|
|
|
|
![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
proof of Sylow theorems
|
(Proof)
|
|
|
We let $G$ be a group of order $p^mk$ where $p\nmid k$ and prove Sylow's theorems.
First, a fact which will be used several times in the proof:
Proof. By induction on $|G|$ . If $|G|=1$ then there is no $p$ which divides its order, so the condition is trivial.
Suppose $|G|=p^mk$ , $p\nmid k$ , and the proposition holds for all groups of smaller order. Then we can consider whether $p$ divides the order of the center, $Z(G)$ .
If it does, then by Cauchy's theorem, there is an element $f$ of $Z(G)$ of order $p$ , and therefore a cyclic subgroup generated by $f$ , $\langle f\rangle$ , also of order $p$ . Since this is a subgroup of the center, it is normal, so $G/\langle f\rangle$ is well-defined and of order $p^{m-1}k$ . By the inductive hypothesis, this group has a subgroup $P/\langle f\rangle$ of order $p^{m-1}$ . Then there is a corresponding subgroup $P$ of $G$ which has $|P|=|P/\langle f\rangle|\cdot|\langle f\rangle|=p^m$ .
On the other hand, if $p\nmid |Z(G)|$ then consider the conjugacy classes not in the center. By the proposition above, since $Z(G)$ is not divisible by $p$ , at least one conjugacy class can't be. If $a$ is a representative of this class then we have $p\nmid |[a]|=[G:C(a)]$ , and since $|C(a)|\cdot[G:C(a)]=|G|$ , $p^m\mid |C(a)|$ . But $C(a)\neq G$ , since $a\notin Z(G)$ , so $C(a)$ has a subgroup of order $p^m$ , and this is also a subgroup of $G$ . 
Proposition 3 The intersection of a Sylow p-subgroup with the normalizer of a Sylow p-subgroup is the intersection of the subgroups. That is, $Q\cap N_G(P)=Q\cap P$ .
Proof. If $P$ and $Q$ are Sylow p-subgroups, consider $R=Q\cap N_G(P)$ . Obviously $Q\cap P\subseteq R$ . In addition, since $R\subseteq N_G(P)$ , the second isomorphism theorem tells us that $RP$ is a group, and $|RP|=\frac{|R|\cdot|P|}{|R\cap P|}$ . $P$ is a subgroup of $RP$ , so $p^m\mid |RP|$ . But $R$ is a subgroup of $Q$ and $P$ is a Sylow p-subgroup, so $|R|\cdot |P|$ is a multiple of $p$ . Then it must be that $|RP|=p^m$ , and therefore $P=RP$ , and so $R\subseteq P$ . Obviously $R\subseteq Q$ , so $R\subseteq Q\cap P$ . 
The following construction will be used in the remainder of the proof:
Given any Sylow p-subgroup $P$ , consider the set of its conjugates $C$ . Then $X\in C\leftrightarrow X=xPx^{-1}=\{xpx^{-1}|\forall p\in P\}$ for some $x\in G$ . Observe that every $X\in C$ is a Sylow p-subgroup (and we will show that the converse holds as well). We let $G$ act on $C$ by conjugation:
This is clearly a group action, so we can consider the orbits of $P$ under it; this remains true if we only consider elements from some subset of $G$ . Of course, if all $G$ is used then there is only one orbit, so we restrict the action to a Sylow p-subgroup $Q$ . Name the orbits $O_1,\ldots, O_s$ , and let $P_1,\ldots, P_s$ be representatives of the corresponding orbits. By the orbit-stabilizer theorem, the size of an orbit is the index of the stabilizer, and under this action the stabilizer of any $P_i$ is just $N_Q(P_i)=Q\cap N_G(P_i)=Q\cap P$ , so $|O_i|=[Q:Q\cap P_i]$ .
There are two easy results on this construction. If $Q=P_i$ then $|O_i|=[P_i:P_i\cap P_i]=1$ . If $Q\neq P_i$ then $[Q:Q\cap P_i]>1$ , and since the index of any subgroup of $Q$ divides $Q$ , $p\mid |O_i|$ .
Proposition 4 The number of conjugates of any Sylow p-subgroup of $G$ is congruent to $1$ modulo $p$
In the construction above, let $Q=P_1$ . Then $|O_1|=1$ and $p\mid |O_i|$ for $i\neq 1$ . Since the number of conjugates of $P$ is the sum of the number in each orbit, the number of conjugates is of the form $1+k_2p+k_3p+\cdots+k_sp$ , which is obviously congruent to $1$ modulo $p$ .
Proposition 5 Any two Sylow p-subgroups are conjugate
Proof. Given a Sylow p-subgroup $P$ and any other Sylow p-subgroup $Q$ , consider again the construction given above. If $Q$ is not conjugate to $P$ then $Q\neq P_i$ for every $i$ , and therefore $p\mid |O_i|$ for every orbit. But then the number of conjugates of $P$ is divisible by $p$ , contradicting the previous result. Therefore $Q$ must be conjugate to $P$ . 
Proposition 6 The number of subgroups of $G$ of order $p^m$ is congruent to $1$ modulo $p$ and is a factor of $k$
Proof. Since conjugates of a Sylow p-subgroup are precisely the Sylow p-subgroups, and since a Sylow p-subgroup has $1$ modulo $p$ conjugates, there are $1$ modulo $p$ Sylow p-subgroups.
Since the number of conjugates is the index of the normalizer, it must be $|G:N_G(P)|$ . Since $P$ is a subgroup of its normalizer, $p^m\mid N_G(P)$ , and therefore $|G:N_G(P)|\mid k$ . 
|
"proof of Sylow theorems" is owned by Henry. [ full author list (2) | owner history (2) ]
|
|
(view preamble | get metadata)
Cross-references: factor, sum, congruent, number, stabilizer, index, orbit-stabilizer theorem, action, subset, orbits, group action, conjugation, act on, converse, remainder, multiple, second isomorphism theorem, addition, normalizer, intersection, divisible, proposition, inductive hypothesis, well-defined, normal, subgroup, generated by, cyclic subgroup, Cauchy's theorem, induction, Sylow p-subgroup, equation, side, right, right hand side, left hand side, class equation, center, conjugacy class, size, divides, proof, Sylow's theorems, order, group
This is version 9 of proof of Sylow theorems, born on 2002-07-22, modified 2006-08-13.
Object id is 3182, canonical name is ProofOfSylowTheorems.
Accessed 8662 times total.
Classification:
| AMS MSC: | 20D20 (Group theory and generalizations :: Abstract finite groups :: Sylow subgroups, Sylow properties, $\pi$-groups, $\pi$-structure) |
|
|
|
|
|
|
Pending Errata and Addenda
|
|
|
|
|
|
|
|
|
|
|
![$\displaystyle \vert G\vert=\vert Z(G)\vert+\sum_{[a]\neq Z(G)} \vert[a]\vert $](http://images.planetmath.org:8080/cache/objects/3182/js/img2.png "$\displaystyle \vert G\vert=\vert Z(G)\vert+\sum_{[a]\neq Z(G)} \vert[a]\vert $")
