Sylow theorems, proof of

We let $G$ be a group of order $p^{m}k$ 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:

 $|G|=|Z(G)|+\sum_{[a]\neq Z(G)}|[a]|$

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 $p|Z(G)$. ∎

Proposition 2.

$G$ has a Sylow p-subgroup

Proof.

By induction on $|G|$. If $|G|=1$ then there is no $p$ which divides its order, so the condition is trivial.

Suppose $|G|=p^{m}k$, $p\nmid k$, and the 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:

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

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 $Q$. 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 $1+k_{2}p+k_{3}p+\cdots+k_{s}p$, 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 $|G:N_{G}(P)|\mid k$. ∎

Title Sylow theorems, proof of SylowTheoremsProofOf 2013-03-22 12:51:02 2013-03-22 12:51:02 Henry (455) Henry (455) 12 Henry (455) Proof msc 20D20 SylowPSubgroup SylowsThirdTheorem