Sylow's theorem: Difference between revisions

Revision as of 00:30, 22 February 2007

This article gives the statement, and possibly proof, of a basic fact in group theory.
View a complete list of basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

This result is related to the theory of Sylow subgroups

Statement

Verbal statement

The Sylow's theorem(s) give(s) information about the existence of $p$ -Sylow subgroups of a finite group, as well as the relation among them. More specifically, given a finite group:

Existence: For any prime $p$ , there exists a $p$ -Sylow subgroup
Conjugacy: Any two $p$ -Sylow subgroups are conjugate in the whole group
Domination: Any $p$ -subgroup is contained inside some $p$ -Sylow subgroup
Congruence: The number of $p$ -Sylow subgroups divides the index of any $p$ -Sylow subgroup and is also congruent to $1$ modulo $p$ .

Symbolic statement

Let $G$ be a finite group and $p$ a prime. A subgroup of $G$ is termed a $p$ -Sylow subgroup if its order is a power of $p$ and its index is relatively prime to $p$ . Then Sylow's theorem states that:

Existence: There exists a $p$ -Sylow subgroup $P$ of $G$
Conjugacy: If $P$ and $Q$ are $p$ -Sylow subgroups of $G$ then there exists $g$ in $G$ such that $g P g^{- 1} = Q$ viz $P$ and $Q$ are conjugate subgroups)
Domination: Let $P$ be a $p$ -Sylow subgroup and $Q$ a $p$ -group. Then there exists a $g$ in $G$ such that $g Q g^{- 1} \subseteq P$ .
Congruence: Let $S y l_{p} (G)$ denote the set of $p$ -Sylow subgroups of $G$ and $n_{p}$ denote the cardinality of $S y l_{p} (G)$ . Then, $n_{p} \equiv 1 mod p$ .

Proof of existence part

Proof by action on subsets

Let $n$ be the order of $G$ . Let $p^{r}$ be the higher power of $p$ that divides the order of $G$ . Clearly, a subgroup of $G$ is a $p$ -Sylow subgroup if and only if it has order $p^{r}$ . If we denote $n / p^{r}$ by $m$ , then the index of any $p$ -Sylow subgroup must be $m$ .

Consider the action of $G$ by left multiplication on the set of subsets of $G$ of size $p^{r}$ . Here are some observations regarding this action:

The size of the set on which $G$ acts is the number of subsets of size $p^{r}$ in $G$ . This is a binomial coefficient, and an appeal to Lucas' theorem reveals that its value is relatively prime to $p$ .
Since the total cardinality of the set is relatively prime to $p$ , there must exist at least one orbit under the action of $G$ which has size relatively prime to $p$ . Let $S$ be the isotropy subgroup for a point in this orbit.
Now, since the size of the orbit is relatively prime to $p$ , the index of $S$ is relatively prime to $p$ , and hence a divisor of $m$ .
On the other hand, given any subset $T$ of size $p^{r}$ , we know that the translates of $T$ cover the whole of $G$ , and therefore there are at least $n / p^{r}$ members in the orbit of $T$ .
Combining these two facts $S$ must have index exactly $m$ -- hence it is a subgroup.

Interestingly, the only nontrivial result we use here (Lucas' theorem) can itself be proved using group theory (although a purely algebraic proof also exists).

Proof by conjugation action

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

Proof of domination

Proof using coset spaces

Let $G$ be a finite group, $P$ a $p$ -Sylow subgroup, and $Q$ a $p$ -group. Suppose $n = p^{r} m$ where $n$ is the order of $G$ adn $m$ is relatively prime to $p$ . Then:

$G$ naturally acts (on the left) on the left coset space of $P$ .
Since $Q$ is a subgroup of $G$ , $Q$ also acts on the left coset space of $P$
The left coset space of $P$ has cardinality $m$ , which is relatively prime to $p$ . Hence, under the action of $Q$ (which is a $p$ -group) on this set, there is at least one fixed point. Let the fixed point be $g P$ .
We have $Q g P = g P$ . This gives us: $(g^{- 1} Q g) P = P$ , and hence $g^{- 1} Q g \subseteq P$ . Thus, a conjugate of $Q$ is contained inside $P$ .

@@ Line 1: / Line 1: @@
 {{basic fact}}
+{{sylow theory result}}
 ==Statement==
@@ Line 20: / Line 22: @@
 * '''Domination''': Let <math>P</math> be a <math>p</math>-Sylow subgroup and <math>Q</math> a <math>p</math>-group. Then there exists a <math>g</math> in <math>G</math> such that <math>gQg^{-1} \subseteq P</math>.
 * '''Congruence''': Let <math>Syl_p(G)</math> denote the set of <math>p</math>-Sylow subgroups of <math>G</math> and <math>n_p</math> denote the cardinality of <math>Syl_p(G)</math>. Then, <math>n_p \equiv 1\mod p</math>.
+==Proof of existence part==
+===Proof by action on subsets===
+Let <math>n</math> be the [[order of a group|order]] of <math>G</math>. Let <math>p^r</math> be the higher power of <math>p</math> that divides the order of <math>G</math>. Clearly, a subgroup of <math>G</math> is a <math>p</math>-Sylow subgroup if and only if it has order <math>p^r</math>. If we denote <math>n/p^r</math> by <math>m</math>, then the index of any <math>p</math>-Sylow subgroup must be <math>m</math>.
+Consider the action of <math>G</math> by left multiplication on the set of ''subsets'' of <math>G</math> of size <math>p^r</math>. Here are some observations regarding this action:
+* The size of the set on which <math>G</math> acts is the number of subsets of size <math>p^r</math> in <math>G</math>. This is a binomial coefficient, and an appeal to [[Lucas' theorem]] reveals that its value is relatively prime to <math>p</math>.
+* Since the total cardinality of the set is relatively prime to <math>p</math>, there must exist at least one orbit under the action of <math>G</math> which has size relatively prime to <math>p</math>. Let <math>S</math> be the isotropy subgroup for a point in this orbit.
+* Now, since the size of the orbit is relatively prime to <math>p</math>, the index of <math>S</math> is relatively prime to <math>p</math>, and hence a divisor of <math>m</math>.
+* On the other hand, given any subset <math>T</math> of size <math>p^r</math>, we know that the translates of <math>T</math> cover the whole of <math>G</math>, and therefore there are at least <math>n/p^r</math> members in the orbit of <math>T</math>.
+* Combining these two facts <math>S</math> must have index exactly <math>m</math> -- hence it is a subgroup.
+Interestingly, the only nontrivial result we use here (Lucas' theorem) can itself be proved using group theory (although a purely algebraic proof also exists).
+===Proof by conjugation action===
+{{fillin}}
+==Proof of domination==
+===Proof using coset spaces===
+Let <math>G</math> be a [[finite group]], <math>P</math> a <math>p</math>-Sylow subgroup, and <math>Q</math> a <math>p</math>-group. Suppose <math>n=p^rm</math> where <math>n</math> is the order of <math>G</math> adn <math>m</math> is relatively prime to <math>p</math>. Then:
+* <math>G</math> naturally acts (on the left) on the left coset space of <math>P</math>.
+* Since <math>Q</math> is a subgroup of <math>G</math>, <math>Q</math> also acts on the left coset space of <math>P</math>
+* The left coset space of <math>P</math> has cardinality <math>m</math>, which is relatively prime to <math>p</math>. Hence, under the action of <math>Q</math> (which is a <math>p</math>-group) on this set, there is at least one fixed point. Let the fixed point be <math>gP</math>.
+* We have <math>QgP = gP</math>. This gives us: <math>(g^{-1}Qg)P = P</math>, and hence <math>g^{-1}Qg \subseteq P</math>. Thus, a conjugate of <math>Q</math> is contained inside <math>P</math>.