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[parent] 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:

Proposition 1   If $ p$ divides the size of every conjugacy class outside the center then $ p$ divides the order of the center.
Proof. This follows from the class equation:
$\displaystyle \vert G\vert=\vert Z(G)\vert+\sum_{[a]\neq Z(G)} \vert[a]\vert $
If $ p$ divides the left hand side, and divides all but one entry on the right hand side, it must divide every entry on the right side of the equation, so $ p\vert Z(G)$. $ \qedsymbol$
Proposition 2   $ G$ has a Sylow p-subgroup
Proof. By induction on $ \vert G\vert$. If $ \vert G\vert=1$ then there is no $ p$ which divides its order, so the condition is trivial.

Suppose $ \vert G\vert=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 $ \vert P\vert=\vert P/\langle f\rangle\vert\cdot\vert\langle f\rangle\vert=p^m$.

On the other hand, if $ p\nmid \vert Z(G)\vert$ 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 \vert[a]\vert=[G:C(a)]$, and since $ \vert C(a)\vert\cdot[G:C(a)]=\vert G\vert$, $ p^m\mid \vert C(a)\vert$. 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$. $ \qedsymbol$

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 $ \vert RP\vert=\frac{\vert R\vert\cdot\vert P\vert}{\vert R\cap P\vert}$. $ P$ is a subgroup of $ RP$, so $ p^m\mid \vert RP\vert$. But $ R$ is a subgroup of $ Q$ and $ P$ is a Sylow p-subgroup, so $ \vert R\vert\cdot \vert P\vert$ is a multiple of $ p$. Then it must be that $ \vert RP\vert=p^m$, and therefore $ P=RP$, and so $ R\subseteq P$. Obviously $ R\subseteq Q$, so $ R\subseteq Q\cap P$. $ \qedsymbol$

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}\vert\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:

$\displaystyle g\cdot X=g\cdot xPx^{-1}=gxPx^{-1}g^{-1}=(gx)P(gx)^{-1} $

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 $ \vert O_i\vert=[Q:Q\cap P_i]$.

There are two easy results on this construction. If $ Q=P_i$ then $ \vert O_i\vert=[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 \vert O_i\vert$.

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 $ \vert O_1\vert=1$ and $ p\mid \vert O_i\vert$ 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 \vert O_i\vert$ 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$. $ \qedsymbol$
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 $ \vert G:N_G(P)\vert$. Since $ P$ is a subgroup of its normalizer, $ p^m\mid N_G(P)$, and therefore $ \vert G:N_G(P)\vert\mid k$. $ \qedsymbol$



"proof of Sylow theorems" is owned by Henry. [ full author list (2) | owner history (2) ]
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See Also: Sylow p-subgroup, Sylow's third theorem


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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 7317 times total.

Classification:
AMS MSC: 20D20 (Group theory and generalizations :: Abstract finite groups :: Sylow subgroups, Sylow properties, $\pi$-groups, $\pi$-structure)
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