JustToThePoint English Website Version
JustToThePoint en español

Maximize your online presence with our exclusive offer: Get a stunning hero banner, the hero you need and deserve, at an unbeatable price! Bew, 689282782, bupparchard@gmail.com

Some applications of Sylow’s Theorems

4th Law. Life, economic collapse, and history are like a roll of toilet paper, the closer it gets to the end, the faster it goes, Apocalypse, Anawim, #justtothepoint.


Cauchy’s Theorem. Let G be a finite group and let p be a prime that divides the order of G. Then, G has an element of order p, so also a subgroup with order p.

Sylow’s First Theorem. Let G be a finite group and p be a prime. If pk | |G| for some integer k ∈ ℤ, then G has a subgroup of order pk, i.e., ∃H ≤ G such that |H| = pk.

Sylow’s Second Theorem. Let G be a finite group and p a prime dividing |G|. Then, all Sylow p-subgroups are conjugates, that is, if P and Q are Sylow p-subgroups, then there exists some g ∈ G such that gPg-1 = Q.

Sylow’s Third Theorem. Let G be a finite group and p a prime dividing |G|. Then, the number of Sylow p-subgroups of G is congruent to 1 (mod p) and divides |G|, np ≡ 1 (mod p), np | |G|


Theorem. Let G be a group. If G has an unique subgroup H (H ≤ G) of a given order, then that subgroup is normal, H ◁ G.


Let H ≤ G be the only subgroup of G of a given order. ∀g ∈ G, [Note 2 below. The conjugate of a subgroup is a subgroup] gHg-1 ≤ G

[Set equivalence of Regular Representations, If S is a finite subset of a group G, then |a·S| = |S| = |S·a| because the left (λa: S → S, ∀x∈S: λa(x) = a·x) and right regular representations are permutations, and therefore bijections, Note 1] Besides, |Ha| = |H|, that is because |Ha| = |aHa-1| = |(aH)·a-1| =[|a·S| = |S| = |S·a| ] |aH| =[All cosets have the same size] |H|

|Ha| = {y∈ G: ∃x ∈ H: y = axa-1} = aHa-1

However, by assumption, H is the only subgroup of G of order |H| ⇒ gHg-1 = H ⇒ [A subgroup N of G is normal if and only if N is self-conjugate, that is, ∀g∈G: g·N·g-1 = N, g-1·N·g = N ] H is normal∎

Note 1: Let G be a group and N be a normal subgroup N ◁ G (∀g ∈ G: g·N = N·g where g·N denotes the subset product of g with N = {a·n: n ∈ N}.

Note 2: Let G be a group, let H ≤ G be a subgroup of G, then the conjugate of H by a is a subgroup, i.e., ∀H ≤ G, a ∈ G: Ha ≤ G where Ha = aHa-1.

Corollary. A Sylow p-subgroup of a finite group G is a normal subgroup of G if and only if it is the only Sylow p-subgroup of G.


⇐) If G has only one Sylow p-subgroup P ⇒ [Let G be a group. If G has an unique subgroup H (H ≤ G) of a given order, then that subgroup is normal, H ◁ G.] P must be normal, P ◁ G.

⇒) Conversely, suppose a Sylow p-subgroup P is normal ⇒ [Second Sylow Theorem. All the Sylow p-subgroups of a finite group are conjugates] since P is normal, it equals its conjugates, so it is the only one.

Theorem. Let G be a group with H, K ◁ G, H ∩ K = {e}, and HK = G. Then, G ≋ H x K.


Since H ◁ G, ∀h ∈ H, k∈ K, khk-1∈ H ⇒ ∃h’∈ H such that khk-1 = h’ ⇒ [*k both sides of the equation] kh = h’k

Similarly, since K ◁ G, ∀h ∈ H, k∈ K, h-1kh ∈ K, ∃k’∈ K such that h-1kh = k’ ⇒ [*h both sides of the equation] kh = hk'

Therefore, kh = h’k = hk’ ⇒ hk’ = h’k ⇒ [*h’-1] (h’)-1hk’ = k ⇒ [*k’-1] (h’)-1h = k(k’)-1

h’-1h = kk’-1 [h’-1h ∈ H and kk’-1 ∈ K] ∈ H ∩ K = {e}

Thus, h’-1h = e ⇒ h’ = h and kk’-1 = e ⇒ k’ = k. Finally, kh = h’k = hk’ = hk, kh = hk.

Then, we can define Φ:H x K → G, Φ((h, k)) = hk. We claim that Φ is a group isomorphism.

Φ((h1, k1))Φ((h2, k2)) = h1k1h2k2 = [∀h ∈ H, k ∈ K, hk = kh] h1h2k1k2

Φ((h, k)) = e ⇒ hk = e ⇒ h = k-1 ∈ H ∩ K ⇒[By assumption, H ∩ K = {e}] h = e ⇒ [hk = e] k = e, i.e., Ker(Φ) = {(e, e)}, and by the First Group Isomorphism Theorem, (H x K)/Ker(Φ) ≋ Φ(H x K).

Theorem. Let G be a group of order pq, p, q be distinct primes such that p < q, then it has a normal subgroup of order q. In particular, no group of order pq can be simple. Besides, if p does not divide q -1 (q ≆ 1 (mod p)), then G is cyclic and G ≋ ℤpq ≋ ℤpxℤq


Let |G| = pq, p < q, nq be the number of Sylow q-subgroups.

By Sylow’s Third Theorem, nq | |G| and nq ≡ 1 (mod q) ⇒ The divisors of pq are 1, p, q, and pq. However, q and pq are ≡ 0 (mod q). And as p < q, p ≇1 (mod q) -0, 1, 2, ···, p (p ≡ p (mod q) ),.·· q-1-. Therefore, nq = 1 ⇒ There is exactly one Sylow q-subgroup of G, say H ⇒[H Sylow p-Subgroup is unique ↭ H ◁ G] H must be normal.

Next, suppose that q ≆ 1 (mod p). Let np be the number of Sylow p-subgroups of G. Again, by Sylow’s Third Theorem, np | pq (so np = 1, q, p, and pq) and np ≡ 1 (mod p). However, q and pq are ≡ 0 (mod q) and the assumption above (q ≆ 1 (mod p)) prevents nq = q, and therefore np =1 ⇒ there is a unique Sylow p-subgroup, and then as a unique Sylow p-subgroup is normal.

Recall that every group of order p prime is a cyclic group and isomorphic to ℤp. Considering this, these Sylow subgroups H, K must be cyclic and isomorphic to ℤp and ℤq respectively.

H ∩ K ≤ H, H ∩ K ≤ K ⇒ |H ∩ K| divides both p and q, and therefore is the trivial subgroup, {e}.

It follows that |HK| = |H||K||H∩K| = pq = |G| ⇒ HK = G. Then, by the previous theorem [Recall. Theorem. Let G be a group with H, K ◁ G, H ∩ K = {e}, and HK = G. Then, G ≋ H x K.], G ≋ ℤpxℤq ≋ [The direct product of two cyclic group of coprime order is itself cyclic] ℤpq


If H, K ≤ G subgroups of a finite group of order 40, K ◁ G, |K| = 5, |H| = 8 ⇒ |H∩K| needs to divide 8 and 5 (it is a subgroup of H and K), so |H∩K| = 1 ⇒ |HK| = |H||K||H∩K| ⇒ G = HK. If H is normal, too, then by the previous theorem G ≋ H x K.

From Cauchy theorem: |G| < ∞, p is prime, p ∣ |G| ⇒ in G exists an element with order p, so also subgroup with order p. In our case, |S| = 2, |P| = 7.

Let’s suppose the case that we have seven Sylow 2-subgroups. G has a unique Sylow 7-subgroup which is isomorphic to ℤ7, and six elements (without considering the identity) of order 7 in G. Therefore, we have that one element in G have order 1, namely the identity, ({e}), 6 have order 7, (ℤ7 - {e}) and 7 (ℤ2- {e}) corresponding to the subgroups of order 2, have order 2.

Let r, |r| = 7, |s| = 2, P be the unique Sylow 7-subgroup ⇒ P is normal and cyclic (P ◁ G, P = ⟨r⟩) ⇒ [P ◁ G] srs-1 ∈ P ⇒ [P = ⟨r⟩] srs-1 = [s=s-1 because s has order 2] srs = rn for some n ∈ ℤ ⇒ ssrss = srns ⇒ [s2 = e] r = srns = [srs=rn⇒ (replace r by rn) srns = (rn)n] = rn2. Therefore, r = rn2, since r has order 7, n2≡1 mod 7 ⇒ n = ± 1.

  1. n = 1, srs = r ⇒ sr = rs ⇒ G ≋ ⟨s⟩ x ⟨r⟩ ≋ ℤ2 x ℤ7 ≋ ℤ14
  2. n = -1 or 6 (1 + 6 ≡ 0 (mod 7), 6 = -1 in ℤ7) ⇒ srs = r-1. Therefore, we have r7=1, s2=1, and sr = r-1s, G = ⟨r, s: r7 = e = s2, sr = r-1s⟩ this is the dihedral group D7.

Thus, there are only two groups, up to isomorphism, of order 14, namely ℤ14 and D7.

Theorem. Let p be a odd prime. Then, there are only two groups of order 2p, up to isomorphism, namely ℤ2p and Dp.


It is the same argument as before, 2p is a semi-prime ⇒ There is a unique, normal Sylow p-subgroup, and cyclic (Every prime group is cyclic), isomorphic to ℤp, and G cannot be simple.

p ≡ 1 (mod 2) because p is an odd number ⇒ n2 = 1 or p. If n2 = 1, G ≋ ℤ2xℤp ≋ ℤ2p. Otherwise, n2 = p, G ≋ Dp as we have shown in the previous example.


This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn (Abstract Algebra), and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. Andrew Misseldine: College Algebra and Abstract Algebra.
Bitcoin donation

JustToThePoint Copyright © 2011 - 2024 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.

This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.