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Math 120 practice final Solution
June 8, 2006
Problem 1
(a) True. Because (g 1 , g 2 )(g′ 1 , g 2 ′) = (g 1 g′ 1 , g 2 g′ 2 ) = (g′ 1 g 1 , g′ 2 g 2 ) = (g 1 ′, g′ 2 )(g 1 , g 2 )
(b) False. Z 2 ⊕ Z 2 is not cyclic.
(c) True. It’s by Cauchy’s Theorem (P.96).
(c) False, Z 2 ⊕ Z 2 ⊕ Z 25 does not contain an element of order 4.
(d) True. If there exists one element has order 4 then G ∼= Z 4. If all elements except the identity have order 2, then pick a, b two distinct elements in G, then ab and ba should be the fourth element in G other than a, b, identity. So the group is commutative.
(e) False. S 3 is not commutative.
(f) False. Let H =< (12) > and G = S 3. First you can check H is not normal in G. And prove directly NG(H) = H. So NG(H) is not normal in this example.
(g) True. There is only one such group : Z 17
(h) False. The number of 5-sylow subgroups should be congruent to 1 mod
Problem 2
(a) First we want find the image of 1. Try to solve k to make 7·k = 1 (mod
- will give you k=43. So φ(1) = φ(7·43) = φ(7)·43 = 6·43 = 3 (mod 15). Then we know for every n in Z 50 , φ(n) = φ(1 · n) = n · φ(1) = 3n (mod 15).
(b) Ker(φ) = {n ∈ Z 50 | 3 n = 0 (mod 15) } = 5 · Z 50
(c) Image=3 · Z 15 Problem 4
Let G/Z(G) =< x > because G/Z(G) is cyclic. Take ˜x in φ−^1 (x). We can write every element in G as ˜xh, h ∈ Z(G). Randomly pick two elements in G, we can write them as ˜xih 1 and ˜xjh 2 where h 1 , h 2 ∈ Z(G).
˜xi^ h ︸︷︷︸ 1 x˜j h 1 ∈Z(G)
h 2 = x ︸︷︷︸˜i^ x˜j x ˜i^ x˜j^ commutes
h ︸︷︷︸ 1 h 2 h 1 ∈Z(G)
= ˜xj^ ︸︷︷︸˜xih 2 h 2 ∈Z(G)
h 1 = ˜xj^ h 2 x˜ih 1
This implies any two elements in G commutes. So G is a commutative group.
Problem 5
No. Take H =< (123) > and G = S 3. H is cyclic and normal in S 3. S 3 /H has order 2 so it’s isomorphic to Z 2 which is commutative. But S 3 is not commutative.
Problem 6
If we write D 8 = {a, b|a^4 = b^2 = 1, bab−^1 = a−^1 } Define a homomorphism φ : D 8 → Z 2 ⊕ Z 2 by sending aibj^ to (i, j). You can check this is an surjective homomorphism and hte kernel is Z(D 8 ) = a^2 , id. By the first isomorphism theorem, D 8 /Z(D 8 ) ∼= Z 2 ⊕ Z 2
Problem 7
Because 45 = 3^2 · 5, every commutative group of order 45 is isomorphic to Z 45 or Z 3 ⊕ Z 3 ⊕ Z 5. 3 is an element Z 45 of order 15. (0,1,1) is an element in Z 3 ⊕ Z 3 ⊕ Z 5 which has order 15.
In Z 4 ⊕ Z 4 it’s possible because (1,0) and (0,1) have order 4 and (2,0) 6 = (0,2). In Z 4 ⊕ Z 2 ⊕ Z 2 , order 4 elements are: (1,*, *) and (3, *, *). But all their square = (2,0,0). So G is not isomorphic to Z 4 ⊕ Z 2 ⊕ Z 2. There is no order 4 element in Z 2 ⊕ Z 2 ⊕ Z 2 ⊕ Z 2 So G is not isomorphic to Z 2 ⊕ Z 2 ⊕ Z 2 ⊕ Z 2. Therefore we can conclude G is isomorphic to Z 4 ⊕ Z 4.
Problem 12
For any h ∈ H and k ∈ K, hkh−^1 k−^1 ∈ H ∩ K because H and K are normal. So hkh−^1 k−^1 = e, hk = kh. So HK is a commutative group.
Problem 13
Let H =< h | h^5 = e >. For any g ∈ G, let ord(g) = k. Because k divides —G—, k has to be an odd number. Because H is normal, ghg−^1 = hi^ for some i ∈ { 0 , 1 , 2 , 3 , 4 }. i may depend on g.
h =gkhg−k^ = gk−^1 (ghg−^1 )g−k+1^ = gk−^1 hig−k+1^ = gk−^2 (ghig−^1 )g−k+ =gk−^2 (ghg−^1 )ig−k+2^ = gk−^2 hi
2 g−k+2^ =... = hi
k
Because h has order 5 and h = hi k , 1 = ik^ (mod 5). If i=2 or 3 or 4, then k must be even. So i=1. Hence ghg−^1 = h, gh = hg. Because we pick arbitrary g, That means h commutes with every g ∈ G. So H ≤ Z(G).
Problem 14
By Sylow’s theorem, if 7-sylow subgroup is not normal then there are at least 8 7-sylow subgroups. Because —G—¡49, 7-sylow subgoup has order 7. They are cyclic, so every element except ideneity is a generator. Therefore the intesection of two distinct 7-sylow subgroup is only the identity. So G should contain at least 8*6+1=49 differnent elements, which is a contradiction.
Problem 15
168 = 8 ∗ · 3 · 7. If there are k 7-sylow subgroups then k divides 24 and k ≡ 1 (mod 7). So k=8.
Problem 16 175 = 5^2 · 7. Using sylow’s theorem one can show 5-sylow subgoup H and 7-sylow subgroup K are normal. And HK=G, H ∩ K = e. By problem 12 G is commutative.
Problem 17
If |Z(G)| = 4, |G/Z(G)| = 15. One can use sylow’s theorem to show every group of order 15 is isomorphic ro Z 15 , which is cyclic. By problem 4, G is commutative. Then |Z(G)| = |G| = 60, contradiction.
Problem 18
By Sylow’s theorem every two p-sylow subgroup are conjugate. So the conjugation gives an isomorphism.
