MATH 241 Homework 10, Fall 2011

Math 241 Problem Set 10 solution manual Exercise. A10.1 i G = S4, then |G| = 4! = 24 = 23.3. A sylow 2-subgroup of S4 is a subgroup of order 8, so we can consider D4. A sylow 3-subgroup of S4 is a subgroup of order 3, so we can consider the subgroup generated by any 3-cycle. eg: < (123) >. ii G = S5, then |G| = 5! = 120 = 23.3.5. A sylow 2-subgroup of S5 is a subgroup of order 8. Notice that the elements of D4 when considered as pemutations in S5 will still form a subgroup whose order would be 8. A sylow 3-subgroup of S5 is a subgroup of order 3, so we can consider the subgroup generated by any 3-cycle. A sylow 5-subgroup of S5 is a subgroup of order 5, so we can consider the subgroup generated by any 5-cycle. iii G = Z100 × Z30, then |G| = 3000 = 23.3.53 A sylow 3-subgroup of Z100 × Z30 is a subgroup of order 3, so we can consider the subgroup < (0, 10) >= {(0, 0), (0, 10), (0, 20)}. A sylow 2-subgroup of Z100 × Z30 is a subgroup of order 8, so we can consider the subgroup < (25, 0), (0, 15) >= {(0, 0), (25, 0), (50, 0), (75, 0), (0, 15), (25, 15), (50, 15), (75, 15)}. A sylow 5-subgroup of Z100 × Z30 is a subgroup of order 125, so we can consider the subgroup < (4, 0), (0, 6) >= {(a, b) | a ≡ 0 mod(4), and b ≡ 0 mod(6)}. iv G = GL2(Z7), then |G| = (72 − 1)(72 − 7) = 2016 = 25.32.7. A sylow 7-subgroup of G is a subgroup of order 7, so we can consider the subgroup: { � 1 0 0 1 � � 1 1 0 1 � � 1 2 0 1 � � 1 3 0 1 � � 1 4 0 1 � � 1 5 0 1 � � 1 6 0 1 � } =< � 1 1 0 1 � >. A sylow 3-subgroup of G is a subgroup of order 9, so we can consider the subgroup: { � a 0 0 b � | a, b ∈ {1, 2, 4}}. A sylow 2-subgroup of G is a subgroup of order 25, so we can consider the subgroup: So consider first the subgroup H′ =< � 2 −4 4 2 � > , H′ is a subgroup of order 16. Then our Sylow subgroup is H = H′ ∪ � 1 0 0 −1 � H′ 1 Exercise. A10.2 a- First we find the normalizer of T. So let g = � i j k l � ∈ GL(2, R), for g to be in the normalizer of T then we must have g. � 1 0 0 2 � .g−1 = � a′ 0 0 b′ � =⇒ {a′, b′} = {1, 2}, reason is because the matrix � a′ 0 0 b′ � must have 1,2 as eigen values. Then we have the following : either g. � 1 0 0 2 � .g−1 = � 1 0 0 2 � which implies that g. � 1 0 0 2 � = � 1 0 0 2 � g =⇒ � i 2j k 2l � = � i j 2k 2l � =⇒ j = k = 0 =⇒ g must be a diagonal matrix =⇒ g ∈ T or g. � 1 0 0 2 � .g−1 = � 2 0 0 1 � which implies that g. � 1 0 0 2 � = � 1 0 0 2 � g =⇒ � i 2j k 2l � = � 2i 2j k l � =⇒ i = l = 0 =⇒ g ∈ � 0 1 1 0 � T This show that N(T) ⊆ T ∪ � 0 1 1 0 � T, Now it is easy to check that the right side is also contained in the normalizer of T. Now we can deduce that index of T in N(T) is 2. Now For the normalizer of U, we do the same, you can start by finding the elements g such that g � 0 −1 1 0 � g−1 ∈ U, and you will get that g must be either in U or in � 1 0 0 −l � U. FInally you can verify that U ∪ � 1 0 0 −l � U ⊂ N(U) and hence N(U) = U ∪ � 1 0 0 −l � U. Then we can deduce that index of U in N(U) is 2. b- Now to find the normalizer for T consider the matrix A =   1 0 . . . . . . 0 0 2 0 . . . 0 0 . . . ... . . . 0 0 . . . 0 ... 0 0 . . . . . . 0 n   , g ∈ N(T) =⇒ gAg−1 =   a1 0 . . . . . . 0 0 a2 0 . . . 0 0 . . . ... . . . 0 0 . . . 0 ... 0 0 . . . . . . 0 an   but this implies that {a1, a2, ..., an} = {1, 2, ..., n} in some order. So ∃ π ∈ Sn such that ai = π(i) ∀i ∈ {1, 2, ..., n}. 2 So gAg−1 =   π(1) 0 . . . . . . 0 0 π(2) 0 . . . 0 0 . . . ... . . . 0 0 . . . 0 ... 0 0 . . . . . . 0 π(n)   = Pπ   1 0 . . . . . . 0 0 2 0 . . . 0 0 . . . ... . . . 0 0 . . . 0 ... 0 0 . . . . . . 0 n   P −1 π where Pπ is a permutation matrix. so we get that g ∈ Pπ.Stab(A) = Pπ.T for some π ∈ Sn. Hence N(T) = ∪ π∈SNPπ.T, and then N(T)/T = { ¯Pπ | π ∈ Sn} ∼= Sn. Section. 37 Exercise. 5 Let G be a group of order 96. G must contain a sylow 2-subgroup of order 25. If G containes 1 sylow 2-subgroup then we know that this subgroup is normal and hence G is not simple. So consider that G containes more than one sylow 2-subgroups H and K. H ∩K is a subgroup if H and of K (H ̸= K), by a counting argument we can see that the order of H ∩K is 16. Moreover, the normalizer of H ∩ K containes both subgroups H, and K, hence its order should be a divisor of 94 which is greater than 32, hence N(H ∩ K) − G, so H ∩ K is normal in G. Section. 18 Exercise. 11 It is easy to see that this set is closed under addition and multicplication, for example (a+b √ 2)(a′+ b′√ 2) = aa′ +2bb′ + √ 2(ab′ +ba′) with aa′ +2bb′, ab′ +ba′ ∈ Z . Moreover, 0=0+0 √ 2 and 1=1+0 √ 2 both belong to the set. Assotiativity and distributativity follows directly from the properties of the usual addition and multiplication in R. So we have a ring. This ring is commutative since usual multiplication is commutative. This ring is not a field since 2 is not a unit, for if it was then ∃ x ∈ R such that 2x = 1, then x = 1/2 contradiction ( 1/2 = a + b √ 2 with a, b ∈ Z then √ 2 = 1/2−a b which is impossibe. Exercise. 12 Similarly as above it is easy to see that this set is a commutative ring. Now let x1 = a + b √ 2 be an non-zero element in the ring , notice that then a2 − 2b2 ̸= 0 since a, b ∈ Q ,and then the element x2 = a a2−2b2 − b a2−2b2 √ 2 is an element in the ring with the property x1.x2 = 1. Then our ring is a field. Exercise. 15 The units in Z × Z are : (1,1), (-1,1), (1,-1), (-1,-1). Exercise. 20 a- The number of elements in M2(Z2) is 16, since each entery of the matrix can be either 0, or 1. 3 b- The invertible elements are: � 1 0 0 1 � , � 0 1 1 0 � � 1 1 0 1 � � 0 1 1 1 � � 1 1 1 0 � � 1 0 1 1 � . Exercise. 38 R is commutative ⇔ ab = ba ∀ a, b ∈ R ⇔ ab−ba = 0 ∀ a, b ∈ R ⇔ a2 −b2 +(ab−ba) = a2 −b2 ⇔ (a − b)(a + b) = a2 − b2. 4