Sylow theorems, proof of

We let G be a group of order pmk where pk and prove Sylow’s theoremsMathworldPlanetmath.

First, a fact which will be used several times in the proof:

Proposition 1.

If p divides the size of every conjugacy classMathworldPlanetmathPlanetmath outside the center then p divides the order of the center.


This follows from the class equationMathworldPlanetmathPlanetmath:


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.

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

Suppose |G|=pmk, pk, 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, f, also of order p. Since this is a subgroupMathworldPlanetmathPlanetmath of the center, it is normal, so G/f is well-defined and of order pm-1k. By the inductive hypothesis, this group has a subgroup P/f of order pm-1. Then there is a corresponding subgroup P of G which has |P|=|P/f||f|=pm.

On the other hand, if p|Z(G)| then consider the conjugacy classes not in the center. By the propositionPlanetmathPlanetmath 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|[a]|=[G:C(a)], and since |C(a)|[G:C(a)]=|G|, pm|C(a)|. But C(a)G, since aZ(G), so C(a) has a subgroup of order pm, and this is also a subgroup of G. ∎

Proposition 3.

The intersectionMathworldPlanetmath of a Sylow p-subgroup with the normalizerMathworldPlanetmath of a Sylow p-subgroup is the intersection of the subgroups. That is, QNG(P)=QP.


If P and Q are Sylow p-subgroups, consider R=QNG(P). Obviously QPR. In additionPlanetmathPlanetmath, since RNG(P), the second isomorphism theorem tells us that RP is a group, and |RP|=|R||P||RP|. P is a subgroup of RP, so pm|RP|. But R is a subgroup of Q and P is a Sylow p-subgroup, so |R||P| is a multipleMathworldPlanetmathPlanetmath of p. Then it must be that |RP|=pm, and therefore P=RP, and so RP. Obviously RQ, so RQP. ∎

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 XCX=xPx-1={xpx-1|pP} for some xG. Observe that every XC is a Sylow p-subgroup (and we will show that the converseMathworldPlanetmath holds as well). We let G act on C by conjugationMathworldPlanetmath:


This is clearly a group actionMathworldPlanetmath, 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 O1,,Os, and let P1,,Ps be representatives of the corresponding orbits. By the orbit-stabilizer theorem, the size of an orbit is the index of the stabilizerMathworldPlanetmath, and under this action the stabilizer of any Pi is just NQ(Pi)=QNG(Pi)=QP, so |Oi|=[Q:QPi].

There are two easy results on this construction. If Q=Pi then |Oi|=[Pi:PiPi]=1. If QPi then [Q:QPi]>1, and since the index of any subgroup of Q divides Q, p|Oi|.

Proposition 4.

The number of conjugates of any Sylow p-subgroup of G is congruentMathworldPlanetmath to 1 modulo p

In the construction above, let Q=P1. Then |O1|=1 and p|Oi| for i1. 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+k2p+k3p++ksp, which is obviously congruent to 1 modulo p.

Proposition 5.

Any two Sylow p-subgroups are conjugate


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 QPi for every i, and therefore p|Oi| 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 pm is congruent to 1 modulo p and is a factor of k


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:NG(P)|. Since P is a subgroup of its normalizer, pmNG(P), and therefore |G:NG(P)|k. ∎

Title Sylow theoremsMathworldPlanetmath, 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