19. p-groups

The first Sylow theorem gives a special case where the answer to theabove question is positive.Theorem 20.2 (The first Sylow theorem) Let G be a finite group, andlet |G| = p n m, where p is a prime and gcd(p, m) = 1. Then for each i,1 ≤ i ≤ n, G contains a subgroup of order p i .Proof. We prove the theorem by induction on the order of G. If |G| = 1there is nothing to prove, so let us assume that |G| > 1.Let, as in Theorem 18.6, C 1 ,. . . , C r be the conjugacy classes of G containingmore than one element, and let x j ∈ C j , 1 ≤ j ≤ r, be representatives ofthese classes. By Theorem 18.4 we have1 < |C j | = [G : N(x j )],and therefore N(x j ) is a proper subgroup of G. If p n divides the order ofN(x j ) for some j, then, by induction, N(x j ) contains a subgroup of order p ifor each i, 1 ≤ i ≤ n. But N(x j ) ≤ G, and hence G contains a subgroup oforder p i for each i, 1 ≤ i ≤ n.Let us now assume that p n does not divide |N(x j )| for any j, 1 ≤ j ≤ r.By Lagrange’s Theorem we have|G| = |N(x j )|[G : N(x j )],and since p n ||G|, we conclude that p divides [G : N(x j )] for all j, 1 ≤ j ≤ r.From the class equation of G:r∑|G| = |Z(G)| + [G : N(x j )]we now conclude that p divides |Z(G)|.Z(G) is an abelian group, and by the Fundamental Theorem for FiniteAbelian Groups, it is isomorphic to a direct sum of cyclic groups of primepower orders. At least one of these summands must be Z p l for some l. Butthen p l−1 Z p l is a subgroup of Z p l isomorphic to Z p . We conclude that Z(G)contains a subgroup N of order p.Since elements of Z(G) commute with all elements of G, it follows thatN ✂ G. Form the quotient group G/N; it has order p n−1 m < |G|. Byinduction G/N contains a subgroup of order p j for every j, 1 ≤ j ≤ n−1. Bythe Correspondence Theorem, every such subgroup of G/N has a form K/N,where K is a subgroup of G containing N. But then |K| = |K/N||N| = p j p =p j+1 . Therefore G contains subgroups of orders p 2 , . . . , p n . G also containsa subgroup of order p, namely N itself, and this completes the proof of thetheorem. j=1

Definition 20.3. If G is a finite group of order p n m, where p is a prime andgcd(p, m) = 1, then every subgroup of order p n is called a Sylow p-subgroupof G.The first Sylow theorem states that a finite group G contains a Sylow p-subgroup for every prime p dividing |G|. The Second Sylow Theorem assertsthat all Sylow p-subgroups are related via conjugation.Theorem 20.4 (The second Sylow theorem) Let G be a finite group andlet p be a prime dividing the order of G. Every conjugate of a Sylow p-subgroup of G is again a Sylow p-subgroup of G. Conversely, every twoSylow p-subgroups are conjugate in G.In order to prove this theorem we need two lemmas.Lemma 20.5. Let P be a Sylow p-subgroup of a finite group G, and letg ∈ G be an element whose order is a power of p. If g −1 P g = P then g ∈ P .Proof. The condition g −1 P g means that g ∈ N(P ), and so 〈g〉 ≤ N(P ).Since P ✂N(P ), we can apply the Second Isomorphism Theorem, which givesTherefore〈g〉PP ∼ = 〈g〉〈g〉 ∩ P .|〈g〉P | = |〈g〉||P ||〈g〉 ∩ P | ,and since both |〈g〉| and |P | are powers of p, so is |〈g〉P |. On the other hand,P has the maximal possible such an order, and therefore we conclude that〈g〉P = P , which is equivalent to g ∈ P . Lemma 20.6. Let G be a group, and let H, T ≤ G. The number of distinctconjugates of H of the form t −1 Ht with t ∈ T is [T : T ∩ N(H)].Proof. This is a generalisation of Theorem 18.4, and the proof is left asan exercise. Proof of the second Sylow theorem. Let |G| = p n m, where p is a primeand gcd(p, m) = 1, and let P be an arbitrary Sylow p-subgroup of G.To prove the first statement, we note that for g ∈ G we have g −1 P g ∼ = Psince conjugation by g induces an automorphism of G (Theorem 17.6). Inparticular, |g −1 P g| = |P | = p n , and hence g −1 P g is a Sylow p-subgroup.

For the converse statement, denote by K = {P = P 1 , P 2 , . . . , P k } the setof all distinct conjugates of P . Assume that there exists a Sylow p-subgroupQ of G which is not in K. Consider conjugation by elements of Q in K. Forany P i ∈ K, choose q ∈ Q such that q ∉ P i ; by Lemma 20.5 we then haveq −1 P i q ≠ P i . Therefore, each P i has more than one conjugate by elementsfrom Q. By Lemma 20.6, the number of conjugates of P i by elements fromQ is equal to [Q : Q ∩ N(P i )]. Since Q is a p-group, this number must bedivisible by p. We conclude that conjugation by elements in Q partitions theset K into the classes, each of which has a multiple of p elements. Thereforek ≡ 0 (mod p).Now consider the conjugation by elements from P . This time P is theonly conjugate of itself, but the remaining elements of K again split intogroups, each having a multiple of p elements. This time we conclude thatk − 1 ≡ 0 (mod p), which is a contradiction with the conclusion of the lastparagraph. Therefore, we conclude that K contains all Sylow p-subgroups ofG, and hence any two such subgroups are conjugate.The third Sylow theorem discusses the number of Sylow p-subgroups.Theorem 20.7 (The third Sylow theorem) Let G be a finite group, andlet p be a prime dividing the order of G. The number of Sylow p-subgroupsof G divides |G| and has the form ps + 1 for s ≥ 0.Proof. Let P be a Sylow p-subgroup of G. By the second Sylow theorem,any other Sylow p-subgroup of G is a conjugate of P . By Theorem 18.4, thenumber of conjugates of P is [G : N(P )], and this divides |G| by Lagrange’stheorem.On the other hand, if K = {P = P 1 , P 2 , . . . , P k } is the set of all Sylowp-subgroups of G, then we can consider the conjugation in K by the elementsof P . As in the proof of the second Sylow theorem, the only conjugate ofP is P itself, and the rest of K is partitioned into blocks, each block havinga multiple of p elements. Therefore, k has the form ps + 1 for some s ≥ 0.Example 20.8. The symmetric group S 4 has order 24 = 2 3 3. Therefore, bythe first Sylow theorem we conclude that S 4 contains Sylow 2-subgroups oforder 8 and Sylow 3-subgroups of order 3. By the third Sylow theorem, thenumber of Sylow 3-subgroups divides 24 and is of the form 3k +1. Therefore,there are either one or four Sylow 3-subgroups. Now, each of eight threecyclesin S 4 generates a subgroup of order 3, and in this way we obtain4 distinct subgroups of order 3 in S 4 . Therefore, S 4 contains 4 Sylow 3-subgroups.

Similarly, the number of Sylow 2-subgroups of S 4 is either 1 or 3, andone may check that there are actually three of them. (The group H fromExample 6.7 is one of them.)21. Applications of the Sylow TheoremsSylow’s Theorems are a powerful tool for investigating finite groups. Weshall use them to classify the groups of order less than 16. First, however,we show how one can use Sylow’s Theorems to prove non-existence of simplegroups of certain orders. (Let us recall that a group is simple if it containsno proper normal subgroups.) The key observation here is the following:Theorem 21.1. Let G be a group and let P be a Sylow p-subgroup of G.Then P is normal in G if and only if P is a unique Sylow p-subgroup of G.Proof. P is normal ⇔ P is closed under conjugation ⇔ P is the onlyconjugate of itself.On the other hand, by the Second Sylow’s Theorem, any two Sylow p-subgroups of G are conjugate, and hence the theorem. Example 21.2. There exists no simple group of order 200. Indeed, let G bea group of order 200 = 2 3 5 2 . By the Third Sylow’s Theorem, the number ofSylow 5-subgroups of G divides 200 and has the form 5k + 1. The divisors of200 are 1, 5, 25, 2, 10, 50, 4, 20, 100, 8, 40, 200. The only number from theabove list that has the form 5k + 1 is 1. Therefore, G has a unique Sylow5-subgroup, which is then normal in G by Theorem 21.1. This proves thatG is not simple.Example 21.3. There exists no simple group of order 12. For, assume tothe contrary that G is a simple group of order 12 = 2 2 3. By the ThirdSylow’s Theorem the number of Sylow 3-subgroups of G is either 1 or 4.Since G is simple, this number must be four because of Theorem 21.1. Eachof these four subgroups has order 3, and is therefore isomorphic to Z 3 . ByLagrange’s Theorem, the intersection of any two of these subgroups must betrivial. Therefore, G has eight elements of order 3. Next, since G is notsimple, it must contain at least two Sylow 2-subgroups H 1 and H 2 , each ofthem having order 4. By Lagrange’s Theorem H 1 ∩ H 2 has either one or twoelements. Hence we conclude that G contains at least five elements of ordera power of 2, which contradicts the fact that |G| = 12.We can also prove the following nice characterisation of nilpotent groups.

Theorem 21.4. A finite group is nilpotent ⇔ it is isomorphic to the directproduct of p-groups.Proof. (Omitted in lectures)(⇐) Every p-group is nilpotent by Theorem 19.6, and the direct product ofnilpotent groups is again nilpotent by Theorem 16.9.(⇒) Let G be a nilpotent group. Our first claim is that, for an arbitraryproper subgroup H ≤ G, we have N(H) ≠ H. Indeed, if we choose i so thatγ i (G) ⊈ H but γ i+1 (G) ⊆ H, then we have[γ i (G), H] ⊆ [γ i (G), G] = γ i+1 (G) ⊆ H.Choose x ∈ γ i (G)\H. For every h ∈ H we have x −1 hxh −1 = [x, h] ∈ H, andhence x −1 hx ∈ Hh = H. Therefore x ∈ N(H)\H, as required.Next we prove that every Sylow p-subgroup P of G is normal by showingthat N(P ) = G. Assume that N(P ) ≠ G. Then N(P ) is properly containedin N(N(P )) by the first claim. Let x ∈ N(N(P ))\N(P ) be arbitrary. ¿Fromx ∉ N(P ) it follows that x −1 P x is a Sylow p-subgroup distinct from P , whilefrom x ∈ N(N(P )) and P ⊆ N(P ) it follows that x −1 P x ⊆ N(P ). Thiscontradicts Lemma 20.5.Assume that |G| = p α 11 . . . p αnn , and let P i a Sylow p i -subgroup of G;by the previous paragraph we have P i ✂ G. By Lagrange’s Theorem wehave P 1 P 2 . . . P n = G and P i ∩ (P 1 . . . P i−1 P i+1 . . . P n ) = {e}. Therefore, byTheorem 11.7 we have G ∼ = P 1 × . . . × P n , a direct product of p-groups. 22. Groups of order pqIt is possible to classify all the groups of order pq, where p and q are distinctprimes: there are at most two such groups, one of them is Z pq , and the other(if it exists) is a non-abelian group obtained by extending Z p by Z q in away similar to direct products. Here, however, we shall not give this fullclassification, but shall rather restrict our attention to two special cases.Theorem 22.1. Let p and q be two distinct prime numbers, with p > q. Ifp − 1 is not divisible by q then Z pq is the only group of order pq.Proof. Let G be a group of order pq. By the Third Sylow Theorem thenumber of Sylow p-subgroups has the form kp + 1 and divides pq. Note thatthe divisors of pq are 1, p, q and pq. Since p > q we must have k = 0, and soG has a unique Sylow p-subgroup H. By Theorem 21.1, H is normal in G;also we have H ∼ = Z p by Theorem 6.10.

Similarly, the number of Sylow q-subgroups is kq + 1 and divides pq.Clearly we cannot have kq + 1 = q or kq + 1 = pq. However, we cannot havekq + 1 = p either, because of the condition that q does not divide p − 1. Sowe are left with the only option of G having a unique Sylow q-subgroup K.As for H we now deduce that K ✂ G and that K ∼ = Z q .By Lagrange’s Theorem, the order of H ∩ K must divide both |H| = pand |K| = q, so that H ∩ K = {e}. Also by Lagrange’s Theorem, the orderof HK is divisible by both p and q, and so is equal to pq, which meansthat HK = G. By Theorem 11.5 we now have G ∼ = H × K, and henceG ∼ = Z p ⊕ Z q∼ = Zpq by Theorem 12.1. Theorem 22.2. Let p > 2 be a prime. The only groups of order 2p are thecyclic group Z 2p and the dihedral group D p .Proof. Let G be an arbitrary group of order 2p. By the Third Sylow’sTheorem, G has a unique Sylow p-subgroup N. This subgroup has order p,and is therefore cyclic. Let a ∈ N be arbitrary, so thatN = {e = a 0 , a, a 2 , . . . , a p−1 }.G also contains at least one element b of order 2. Clearly b ∉ N, and since[G : N] = 2, it follows that N and Nb are the only cosets of N in G. ThereforeG = {e = a 0 , a, a 2 , . . . , a p−1 , b, ab, a 2 b, . . . , a p−1 b}.The number of Sylow 2-subgroups of G is either 2 or p. Let us consider thesetwo possibilities separately.Case 1: G has a unique Sylow 2-subgroup K. By Theorem 21.1 we haveK ✂G. By Lagrange’s Theorem we must have N ∩K = {e} and NK = G, sothat G ∼ = N × K by Theorem 11.5. Since N ∼ = Z p and K ∼ = Z 2 we concludethat G ∼ = Z p ⊕ Z 2∼ = Z2p by Theorem 12.1.Case 2: G has p Sylow 2-subgroups. This means that G has p elementsof order 2. On the other hand none of the elements of N has order 2, so thatall the elements from G\N have order 2. In particular, b 2 = e. But thenfrom a i ba i b = e we conclude that ba i = a p−i b, so that{ a(a i b j )(a k b l ) =i+k b l if j = 0a i−k+p b l+1 if j = 1.This gives an explicit formula for the multiplication in G, and therefore G isuniquely determined. On the other hand D p is a non-abelian group of order2p, and so G ∼ = D p .

23. Groups of small ordersIn this section we shall classify all the groups of order less than 16. In doingthis we shall make use of the following general classification theorems provedso far:• Z p is the only group of order p (Theorem 6.10);• Z p 2 and Z p ⊕ Z p are the only two groups of order p 2 (Theorem 19.4);• Z pq is the only group of order pq if p − 1 is not divisible by q (Theorem22.1);• Z 2p and D p are the only two groups of order 2p (Theorem 22.2);(p and q denote primes, with p > q).Actually, the above results cover all the orders less than 16, apart from 8and 12. We now deal with these two cases.Theorem 23.1. There are five non-isomorphic groups of order 8, namelyZ 8 , Z 4 ⊕ Z 2 , Z 2 ⊕ Z 2 ⊕ Z 2 , Q 8 and D 4 .Proof. By the Fundamental Theorem for Finite Abelian Groups, everyabelian group of order 8 is isomorphic to one of the groups Z 8 , Z 4 ⊕ Z 2 orZ 2 ⊕ Z 2 ⊕ Z 2 . So let us assume that G is a non-abelian group of order 8.By Lagrange’s Theorem, the possible orders for a non-identity elementof G are 2, 4 and 8. However, G cannot contain an element of order 8, asG is not abelian, let alone cyclic. Also, a group in which every non-identityelement has order 2 is abelian; see Tutorial 1, Question 2. We concludethat G contains an element a of order 4. The subgroup N = {e, a, a 2 , a 3 }generated by a has index 2, and is therefore normal in G.Let now b ∈ G\N be arbitrary. Then N and Nb are the cosets of N inG, and soG = {e, a, a 2 , a 3 , b, ab, a 2 b, a 3 b}.Since N ✂ G, we must have b −1 ab ∈ N. We also know that b −1 ab musthave the same order as a does; see Tutorial 2, Question 1. Therefore, eitherb −1 ab = a or b −1 ab = a 3 . The former possibility leads to ab = ba, which is acontradiction with the assumption that G is non-abelian. We conclude thatb −1 ab = a 3 , i.e. ba = a 3 b. Now we have⎧⎨ a i+k b l if j = 0(a i b j )(a k b l ) = a i+3k b if j = 1, l = 0⎩a i+3k b 2 if j = l = 1.

We see that the multiplication in G is uniquely determined once we know b 2 .Now, b 2 cannot be equal to a i b, as this would imply b = a i . Next, we cannothave b 2 = a, as this would imply G = 〈b〉. Finally, we cannot have b 2 = a 3 ,because it implies b 6 = a 9 = a, and so G = 〈b〉. Therefore b 2 is equal eitherto e or to a 2 , giving two possible multiplications on G. We conclude thatthere are at most two non-abelian groups of order 8.On the other hand, the groups Q 8 and D 4 are non-abelian of order 8.Moreover, one may verify that they are not isomorphic just by looking at theorders of elements (see Tutorial 5, Question 5). Hence the theorem. Now we turn our attention to groups of order 12. We already know thatD 6 and A 4 are non-abelian of this order. The following example exhibitsanother group of order 12.Example 23.2. Let T be the subgroup of S 12 generated by the permutations(1 2 3 4 5 6)(7 8 9 10 11 12) and (1 7 4 10)(2 12 5 9)(3 11 6 8). By using thestandard method for calculating a group given by its generators we can seethatT = {id, (1 2 3 4 5 6)(7 8 9 10 11 12), (1 3 5)(2 4 6)(7 9 11)(8 10 12),(1 4)(2 5)(3 6)(7 10)(8 11)(9 12), (1 5 3)(2 6 4)(7 11 9)(8 12 10),(1 6 5 4 3 2)(7 12 11 10 9 8), (1 7 4 10)(2 12 5 9)(3 11 6 8),(1 8 4 11)(2 7 5 10)(3 12 6 9), (1 9 4 12)(2 8 5 11)(3 7 6 10),(1 10 4 7)(2 9 5 12)(3 8 6 11), (1 11 4 8)(2 10 5 7)(3 9 6 12),(1 12 4 9)(2 11 5 8)(3 10 6 7)}.The orders of elements of T are 1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 6, 6. Similarly, theorders of elements of D 6 and A 4 are 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 6, 6 and 1, 2,2, 2, 3, 3, 3, 3, 3, 3, 3, 3 respectively. Therefore, we see that T is a group oforder 12 which is not isomorphic to D 6 and A 4 .Theorem 23.3. The only groups of order 12 are Z 12 , Z 2 ⊕ Z 2 ⊕ Z 3 , D 6 , A 4and T .Sketch of proof. By the Fundamental Theorem for Finite Abelian Groups,the only abelian groups of order 12 are Z 2 ⊕ Z 2 ⊕ Z 3 and Z 4 ⊕ Z 3∼ = Z1 2.So let G be a non-abelian group of order 12, let s 2 be the number of Sylow2-subgroups of G, and let s 3 be the number of Sylow 3-subgroups of G. Bythe Third Sylow Theorem we have s 2 = 1 or s 2 = 3, and also s 3 = 1 ors 3 = 4.

If s 2 = s 3 = 1, then G has a unique Sylow 2-subgroup N and a uniqueSylow 3-subgroup K. They are both normal in G, and a standard argumentproves that G ∼ = N × K. However, |N| = 4 and |K| = 3, and hence they areboth abelian. But then so is G, which is a contradiction.If s 2 = 3 and s 3 = 4, then G would have eight elements of order three,and a further six elements of order power of 2 (see Example 21.3). Thiscontradicts the fact that |G| = 12.If s 2 = 1 and s 3 = 4, then G has a unique Sylow 2-subgroup N of order 4.N is isomorphic either to Z 4 or to K 4 . The former case, however, leads to acontradiction, and so N ∼ = K 4 . By a standard argument one can now attemptto reconstruct the multiplication for G. One obtains two possibilities, butthey turn out to define isomorphic groups. Therefore, in this case we obtainat most one group.Finally, if s 2 = 3 and s 3 = 1, one obtains further two possible multiplicationson G, thus giving a total of at most three non-abelian groups of order12.To complete the proof, one has to show that D 6 , A 4 and T are nonisomorphic,which can be done by calculating the orders of their elements.We have completed our classification of groups of order less than 16. Theresults are summarised in Table 4.2. Let us mention that the number ofgroups of order 16 is 14, and the number of groups of order 32 is 51. Thereis no known formula giving the number of groups of order n.

order groups reference1 〈e〉2 Z 2 Theorem 6.103 Z 3 Theorem 6.104 Z 4 , Z 2 ⊕ Z 2 Theorem 19.45 Z 5 Theorem 6.106 Z 6 , D 3 (= S 3 ) Theorem 22.27 Z 7 Theorem 6.108 Z 8 , Z 4 ⊕ Z 2 , Z 2 ⊕ Z 2 ⊕ Z 2 Theorem 23.19 Z 9 , Z 3 ⊕ Z 3 Theorem 19.410 Z 10 , D 5 Theorem 22.211 Z 11 Theorem 6.1012 Z 12 , Z 2 ⊕ Z 2 ⊕ Z 3 , D 6 , A 4 , T Theorem 23.313 Z 13 Theorem 6.1014 Z 14 , D 7 Theorem 22.215 Z 15 Theorem 22.1.Table 4.2: groups of order ≤ 16

19. p-groups

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