Algebra Homework Set 2 Hung Tran. P198, 3 If G

Algebra Homework Set 2 Hung Tran. P198, 3 If G is nilpotent then it is isomorphic to the direct product of its Sylow subgroups. So every product of normal subgroup of Sylow subgroups is also a normal subgroups of G. But each Sylow subgroup P is a p-group for some prime p and by theorem 6.1.1, each p-group has a normal subgroup of order pb dividing |P |. Therefore, G has a normal subgroup of each order dividing |G|. Now if G has a normal subgroup of each order dividing |G| then in particular, it has normal Sylow subgroups. By the Sylow theorems, all Sylow subgroups are conjugate, thus all Sylow subgroups of G are normal. By theorem 6.1.3, G is nilpotent. If G is cyclic, then it has a generator a of order |G|. For each n divides |G|, then for t = |G|/n, T =< at > is a subgroup of order n. Suppose D is any subgroup of order n. By theorem 2.3.7, D is also a cyclic group, D =< ad > for some d. adn = 1 ⇒ |G| divides dn, or t divides d. So D is a subgroup of T ⇒ D = T as |D| = |T |. If G has a unique subgroup of each order dividing G then each Sylow subgroup is normal in G as any conjugation of a subgroup would result in a subgroup of the same order. So by 6.1.3, G is nilpotent and G is isomorphic to a direct product of its Sylow subgroups. Let P be one Sylow subgroup corresponding to p a prime dividing |G|. By 6.1.1 P has a normal subgroup of order pb for each pb dividing |P | and by the hypothesis it must be the unique subgroup in G. Let c(g) counts the number of elements of order g in G. c(p) = p − 1 as there is an unique subgroup of order p. Since each unique group or order pi+1 has at least 1 subgroup of order pi and at most 1 such a subgroup, it has exactly one. So every element in the p-group of order pi+1 and outside of that p-subgroup of order pi can not be contained in any p-subgroup of order less than or equal pi since any unique subgroup or order less than pj+1 is already contained in that p-group of order pi . Therefore, those i elements must be of order pi+1 . Thus, c(pi+1 ) = pi+1 − pi . Let α be the greatest integer such that pα divides |G| then c(pα ) > 1 ⇒ P is cyclic. The direct product of cyclic groups whose orders are relatively primes is then a cyclic group.QED P198, 4 If G is a finite nilpotent group then G is isomorphic to the direct product of its Sylow subgroups by 6.1.3. If M is a maximal subgroup of G then M is isomorphic to a direct product of subgroups, each Mi is a subgroup of the Sylow subgroup Pi . At most one of them is properly included in its Sylow subgroup since otherwise let M’ isomorphic to the same subgroups of Sylow subgroups as for M except for one, replacing Mi by Pi if Mi is properly contained in Pi . M’ is then a proper subgroup of G, contradicting the maximality of G. Since M is a proper subgroup of G, exactly one component is properly contained in its Sylow subgroup Pi . That component is also a maximal subgroup in Pi since otherwise by theorem 6.1.1.4, we can construct M’ by replacing that component by the maximal group containing it and obtain a proper subgroup of G properly containing M. Also by 6.1.1.5, that maximal group in Pi is of index pi so M is of index pi . QED 1

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P198,5 Part 2, we prove by induction on the nilpotence class c of G. If c = 1 ⇒ Z1 = G then G is abelian and the result follows. Suppose it is true for any c < n and G is a nilpotent group of class of n. Suppose Z, center of G, and N a normal subgroup of N intersects trivially then consider K = N Z then N E K and Z E K and N ∩ Z = 1. Thus K is a direct product of N and Z. Then, K x = Z x × N x = Z × N , so K is normal in G ⇒ K/Z is normal in G/Z. Since G/Z is nilpotent of class of n-1, we can apply the induction hypothesis to conclude that K/Z intersects Z(G/Z) nontrivially, i.e., there exists Zy a coset of K, different from Z, that commutes with coset Zx for any x in G. Since K is a direct product of N and Z, Zy can be represented such that y ∈ N . ZyZx = ZxZy ⇒ yxy −1 x−1 ∈ Z. But y −1 x−1 ∈ N as N is normal so yxy −1 x−1 ∈ N ∩ Z or yxy −1 x−1 = 1 ⇒ xy −1 x−1 = y −1 ∀x ∈ G ⇒ y ∈ Z. But Zy is different from Z, y is not the identity, leading to a contradiction. QED Part 4. We also use induction on the nilpotence class c of G. If c = 1 ⇒ Z1 = G then G is abelian and the result follows. Suppose it is true for any c < n and G is a nilpotent group of class of n. Suppose H = NG (H) then since Z=Z(G) normalizes H, Z ∈ H ⇒ H/Z < NG/Z (H/Z) as G/Z has nilpotent class of n-1. Therefore, there exists t ∈ / H such that for any h ∈ H: ZtZhZt−1 = Zh1 for some −1 −1 h1 ∈ H ⇒ tht h1 ∈ Z ⊂ H ⇒ tht−1 ∈ H or t normalizes H; thus, H < NG (H) contradiction. QED P198, 9 If G is nilpotent and finite, then by 6.1.3.4 G is identified with a direct product of its Sylow subgroups, each element is identified with a tuple of elements in the Sylow subgroups. Let a, b in G such that (|a|, |b|) = 1 then we compare the two tuples componentwise. For each prime pi , if the i-element is different from the identity then pi divides the order of the tuple. Therefore, one component in the tuple of a is nontrivial iff the corresponding element of b is trivial. By the property of direct product, a commutes with b (proposition 5.1.2). Now suppose whenever (|a|, |b|) = 1 ⇒ ab = ba. Let P1 , P2 , ...Pk be Sylow subgroups corresponding to p1 , p2 , ...pk , distinct primes dividing |G|. Obviously |G| = |P1 ||P2 |...|Pk | (1). For each ai ∈ Pi and aj ∈ Pj for i 6= j it is clear that (|ai |, |aj |) = 1 ⇒ ai aj = aj ai . That is each element of Pi commutes with each element of Pj for i 6= j so P1 P2 ...Pk is a subgroup of G and furthermore since Pi ∩ Pj = (1),|P1 P2 ...Pk | = |P1 ||P2 |...|Pk | = |G|. So every element in G can be presented as a product of elements in the Sylow group, any two of them commute with each other. So it follows immediately G is isomorphic to the external direct product of its Sylow subgroups; hence, G is nilpotent. QED P198, 12, 13. The number of conjugates of an element in the center of the group is exactly 1. By Proposition 4.3.11, two elements of Sn are conjugate iff they have the same cycle type. Therefore, for n > 2, the center of Sn is trivial. So the upper central series for Sn , n ≥ 3 is just (1) ≤ (1).... By direct calculation, Z(A4 ) = (1) and since for n ≥ 5, An is simple ⇒ Z(An ) = (1).

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Therefore the upper central series for An , n ≥ 4 is just (1) ≤ (1).... To calculate lower central series, first we claim that the commutator subgroup of Sn , K, is An . Since each commutator is an even permutation, certainly K ≤ An . Now let (a,b) be any transposition then < (a, b), K > is normal in Sn as Sn /K is abelian (property of the commutator subgroup). Then, by 4.3.11, all transpositions are conjugate and by 3.5, every element of Sn can be written as a product of transpositions. Therefore, < (a, b), K >= Sn ⇒ |K| ≥ |Sn |/2 = |An | ⇒ K = An Now for n ≥ 5, if L = [Sn , An ], then L E Sn and L ⊂ An since An is normal in Sn . An is simple (4.6.24) then [Sn , An ] is either trivial or An . Clearly it is not trivial since (13)((12)(34))(13)((12)(34)) = (13)(24) so [Sn , An ] = An . The simplicity of An also implies that [An , An ] = An . The lower central series for Sn , An are: Sn ≥ An ≥ An ... An ≥ An ≥ An ... Then by direct calculation [A4 , A4 ] = V = ((1), (12)(34), (13)(24), (14)(23)); [S4 , A4 ] = A4 and [A4 , V ] = V , therefore, the lower central series for S4 , A4 are: S4 ≥ A4 ≥ A4 ... A4 ≥ V ≥ V... QED P198, 18 We can assume G000 = 1 since otherwise we just consider G/G000 , G0 /G000 = (G/G000 )0 , G00 /G000 = (G/G000 )00 . So G00 = G000 is equivalent to show G00 = 1 given G000 = 1. Consider the action by conjugation of the group G on the set G00 . By proposition 4.4.13, this action induces an homomorphism from G into Aut (G”) with kernel CG (G00 ). Since G00 is cylcic, Aut(G”) is abelian, therefore G/CG (G00 ) is abelian. Since G/G0 is the largest abelian quotient of G by Proposition 5.4.7, G0 ≤ G/CG (G00 ) ⇒ G00 ≤ Z(G0 ) ⇒ G0 /Z(G0 ) ia cyclic. Then if Z(G0 )a is the generator of G0 /Z(G0 ) then for any x ∈ G, Z(G0 )x = Z(G0 )at or x = zat for some z ∈ Z(G0 ). Thus, x1 x2 = z1 at1 z2 at2 = z2 at2 z1 at1 = x2 x1 . Thus, G0 is abelian making G00 = 1. QED P198. 19 Suppose there is such a subgroup G0 = S4 then by problem 12,13 we have G00 = A4 , G000 = N (using the same notation from problem 12, 13). Since each quotient groups G0 /G00 , G00 /G000 are of prime order (2,3), they are cyclic. By problem 12, we obtain G00 = G000 or A4 = N , a contradiction. QED P199, 24. Suppose a is a non-generator then for every maximal subgroup M, < x, M > is a proper subgroup of G. Since M is maximal M =< x, M > or x ∈ M . So x is in any of maximal subgroup of G; hence, the set of nongenerators are in φ(G) (1). Now suppose b ∈ φ(G) we need to show that b is a nongenerator. Suppose not, then there exists a proper subgroup H such that < H, b >= G. Clearly, H is not a maximal subgroup since otherwise b ∈ H ⇒< H, b >= H < G. So there exists H1 a proper subgroup of G that properly contains H. b ∈ / H1 since otherwise < H, b >≤< H1 , b >< G. So similarly H1 is not a maximal group of G. Repeating the argument we obtain a partially ordered (by inclusion) set H < H1 < H2 .... We show that any chain in the set has an upper bound. Let

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Hi1 < Hi2 < Hi3 ... be a chain and K = ∪Hij . Then K is a subgroup of G since if x, y ∈ K then there exists in such that x, y ∈ Hin then their product and inverses are in K. Furthermore, Hij < K and b ∈ / K ⇒ K < G. So K is actually an upper bound for that chain. Since any chain there is an upper bound, by Zorn’s lemma, there is a maximal element L of the set H < H1 < H2 .... Obviously, L is a maximal subgroup of G not containg b ⇒ contradiction with the definition of b. Therefore, b is nongenrator. (2) (1) and (2) imply that φ(G) is the set of non-generators. QED P 199, 25. First, we’ll show that φ(G) is a characteristic subgroup of G. Let δ ∈ Aut(G) then δ isTalso an automorphism for T the set of maximal subgroups in G. Thus, φ(G) = M maximalinG δ(M ) = δ( M maximalinG M ) = δ(φ(G)). Since δ is arbitrary, φ(G) is a characteristic subgroup of G. In particular, φ(G) C G Let P be a Sylow subgroup of φ(G) for some prime p, dividing |G|. Then by Frattini’s argument, G = φ(G)NG (P ) =< φ(G), NG (P ) >. Since each element of φ(G) is a non-generator, we throw out elements of φ(G) one at a time and the generated group is still G. Since G is finite, φ(G) is also finite, there is a stop to the process and we obtain G =< NG (P ) >= NG (P ) or P is normal in G. In particular, P is normal in φ(G) so by thereom 6.1.3, φ(G) is nilpotent. QED Additional problems 1. a. Since K = P ∩ N is a p-subgroup of N, by the Sylow’s theorem (about conjugates), there exists Q, a Sylow p-subgroup of N such that K ≤ Q. (1) Now Q is a p-subgroup of G, by the same argument there exists H a Sylow psubgroup of G such that Q ≤ H. Let g be the element that conjugates P to H (the existence is guarenteed by Sylow’s theorem also) then Qg ∈ P and since N is normal Qg ∈ N ⇒ Qg ≤ K. (2) But |Qg | = |Q|, (1) and (2) imply that Q = K or P ∩ N ∈ Sylp (N ). QED b. Let G = S4 N = ((1), (13), (14), (34), (134), (143)) and P = ((1), (123), (132)) then P ∈ Syl3 (G) but P ∩ N = (1) < ((1), (134), (143)) = Syl3 (N ). QED c. We denote Ap = pα with α the greatest integer such that pα divides number A. |G| | Then, since N ∩ P ∈ Sylp (N ), |N |pp = |N|P∩P | . Since G/N is a p-group, |G|/|N | is a power of p. Therefore,

|G|p |N |p

=

|G| |N |

P| However, by the 2nd Isomorphism theorem, |N |N | = NP is a subgroup of G, that implies G=NP. QED

|P | |N ∩P | .

Thus |G| = |N P |. Since

2. We’ll prove the statement for the case that G is a finite group. Let N, M be the two normal subgroups that generate G. Since G is finite, both M and N are finite. Now since G is finite, G has a Sylow subgroup P for some prime p. By exercise 2, A = (P ∩ N ) ∈ Sylp (N ) and B = (P ∩ M ) ∈ Sylp (M ). Using the same notation as in the previous exercise, we have |A| = |N |p and |B| = |M |p . (1) Since N, M are nilpotent, A and B are the unique Sylow p-subgroups in N, M respectively; hence, they are also their characteristic subgroups. Consequently, A and B must be normal in G as they are invariant under the conjugation action.

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Furthermore, we can apply the 2nd isomorphism theorem to obtain: |A||B| |AB| = |A∩B| Since (A ∩ B) ≤ (N ∩ M ),A and B are p-subgroups, using the same notation as in |A||B| |A||B| the previous problem, |A ∩ B| ≤ |N ∩ M |p . Therefore, |AB| = |A∩B| ≥ |N ∩M | .(2) On the other hand since M, N are normal subgroups of G we can also apply the 2nd isomorphism again noticing that MN=G: |M |p ||N |p |A||B| |||N | |G| = |M N | = |M |M ∩N | ⇒ |G|p = |M ∩N |p = |N ∩M | (follow (1)). Thus, |G|p ≤ |AB| (follow (2)). However, |G|p = |H|, so H = AB as AB is already a subgroup of H. Since A and B are normal in G, AB is certainly normal in G. As H is arbitrary, any Sylow subgroup of G is normal in G; hence, G is nilpotent. QED 3. Suppose G have a Hall π-subgroup for π = {2, 5} or {3, 5}, that means G has a subgroup of order 20 or 15. Let H be that subgroup and consider the action of G on the set A of left cosets of H in G as describe in section 4.2. By section 1.7, there is a homomorphism from G to SA . Since G is simple, the kernel of that homomorphism must be either (1) or G. Since the action is nontrivial, the kernel can not be G, so it must be 1; thus there is an injective homomorphism from G to SA . However since |A| = 3 or 4, |SA | = 6 or 24. But |G| = 60, there is a contradiction. That shows that a simple group of order 60 does not have these Hall π-subgroups. QED