# 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^{}:

$$|G|=|Z(G)|+\sum _{[a]\ne Z(G)}|[a]|$$ |

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 $f$, $\u27e8f\u27e9$, also of order $p$. Since this is a subgroup^{} of the center, it is normal, so $G/\u27e8f\u27e9$ is well-defined and of order ${p}^{m-1}k$. By the inductive hypothesis, this group has a subgroup $P/\u27e8f\u27e9$ of order ${p}^{m-1}$. Then there is a corresponding subgroup $P$ of $G$ which has $|P|=|P/\u27e8f\u27e9|\cdot |\u27e8f\u27e9|={p}^{m}$.

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 $C(a)\ne G$, 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, $Q\mathrm{\cap}{N}_{G}\mathit{}\mathrm{(}P\mathrm{)}\mathrm{=}Q\mathrm{\cap}P$.

###### 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
$X\in C\leftrightarrow X=xP{x}^{-1}=\{xp{x}^{-1}|\forall p\in P\}$ 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^{}:

$$g\cdot X=g\cdot xP{x}^{-1}=gxP{x}^{-1}{g}^{-1}=(gx)P{(gx)}^{-1}$$ |

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 ${O}_{1},\mathrm{\dots},{O}_{s}$, and let ${P}_{1},\mathrm{\dots},{P}_{s}$ 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 $Q\ne {P}_{i}$ 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 $\mathrm{1}$ modulo $p$

In the construction above, let $Q={P}_{1}$. Then $|{O}_{1}|=1$ and $p\mid |{O}_{i}|$ for $i\ne 1$. Since the number of conjugates of $P$ is the sum of the number in each orbit, the number of conjugates is of the form $1+{k}_{2}p+{k}_{3}p+\mathrm{\cdots}+{k}_{s}p$, 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 $Q\ne {P}_{i}$ 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 $\mathrm{1}$ 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 |
---|---|

Canonical name | SylowTheoremsProofOf |

Date of creation | 2013-03-22 12:51:02 |

Last modified on | 2013-03-22 12:51:02 |

Owner | Henry (455) |

Last modified by | Henry (455) |

Numerical id | 12 |

Author | Henry (455) |

Entry type | Proof |

Classification | msc 20D20 |

Related topic | SylowPSubgroup |

Related topic | SylowsThirdTheorem |