Advanced Algebra I
Sylow Theorems
We are now ready to prove Sylow theorems. The first theorem re- gards the existence of p-subgroups in a given group. The second theo- rem deals with relation between p-subgroups. In particular, all Sylow p-subgroups are conjugate. The third theorem counts the number of Sylow p-subgroups.
Theorem 0.1 (First Sylow theorem). Let G be a finite group of order pnm (where (n, m) = 1). Then there are subgroups of order pi for all 0 ≤ i ≤ n.
Furthermore, for each subgroup Hi of order pi, there is a subgroup Hi+1 of order pi+1 such that HiC Hi+1 for 0 ≤ i ≤ n − 1.
In particular, there exist a subgroup of order pn, which is maximal possible, called Sylow p-subgroup. We recall the useful lemma which will be used frequently.
Lemma 0.2. Let G be a finite p-group. Then
|S| ≡ |S0| (mod p).
proof of the theorem. We will find subgroup of order pi inductively. By Cauchy’s theorem, there is a subgroup of order p. Suppose that H is a subgroup of order pi. Consider the group action that H acts on S = G/H by translation. One show that xH ∈ S0 if and only if x ∈ NG(H). Thus |S0| = |NG(H)/H|. If i < n, then
|S0| ∼= |S| ∼= 0 (mod p).
By Cauchy’s theorem, the group NG(H)/H contains a subgroup of or- der p. The subgroup is of the form H1/H, hence |H1| = pi+1. Moreover, H C H1.
¤ Example 0.3. If G is a finite p-group of order pn, then one has a series of subgroups {e} = H0 < H1 < ... < Hn = G such that |Hi| = pi and HiC Hi+1, Hi+1/Hi ∼= Zp. In particular, G is solvable.
Definition 0.4. A subgroup P of G is a Sylow p-subgroup if P is a maximal p-subgroup of G.
If G is finite of order pnm then a subgroup P is a Sylow p-subgroup if and only if |P | = pn by the proof of the first theorem.
Theorem 0.5 (Second Sylow theorem). Let G be a finite group of order pnm. If H is a p-subgroup of a G, and P is any Sylow p-subgroup of G, then there exists x ∈ G such that xHx−1 < P .
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Proof. Let S = G/P and H acts on S by translation. Thus by the Lemma, one has |S0| ≡ |S| = m(mod p). Therefore, S0 6= ∅. One has
xP ∈ S0 ⇔ hxP = xP ∀h ∈ H ⇔ x−1hx < P.
¤ An immedaitely but important consequence is that two Sylow p- subgroups are conjugate.
Theorem 0.6 (Third Sylow theorem). Let G be a finite group of order pnm. The number of Sylow p-subgroups divides |G| divides |G| and is of the form kp + 1.
Proof. Let S be the conjugate class of a Sylow p-subgroup P (this is the same as the set of all Sylow p-subgroups). We consider the action that G acts on S by conjugation, then the action is transitive. Hence
|S| | |G|.
Furthermore, we consider the action P × S → S by conjugation.
Then
Q ∈ S0 ⇔ xQx−1 = Q ∀x ∈ P ⇔ P < NG(Q).
Both P, Q are Sylow p-subgroup of NG(Q) and therefore conjugate in NG(Q). However, Q C NG(Q), Q has no conjugate other than itself.
Thus one concludes that P = Q. In particular, S0 = {P }. By the
Lemma, one has |S| = 1 + kp. ¤
Example 0.7. Group of order 200 must have normal Sylow subgroups.
Hence it’s not simple. (let rp := number of Sylow p-subgroups. Then r5 = 1).
Example 0.8 (Classification of groups of order 2p (p 6= 2)). Let G be a group of order 2p. If it’s abelian, then it’s cyclic by fundamental the- orem of abelian groups plus Chinese remainder theorem. Let’s suppose that it’s non-abelian.
There are elements a, b of order p, 2 respectively. By Sylow theorem, rp = 1, hence the subgroup < a > is normal. Then one notices that G =< a >< b > for < a > ∩ < b >= {e}. Moreover, bab−1 = ak for some k. One has
a = b2ab−2 = ak2.
It follows that k = 1, −1. If k = 1, then G is abelian. Thus we assume that k = −1. This gives the group Dp :=< a, b|ap = b2 = e, ab = ba−1 >.
Proposition 0.9. If H, K C G and H ∩ K = {e}, HK = G ,then G ∼= H ⊕ K.
Proposition 0.10. Let G be a group of order pq, with p > q distinct primes. If q - p − 1, then G is cyclic. If q | p − 1 then either G is cyclic or there is a unique model of non-abelian group up to isomorphism.
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(Which is a ”semi-direct” of two cyclic groups, or called a metacyclic groups in this case)
