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 subgroup 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, 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 =< 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 Pi be one Sylow subgroup corresponding to pi a prime dividing |G|. By 6.1.1 Pi has a normal subgroup of order pbi for each pbi dividing |Pi | and by the hypothesis it should be the unique subgroup in G. Let c(g) counts the number of elements of order g in G. c(pi ) = pi − 1 as there is an unique subgroup of order pi . Since each group or order pj+1 has at least 1 subgroup of order pji and i at most 1 such a subgroup, it has exactly one. So c(pj+1 ) = pj+1 − pji . Let α be i i α α the greatest integer such that pi divides |G| then c(pi ) > 1 ⇒ Pi is cyclic. The direct product of cyclic groups of relatively prime orders 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 its nomalizer in Pi 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 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 1

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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 the 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|) = 1ab = 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 i 6= j it is clear that (|ai |, |bi |) = 1 ⇒ ai bi = bi ai . That is each element of Pi commutes with each element of Pj for i 6= j. Therefore |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 that each Sylow group is normal in G; hence, G is nilpotent. QED P198, 12, 13. By Proposition 4.3.6, 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 ≥ 5An is simple ⇒ Z(An ) = (1). 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 Then by direct calculation [A4 , A4 ] = N = ((1), (12)(34), (13)(24), (14)(23)); [S4 , A4 ] = A4 and [A4 , N ] = N , therefore, the lower central series for S4 , A4

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By problem 11 proved previously, M is abelian. Since M is finite, there exists an element u of order p for some prime p. Consider K = (x ∈ M |xp = 1) and it is easily seen that K is a subgroup of M (since M is abelian). Now let π ∈ Aut(M ) then consider π(x) for x ∈ K: π(x)p = π(x)...π(x) = π(xp ) = π(1) = 1 ⇒ π(x) ∈ K. That implies K is a characteristic subgroup of G; hence, K is a normal subgroup of G by an argument presented in problem 12. Because of the minimality of M and nontriviality of K (u ∈ K), K=M. So M is abelian and every nontrivial element has order p implying that M is an elementary abelian p-group for some prime p. QED P200, 32 We prove by induction on G. If G has order 1,2, or 3, the statement follows immediately. Now suppose it is true for any finite solvable group of order less than g and |G| = g. Let H be a maximal subgroup of G and M is a minimmal normal subgroup of G. If M is contained in H then by the 4th isomorphism theorem, it corresponds to H ∗ a maximal subgroup in G/M . We can apply induction on G/M and by the 3rd isomorphism theorem |G : H| = |G/M : H ∗ |, we can conclude that H has prime power index. Now suppose M is not a subgroup of H then MH properly contains H. Since M is normal in G, MH is also a subgroup of G (easily checked, for example, m1 h1 m2 h2 = m1 h1 m2 h−1 1 h1 h2 = m3 h3 ). So the maximility of H implies that MH=G. Then by the 2nd isomorphism theorem: |H| |G| |M | |G| HM/M ∼ = H/(H ∩ M ) ⇒ |M | = |H∩M | ⇒ |H| = |M ∩H| Since M is a p-group and M ∩ H is a subgroup of M, H must have p power index. QED Additional Problems 1. Lemma: Let G be an Ω-group and suppose that N,U, V ⊂ G are Ω-subgroup with N E G and U E V and V/U is Ω-simple then a. U N E V N and V N/U N is either trivial and Ω-simple b. (U ∩ N ) E (V ∩ N ) and (V ∩ N )/(U ∩ N ) is either trivial and Ω-simple. Proof: Direct application of the 2nd or Diomond Isomorphism theorem. QED Now let G = G0 D G1 ... D Gn = (1) be the Ω-composition series and consider Ki = H ∩ Gi then, by part b of the lemma, H = K0 D K1 ... D Kn = (1) and each of the factor Ki /Ki+1 is Ω-simple.Now we define Ti = HGi and by part a of the lemma, we have G = T0 D T1 ... D Tn = H and Ki and Ti together constitute an Ω-series for G. Since each factor is either trivial or Ω-simple we can delete some terms in the series to make an Ω-composition series. The Ω-composition series for H and G/H are constructed from Ki and Ti /H similarly. QED 2. Since H is one of the terms in a composition series for G, and the constructions of composition series of H and G/H are corresponding with the contruction of a composition series of G we have: l(G) = l(H) + l(G/H). Since H1 and H2 are normal in G we have H1 normalizes H2 and we can apply the 2nd isomorphism theorem: H1 H2 /H1 ∼ = H2 /(H1 ∩ H2 ) to obtain: l(H1 H2 ) − l(H1 ) = l(H2 ) − l(H1 ∩ H2 ). QED

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Now if L is a linear vector space than L is Ω-groups if addition is group operation and Ω contains multiplication by scalars. Furthermore, l(L) = dim(L). So if S, T are two linear subspaces of a linear space V then by l(ST ) = l(S) + l(T ) − l(S ∩ T ) we translate to the dimensional language to obtain: dim(S + T ) = dim(S) + dim(T ) − dim(S ∩ T ). QED 3. In the proof given in class, the two composition series can be refined by each other using the Schreier’s refinement scheme. That refinement yields 1 1 (G1i,j /G1i,j+1 ) ∼ /Hj,i+1 ) for 0 < i, j < n = (Hj,i So the permuation (i, j) 7→ (j, i) that sends nontrivial factors of Gi,j+1 /Gi,j to nontrivial factors of Hj,i+1 /Hj,i also sends a factor from (Gi ) to a factor in (Hj ). Thus, we can define the permutation π as π(i) = j iff (Gi (Gi+1 ∩ Hj ))/(Gi (Gi+1 ∩ Hj+1 )) is nontrivial. QED 4. a. Let z = xyx−1 y −1 be any generating element of the commutator subgroup and δ be any endomorphism. It suffices to show that δ(z) is also in the commutator subgroup. We have: 1 = δ(xx−1 ) = δ(x)δ(x−1 ) ⇒ δ(x)−1 = δ(x−1 ) and, similarly, δ(y)−1 = δ(y −1 ). Therefore, δ(xyx−1 y −1 ) = δ(x)δ(y)δ(x−1 )δ(y −1 ) = δ(x)δ(y)δ(x)−1 δ(y)−1 also belongs to the commutator subgroup. QED b. False, for example let G = S3 × Z2 then the center of G is Z2 . Now let π be the natural projection from G onto Z2 and γ be the map Z2 7→ S3 by 1 7→ 1 why the other element is mapped to (1, 2) ∈ S3 . Finally take ι to be the inclusion map from S3 into G. Now δ: G 7→ G is defined as δ(g) = ι(γ(π(g))) then it is clear that δ is an endormorphism but δ(Z(G)) 6= Z(G). QED 5. a. Let G = G0 D G1 ... D Gn D (1) be any Ω-series of G. We refine each slice of the series. Let Gn,i+1 be a proper Ω-normal subgroup of Gn containing Gn,i with Gn,0 = 1 then we obtain a sequence of Ω-subgroups Gn,i with Gn,i+1 E Gn,i . Since G satisfies ACC the sequence must stop and we obtain G1n as the last term. Now we similarly construct a chain of subgroup in between G1n and (1) we obtain G2n . By repeating the process we obtain a sequence of Ω-subgroup such that Gn DG1n DG2n .... Since G also satisfies DCC, the sequence must also stop at a point k. The construction guarantees that each Gi+1 is a maximal Ω-normal subgroup of Gin implying n that the slice is refinely completely. Since we only have a finite number of slices we can refine the Ω-series of G into an Ω-composition series of G. QED b. It is obvious that each f i (G) is a Ω-subgroup as f is an Ω-endomorphism. Now we show that f i+1 (G) ⊂ f i (G) for any non-negative i. We have: f i+1 (g) = f i (f (g)) ∈ f i (G) since f (g) ∈ G. So it is proved. Therefore the sequence (f i (G)) is a descending chain of Ω-subgroup of G ordered by inclusion. Since G satisfies DCC, the chain is finite and supposed it stops at f n (G) = H. Then f n+1 (G) = H = f (H) so f is an automorphism of H. Thus, f n is also an automorphism of H. Therefore if a ∈ N ∩ H then f n (a) = 1 since a ∈ N .

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But since f n is also an automorphism of H, a = 1. Thus, N ∩ H = 1 (1). Now f (H) = H ⇒ f n (H) = H ⇒ f 2n (G) = H. Let g be any element of G and f n (g) = h and let s be the element of G such that f 2n (s) = h. Then f n (gf n (s−1 )) = f n (g)f 2n (s−1 ) = hh−1 = 1 ⇒ gf n (s−1 ) ∈ N . Furthermore, f n (s) ∈ H and g = gf n (s−1 )f n (s) ⇒ g ∈ N H. Therefore G=NH(2). (1) and (2) and normality of N ⇒ that G is a semi-direct product of N and H. QED c. G can always be decomposed as a semidirect product as in part b so it is not decomposable in a non-trivial way iff H = 1 or H = G. The former leads to the fact that any endomorphism f is nilpotent while the later implies that any endomorphism f is an automorphism. QED