of order not exceeding
, it follows from Lagrange's Theorem that
has order at least
. Thus
; since
is a
-subgroup of
, it follows that it is a Sylow
-subgroup of
.
Note that this is not true in general if is not normal. For instance
is a Sylow 2-subgroup of the symmetric group
and
is a subgroup of
, but their intersection,
, is evidently not a Sylow 2-subgroup of
.
Corollary 11. Let be a normal subgroup of
, let
be the canonical homomorphism. Then the image of every Sylow
-subgroup of
under
is a Sylow
-subgroup of
; furthermore, every Sylow
-subgroup of
is the image of a Sylow
-subgroup of
under
.
Proof. Let be a Sylow
-subgroup of
; let the order of
be
, where
is a positive integer not divisible by
. By Corollary 9,
is a group of order
; it follows that
is isomorphic to
, a group of order
, so
is a Sylow
-subgroup of
. Now, let
be a Sylow
-subgroup of
; then there exists
such that
. Let
be such that
; then
.